Static friction question

[Tokspor]

A 10-kg block is at rest on an inclined plane. The coefficient of static friction is 0.75. The force of static friction is 17 N. Which of the following would NOT cause the force of static friction to increase?

A. Increasing the mass of the block.
B. Increasing the angle of incline.
C. Increasing the acceleration due to gravity.
D. Increasing the coefficient of static friction.

I chose B, but the correct answer is D. I kind of understand how B would be wrong, but I'm not absolutely sure here. Originally, I thought that since F(friction) = umgcos(theta), if you increase theta, the cos(theta) will decrease and so will F(friction). But apparentely, F(friction) would not decrease--if it did, it would be the correct answer for this question.

So does this mean the coefficient of static friction will increase as cos(theta) decreases to match the force being applied on the object?

As for answer D, the only way I can see this being the correct answer if it is actually referring to the maximum coefficient of static friction it could potentially be given an applied force to counteract. In that case, the applied force will remain the same, as will the force of static friction. Can someone help with or verify my reasoning for B and D? Thanks.

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[ezsanche]

on second thought

F(Friction) = mgsin(theta)

The only thing not included is the coefficient of static friction.

The coefficient of static friction only depends on the surface of the two objects that are in contact with each other.

 

[ezsanche]

tricky question. let me guess this was a kaplan question?

 

[wanderer]

Static friction is less than or equal to the coefficient times the weight. If the opposing force is a lot less than mu*m*g, increasing mu doesn't change anything. The common misconception is that Fs = muS*m*g. It usually doesn't.

 

[RogueUnicorn]

while it is true the maximum static friction decreases as you increase the angle. HOWEVER, you must remember that static friction is a reactive force and as such must remember that as you increase the angle, you increase the component of gravity pulling the box down. so, the ACTUAL static friction force increases, until the decreasing maximum and increasing pulling force intersect and the box starts moving.


edit: this is essentially what wanderer said. ha. remember that Fs=<mu*mg*cos(Th)

 

[Tokspor]

So the "force of static friction" in this question is whatever magnitude of force is needed to oppose the applied force, is that correct?

Because the way I interpret question D is that it's referring to the greatest theoretical value the static friction could have, not the actual static friction being applied in response to the applied force. If I'm interpreting it right, then this makes sense. Is this correct?

And yes--it is a Kaplan question indeed.

 

[RogueUnicorn]

So the "force of static friction" in this question is whatever magnitude of force is needed to oppose the applied force, is that correct?

Because the way I interpret question D is that it's referring to the greatest theoretical value the static friction could have, not the actual static friction being applied in response to the applied force. If I'm interpreting it right, then this makes sense. Is this correct?

And yes--it is a Kaplan question indeed.

correct.

 

[ezsanche]

Static friction is less than or equal to the coefficient times the weight. If the opposing force is a lot less than mu*m*g, increasing mu doesn't change anything. The common misconception is that Fs = muS*m*g. It usually doesn't.

Right!

The opposing force is a a component of the object weight

hence that is why you need to use

the equation

F(Friction) = mgsin(theta)

Funny- how I was able to tell that it was a Kaplan question. The questions on AAMC are more straightforward but then it again the exam isn't as curved as the kaplan exams

 

[RogueUnicorn]

cosine, dude... cosine...

 

[BerkReviewTeach]

while it is true the maximum static friction decreases as you increase the angle.

Actually, it doesn't decrease. The normal force decreases, but static friction must increase to offset the increasing opposing force (sin component of weight).

cosine, dude... cosine...

Again, no! ezsanche is correct in that equation. According to the force diagram, the weight of the stationary object is offset by the normal force and the static frictional force. The normal force cancels out the cosine component, so the static friction must be offsetting the sine component.

You are thinking that Fstatic = mustaticN = mustaticmgcos(theta). What ezsanche wrote does NOT include mustatic, so before telling him he's wrong, you should look more carefully.

The end result is:
mustaticmgcos(theta) = mgsin(theta), which leads to the conclusion that mustatic = sin(theta)/cos(theta) = tan(theta). The value of mustatic increases with the angle of inclination until it breaks free, at which time we conclude that mustaticmax = tan(theta)breakaway

 

[wanderer]

Actually, it doesn't decrease. The normal force decreases, but static friction must increase to offset the increasing opposing force (sin component of weight).

The maximum force does decrease (fs is proportional to n), but the actual force usually will increase if the angle is increased (unless it's about to slip).

 

[ezsanche]

I think Bleargh meant to say "cosign dude cosign" as to state that he agreed with what I wrote

 

[JohnnyBravo]

The maximum force does decrease (fs is proportional to n), but the actual force usually will increase if the angle is increased (unless it's about to slip).

Actually, BRT is correct. Here is an example. Let's say that the static friction is 1 Let's say that I have an object that has a mass of 5kg. Let's assume that I'm at an angle of 30 for simplicity. The force due to gravity is mgsine(theta)=5*10*.5= 25 Newtons. Now, what will force static display? Obviously 25 Newtons. In order to do this, the mu static is .57 which is less than 1!!! The answer is NOT .4. The answer is mgcos(theta)*mu stati= mgsine theta

Well you know mg cancels and mu static=sine(theta)/cos(theta). So, mu static=tan(theta). However, this isn't the maximal mu static. The Maximal mu static is constant. However, the maximal mu static isn't necessarily given. This is something we don't realize. The mu static increases as an angle increases up until it reaches it's maximal value? Why, because if the object isn't accelerating, then the force static and force due to gravity have to be equal.

Now, continuing my example. Let's say that 45 is the last degree before it starts sliding. Well, at this point mgsin theta= 35.35 is the force. Well, when you solve for this, you get mu=1. Why, because cosine of 45 are equal so the mu must be 1. What does this example display, that due to the inherent rules of the incline, mu static must change as you go up or down an incline. However, it's maximal Value doesn't change.

 

[BerkReviewTeach]

The maximum force does decrease (fs is proportional to n), but the actual force usually will increase if the angle is increased (unless it's about to slip).

The actual static force increases until it reaches the maximum, at which time it breaks free once the maximum has been exceeded. The maximum static friction does not decrease. It's a threshold value that depends on the materials that are interacting.

 

[ezsanche]

I'm not buying the whole argument that the coefficient of static friction increases as the angle of elevation increases. the coefficient of static friction is an instrinic quality that depends on the surfaces of the two objects that are in contact.

 

[RogueUnicorn]

Actually, it doesn't decrease. The normal force decreases, but static friction must increase to offset the increasing opposing force (sin component of weight).



Again, no! ezsanche is correct in that equation. According to the force diagram, the weight of the stationary object is offset by the normal force and the static frictional force. The normal force cancels out the cosine component, so the static friction must be offsetting the sine component.

You are thinking that Fstatic = mustaticN = mustaticmgcos(theta). What ezsanche wrote does NOT include mustatic, so before telling him he's wrong, you should look more carefully.

The end result is:
mustaticmgcos(theta) = mgsin(theta), which leads to the conclusion that mustatic = sin(theta)/cos(theta) = tan(theta). The value of mustatic increases with the angle of inclination until it breaks free, at which time we conclude that mustaticmax = tan(theta)breakaway

you're right on the second part, my mistake for a quick looksie. you're wrong on the first. period. read what i said. (ironic.) MAXIMUM. you imply the same with your last line.

 

[RogueUnicorn]

The actual static force increases until it reaches the maximum, at which time it breaks free once the maximum has been exceeded. The maximum static friction does not decrease. It's a threshold value that depends on the materials that are interacting.

the coefficient is a constant - the maximum force is not. there's a formula, one which you've conveniently supplied.

 

[JohnnyBravo]

the coefficient is a constant - the maximum force is not. there's a formula, one which you've conveniently supplied.

The MAXIMAL COEFFICIENT IS CONSTANT. The actual mu static is a function of the angle as BRT said. If at some angle there is a force static that keeps it from moving, then at that angle mgsine=mgcosine(theta)*mu static. This we all agree.

However, the mu static that goes into that equation is NOT the MAXIMAL COEFFICIENT. This is what BRT was trying to say. As the angle increases, the mu exerted increases until it reaches the MAXIMAL value and then the block slides. You are INCORRECT in saying that as the angle increases, the MAXIMAL STATIC FRICTIONAL FORCE DECREASES. This is false, since you don't reach the maximum static force until you reach the highest angle.

BRT statement of this is key, mu static=tangent(theta). This mu static increases to a maximal value as the angle increases.

The same is true of the flat case. If we have a Maximal Mu Static of .4 and a box of 100kg then Maximal Mu Static Force is 400. If I apply a force of 200 then the mu static is 200. However, what is the MU STATIC APPLIED? It's not .4. It is 200=mgcosine(theta)*mu static. In this case then Mu static=.2 and not .4 the Maximal value

Again, the reason that force static supplied changes is because MU STATIC SUPPLIED changes. In closing, the Maximal Mu which is a function of the material doesn't change. However, it is not exerted unless needed. Not only is the static force exerted generally less than the MAXIMAL STATIC FORCE but the mu static exerted is also generally less than the MAXIMAL MU STATIC by necessity. Good discussion all. Bleargh, I am not bolding to belittle you, it's so anyone else reading can pick up on the nuances. I actually am amazed at your encyclopedic knowledge of all that is MCAT. I had to go back to my old textbooks to get this right. I know I'm right because I remember an example in physics class on this very concept and 90% of the class got it wrong.

 

[BerkReviewTeach]

I'm not buying the whole argument that the coefficient of static friction increases as the angle of elevation increases. the coefficient of static friction is an instrinic quality that depends on the surfaces of the two objects that are in contact.

How do you explain the following. A box on a flat surface has no static friction force, as long as no lateral force is applied. This means that mustaticN = 0. Clearly, N cannot be 0, so mustatic must be equal to zero. Once you raise the incline, assuming the box does not slip, there must be an increasing REACTIONARY force holding it in place. The normal force is decreasing, so the only term to explain it is the mustatic increasing. It increases from zero until it's maximum value.

They wouldn't add the caveat maximum, if there wasn't a minimum.

 

[BerkReviewTeach]

you're right on the second part, my mistake for a quick looksie. you're wrong on the first. period. read what i said. (ironic.) MAXIMUM. you imply the same with your last line.

You said, "while it is true the maximum static friction decreases as you increase the angle." This is absolutely beyond any sense of the imagination wrong. You are making some very fundamental errors here, based on a lack of understanding of the mustatic term. It varies, because it is a reactionary term. See my previous post about the box on the flat surface. It will help you understand this better.

the coefficient is a constant - the maximum force is not. there's a formula, one which you've conveniently supplied.

It is NOT a constant, much like the rate constant in chemical kinetics in not a constant. You have to let go of your preconceived notion that mustatic is a constant, then you'll actually get this. It is equal to tan(theta), which means it varies with angle of incline. In more general terms, it varies with applied force.

Consider pushing a box on a carpeted surface. It takes a threshold force to move the box. The harder you push, the harder it pushes back. The normal force is constant and static friction is increasing, so it has to be that mustatic is increasing.

Listen to Johnny Bravo here, because he's spot on with this.

 

[RogueUnicorn]

The MAXIMAL COEFFICIENT IS CONSTANT. The actual mu static is a function of the angle as BRT said. If at some angle there is a force static that keeps it from moving, then at that angle mgsine=mgcosine(theta)*mu static. This we all agree.

However, the mu static that goes into that equation is NOT the MAXIMAL COEFFICIENT. This is what BRT was trying to say. As the angle increases, the mu exerted increases until it reaches the MAXIMAL value and then the block slides. You are INCORRECT in saying that as the angle increases, the MAXIMAL STATIC FRICTIONAL FORCE DECREASES. This is false, since you don't reach the maximum static force until you reach the highest angle.

BRT statement of this is key, mu static=tangent(theta). This mu static increases to a maximal value as the angle increases.

The same is true of the flat case. If we have a Maximal Mu Static of .4 and a box of 100kg then Maximal Mu Static Force is 400. If I apply a force of 200 then the mu static is 200. However, what is the MU STATIC APPLIED? It's not .4. It is 200=mgcosine(theta)*mu static. In this case then Mu static=.2 and not .4 the Maximal value

Again, the reason that force static supplied changes is because MU STATIC SUPPLIED changes. In closing, the Maximal Mu which is a function of the material doesn't change. However, it is not exerted unless needed. Not only is the static force exerted generally less than the MAXIMAL STATIC FORCE but the mu static exerted is also generally less than the MAXIMAL MU STATIC by necessity. Good discussion all. Bleargh, I am not bolding to belittle you, it's so anyone else reading can pick up on the nuances. I actually am amazed at your encyclopedic knowledge of all that is MCAT. I had to go back to my old textbooks to get this right. I know I'm right because I remember an example in physics class on this very concept and 90% of the class got it wrong.

what i'm confused about is how what you're saying is in any way different to what i am saying. i'm not talking about maximum exerted, but rather maximum theoretical as calculated by the formula involving cosine. with your example, the static friction at that moment in 200N. i agree. however, i can apply 200N of pulling force on it and it will still not move. do you not agree? the maximum static friction force is 400N whether you actually "need" it or not

 

[RogueUnicorn]

It is NOT a constant, much like the rate constant in chemical kinetics in not a constant. You have to let go of your preconceived notion that mustatic is a constant, then you'll actually get this. It is equal to tan(theta), which means it varies with angle of incline. In more general terms, it varies with applied force.

uh... ok? you're saying to coefficient of friction is NOT a constant? really? this would absolutely be news to me.

http://en.wikipedia.org/wiki/Friction#Coefficient_of_friction

"The coefficient of friction (COF), also known as a frictional coefficient or friction coefficient and symbolized by the Greek letter &#956;, is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. The coefficient of friction depends on the materials used;"

 

[Beta Cell]

How do you explain the following. A box on a flat surface has no static friction force, as long as no lateral force is applied. This means that mustaticN = 0. Clearly, N cannot be 0, so mustatic must be equal to zero. Once you raise the incline, assuming the box does not slip, there must be an increasing REACTIONARY force holding it in place. The normal force is decreasing, so the only term to explain it is the mustatic increasing. It increases from zero until it's maximum value.

They wouldn't add the caveat maximum, if there wasn't a minimum.

Excellent point.

 

[JohnnyBravo]

while it is true the maximum static friction decreases as you increase the angle. HOWEVER, you must remember that static friction is a reactive force and as such must remember that as you increase the angle, you increase the component of gravity pulling the box down. so, the ACTUAL static friction force increases, until the decreasing maximum and increasing pulling force intersect and the box starts moving.


edit: this is essentially what wanderer said. ha. remember that Fs=<mu*mg*cos(Th)

We all agree that the Force static friction applied increases as the angle increases However, you said that the Maximum static friction decreases as you increase the angle, which isn't true. I thought it was, and then after consulting my book realized it wasn't. So, that's why I bolded that part so anyone reading can see why it's incorrect. The Maximum static friction is constant and the mu static supplied actually increases as the angle is increased until it reaches Maximal MU static at which point it gives off the Maximal Force Static

 

[RogueUnicorn]

dp

 

[RogueUnicorn]

We all agree that the Force static friction applied increases as the angle increases However, you said that the Maximum static friction decreases as you increase the angle, which isn't true. I thought it was, and then after consulting my book realized it wasn't. So, that's why I bolded that part so anyone reading can see why it's incorrect. The Maximum static friction is constant and the mu static supplied actually increases as the angle is increased until it reaches Maximal MU static at which point it gives off the Maximal Force Static

the only way this can be true, for maximum static friction to be constant while increasing the angle, provided the formula Fs =< (coeff. stat. fric) * mgcos(theta), is if the coefficient of static friction has a corresponding, equal in magnitude increase with the angle.

what you highlight would be the case for a STATIC angle theta with increasing force applied in the direction parallel to the surface of the incline

 

[JohnnyBravo]

uh... ok? you're saying to coefficient of friction is NOT a constant? really? this would absolutely be news to me.

http://en.wikipedia.org/wiki/Friction#Coefficient_of_friction

"The coefficient of friction (COF), also known as a frictional coefficient or friction coefficient and symbolized by the Greek letter &#956;, is a dimensionless scalar value which describes the ratio of the force of friction between two bodies and the force pressing them together. The coefficient of friction depends on the materials used;"

Bleargh. We agree with you. What we're saying is that the box doesn't always give off the Maximal Mu. Look at my example. Like you said, if Mu is .4 for the 100 Kg box, then the Maximal Staic Friction is 400 when Mu has its MAXIMUM STATIC Value. However, if I push 200, then we all know that static friction is 200.

Now, here is where the argument/confusion lies. At the point that it is giving out 200 N, what is the Mu static? It is 200=100(10)cos(0)(MU static).

Mu static here is .2 and NOT .4. No one is arguing that the Maximum value is .4. However, that value isn't always exerted if the force applied is less than Mu static. How does the box "Only" apply 200? It's because it lowers its mu static to .2. As you "Push harder" the box also "pushes back harder." How does it do this, but increasing Mu "Proportionally to you until it reaches its maximal value of .4). At this point, the box begins to slide.

 

[JohnnyBravo]

the only way this can be true, for maximum static friction to be constant while increasing the angle, provided the formula Fs =< (coeff. stat. fric) * mgcos(theta), is if the coefficient of static friction has a corresponding, equal in magnitude increase with the angle.

what you highlight would be the case for a STATIC angle theta with increasing force applied in the direction parallel to the surface of the incline

Yes, that is what I was assuming.

 

[RogueUnicorn]

Yes, that is what I was assuming.

i have never heard of the coefficient of static (or any) friction changing based on angle. if you can show me some place i can find this info i'd be much obliged

 

[RogueUnicorn]

Bleargh. We agree with you. What we're saying is that the box doesn't always give off the Maximal Mu.

this doesn't, i think, have much to do with my assertion that as you increase incline angle, the maximum force of static friction possible must decline due to a decrease in normal force

 

[JohnnyBravo]

this doesn't, i think, have much to do with my assertion that as you increase incline angle, the maximum force of static friction possible must decline due to a decrease in normal force

Yes it does, because as you increase the angle, Mu static supplied is increasing and eventually you reach the Maximal Static Force.

Lets say we have a car with super tires in the middle of a 2 meter long board. At rest there is no static friction force. Now, as I lift the board it still doesn't slide. There must be static friction. You're saying that as I increase the angle, the Maximal static friction possible decreases. How can this be? We all agree that the static friction force applied increases. However, the maximal static frictional force CAN'T decrease as you increase the angle because the STATIC FRICTIONAL FORCE SUPPLIED INCREASES.

As I continue to raise it to say 50 it stays and then begins to slide. At this point, I've reached the Maximal Frictional Force and thus the Maximal Mu static.

At 25 degrees there was a force static. This force static is obviously less than the Maximal force static. So, please explain to me, how the maximal static frictional force decreases as the angle increases? Let's say that the max force static is 50 at the 50 degree angle. Let's say it's 25 at the 25 degree angle for simplicity. The force static supplied at 25 degrees is less THAN THE MAXIMAL STATIC FORCE which isn't supplied until a higher angle is reached. How does increasing the angle lower THE MAXIMAL STATIC FORCE. The answer is that it doesn't. The Maximum MU static isn't used until it's necessary. The force static increases as you increase the angle just like it does as you increase the force on a box on a flat bed.

If what you're saying is true, then that means as I push harder on a flat box and it gives out a greater static force, that the Maximal Static force that the box can give out decreases. This isn't true. What decreases is the amount of force static that the Force static can give out before the box starts to slide.

 

[RogueUnicorn]

the maximal static force decreasese because of the decrease in normal force due to the increase in inclination i don't know how else to put this.

the maximum force and the supplied force can change independently up to the point which they equal each other

 

[ezsanche]

I get what you are saying now. You are saying that the coefficient increases not the max coefficient. The max coeff is a constant and the rate constant is a constant as long as temperature is constant.

 

[JohnnyBravo]

the maximal static force decreasese because of the decrease in normal force due to the increase in inclination i don't know how else to put this.

the maximum force and the supplied force can change independently up to the point which they equal each other

I get what you're saying, and it's wrong.

That isn't the maximal static friction though. This is what we are arguing. The maximum static friction is constant. It doesn't change. You're saying that the maximum static friction decreases because fnormal decreases. That makes no sense. Why is the Force friction supplied increasing?

Again, mgcos(theta)*ms=fstatic
You're saying that as we increase the angle, the MAXIMAL STATIC FRICTIONAL FORCE DECREASES. Not true. The Maximal Static Frictional Force has ONE value that isn't reached until you reach the MAXIMAL ANGLE before the box slides. That value doesn't decrease.

You never answered my question. mgcos(theta)*ms=fstatic You state and we agree, that as you raise the angle force static supplied increases. The only way for this to happen is that mu static supplied increases over the smaller normal force. You're saying (incorrectly), that because the normal force component decreases, that the MAXIMAL FRICTIONAL FORCE STATIC decreases. This is wrong on two levels. The first is that the force static increases and it does because Mu static increases as the angle increases overcoming the smaller normal force. The second reason is that you're saying that the MAXIMUM FRICTIONAL FORCE CHANGES. It doesn't. It has only one value. If anything, what you say would make more sense that the Maximal Static Force increases with increasing angle. Why? The MU static value increasing is greater than the decrease in normal force as angle increases. This increases the static force supplied and would make more sense than your current one. It would still be invalid, because there is only One Maximal friction force BEFORE the block moves. What you're saying is that as I raise the level of the board, the static force needed to overcome in order to move decreases which we've shown isn't true at all.

the maximum force and the supplied force can change independently up to the point which they equal each other
Again, that's like saying that for the 400 Maximal Static Friction example I gave, that as I increase my force from 200 to 300, and the force static supplied increases from 200 to 300, that the Maximal potential energy decreases to 350. You realize this is what you are saying right? The Maximal frictional force is 400 and has only one value. The frictional force supplied decreases.

Nothing decreases. Why is the frictional force supplied increasing? It's increasing because the force supplied is moving closer to the Maximal Force supplied due to an INCREASING MU. The Maximal force supplied has an INHERENT MAXIMAL VALUE that DOESN'T DECREASE because you're increasing the angle.

I took Calc Based physics and a couple elective classes and what you're saying has never been broached. It doesn't make sense. I get what you're saying as did BRT, and we stated why what you said doesn't make sense. What would the Maximal Static Force decrease with the angle? You're stating that the decreasing Normal force with an increasing angle causes the Maximal Static Force to decrease with an increasing angle. BRT carefully showed why this isn't the case. If this were true, than static friction WOULDN'T INCREASE since the DIMINISHING NORMAL FORCE would overcome the INCREASING MU with angle.

 

[ezsanche]

ok but here is what I am thinking since now my mind is addled by reading both sides of the arguments.

Let say we have a 50 newton object that is place on a flat table.
It takes 50 newtons to make it move from rest. (Lets say the coefficient of static friction is 1.)

It would seem to me that the max static frictional force is 50.

Now if you place the object on a ramp and increase the angle it will remain at rest. But if you increase the the ramp to 45 degrees it will be on the verge of slipping. The maxi static force is now a fraction than it was before when it was lying flat on the table.


Seems to me that the max static frictional force has changed.
So confused right now

 

[BennieBlanco]

I get what you're saying, and it's wrong.

That isn't the maximal static friction though. This is what we are arguing. The maximum static friction is constant. It doesn't change. You're saying that the maximum static friction decreases because fnormal decreases. That makes no sense. Why is the Force friction supplied increasing?

Again, mgcos(theta)*ms=fstatic
You're saying that as we increase the angle, the MAXIMAL STATIC FRICTIONAL FORCE DECREASES. Not true. The Maximal Static Frictional Force has ONE value that isn't reached until you reach the MAXIMAL ANGLE before the box slides. That value doesn't decrease.

You never answered my question. mgcos(theta)*ms=fstatic You state and we agree, that as you raise the angle force static supplied increases. The only way for this to happen is that mu static supplied increases over the smaller normal force. You're saying (incorrectly), that because the normal force component decreases, that the MAXIMAL FRICTIONAL FORCE STATIC decreases. This is wrong on two levels. The first is that the force static increases and it does because Mu static increases as the angle increases overcoming the smaller normal force. The second reason is that you're saying that the MAXIMUM FRICTIONAL FORCE CHANGES. It doesn't. It has only one value. If anything, what you say would make more sense that the Maximal Static Force increases with increasing angle. Why? The MU static value increasing is greater than the decrease in normal force as angle increases. This increases the static force supplied and would make more sense than your current one. It would still be invalid, because there is only One Maximal friction force BEFORE the block moves. What you're saying is that as I raise the level of the board, the static force needed to overcome in order to move decreases which we've shown isn't true at all.

the maximum force and the supplied force can change independently up to the point which they equal each other
Again, that's like saying that for the 400 Maximal Static Friction example I gave, that as I increase my force from 200 to 300, and the force static supplied increases from 200 to 300, that the Maximal potential energy decreases to 350. You realize this is what you are saying right? The Maximal frictional force is 400 and has only one value. The frictional force supplied decreases.

Nothing decreases. Why is the frictional force supplied increasing? It's increasing because the force supplied is moving closer to the Maximal Force supplied due to an INCREASING MU. The Maximal force supplied has an INHERENT MAXIMAL VALUE that DOESN'T DECREASE because you're increasing the angle.

I took Calc Based physics and a couple elective classes and what you're saying has never been broached. It doesn't make sense. I get what you're saying as did BRT, and we stated why what you said doesn't make sense. What would the Maximal Static Force decrease with the angle? You're stating that the decreasing Normal force with an increasing angle causes the Maximal Static Force to decrease with an increasing angle. BRT carefully showed why this isn't the case. If this were true, than static friction WOULDN'T INCREASE since the DIMINISHING NORMAL FORCE would overcome the INCREASING MU with angle.

Lol. There is physics and thinking, then there are people arguing over wording. Both parties understand the concept correct (obviously, as bleargh got 15 in PS).

Moving to Linguistics Forum

that the MAXIMAL FRICTIONAL FORCE STATIC decreases. This is wrong on two levels. The first is that the force static increases and it does because Mu static increases as the angle increases overcoming the smaller normal force. The second reason is that you're saying that the MAXIMUM FRICTIONAL FORCE CHANGES. It doesn't. It has only one value.

This is really a dumb argument. The maximum force is what the force is at the moment. Theoretically it can change, SURE, but at any moment the maximum force is what it is. You may say that it can reach 400 N but it could even reach 500 N or maybe only 0 N, it depends on where the box is in time and space.

You're argument shows that you understand reading a book but you don't understand physics.

mgcos(theta)*ms=fstatic is an equation physicists developed because it holds true on earth in all the experimental situations tested. It is no more a law that any laws made by a government, it can be broken and manipulated. We use equations to describe things that happen in our experience. Gravity as we know it isn't even applicable in the entire universe, but it allows us to describe things on earth and in our solar system.

It is funny to argue over such a simple concept just because of wording.

the Maximal frictional force is 400 and has only one value.

If I move the box next to a black hole then this isn't true. So you are wrong. See, anyone can play the language game. I will buy your statement that it is maxed at 400 just as I will buy bleargh's statement that the maximum is changing. Either is fine. Everyone is right, or maybe everyone is wrong. Physics isn't about absolutes, it is about testing what is applicable in the universe in a situation.

Newton's laws are not laws, they just describe motion/forces in most of our solar system.

 

[BennieBlanco]

Sorry, just too funny.

the maximal static force decreasese because of the decrease in normal force due to the increase in inclination i don't know how else to put this.

You're saying that the maximum static friction decreases because fnormal decreases. That makes no sense. Why is the Force friction supplied increasing?

Again, mgcos(theta)*ms=fstatic
You're saying that as we increase the angle, the MAXIMAL STATIC FRICTIONAL FORCE DECREASES. Not true. The Maximal Static Frictional Force has ONE value that isn't reached until you reach the MAXIMAL ANGLE before the box slides. That value doesn't decrease.

JohnnyBravo, stop being so nerdy. He stated the TOTAL force that is exerted on the box is going down as the incline increases.

You say the max has one value and he stated the max is the force exerted at a moment, either is fine.

You understand this, he understands it, and even C students in physics understand this idea. Stop mincing words.

But. But. But. pg. 243 says...

 

[JohnnyBravo]

This is really a dumb argument. The maximum force is what the force is at the moment. Theoretically it can change, SURE, but at any moment the maximum force is what it is. You may say that it can reach 400 N but it could even reach 500 N or maybe only 0 N, it depends on where the box is in time and space.

You're hilarious. I got a 14 on PS so Bleargh is clearly more qualified. The irony is beyond me. The MCAT tests basic concepts and has no bearing on the true conceptual knowledge of many concepts. I'm the one using examples and providing concrete proof. Funny, BRT agreed with me and provided an example showing what Bleargh stated was incorrect.

You're the one who is lacking understanding of this concept. The Static Frictional force that is given out isn't always MAXIMAL. So, you're saying that if I have a block that doesn't slide until I raise it to a level of 50 degrees, that every static force exerted before that is MAXIMAL? That's hilarious and so wrong.

You mocked me in your post by saying the following: The maximum force is what the force is at the moment. No, this is incorrect and was the source of confusion in the original poster. The static force supplied isn't always equal to the MAXIMUM STATIC FRICTIONAL FORCE. That is the whole point of this argument. Don't interject if you don't understand what's going on. I'm not arguing Bleargh over wording. The fact that you said that tells me all I need to know.

The maximal static friction for a level system is equal to fs max=Fnormal*mu static maximal. Let's say the road has a static friction of .4 which is actually a decent value. I have a car that weighs 100 kg. The Maximum static frictional force is 400 Newtons. If I push with a force of 300 Newtons, the Car exerts a Static Friction Force of 300 Newtons back on me. Is this Static Friction maximal? Anyone who understands this concept would say no. If I apply a force of 400, then obviously it will apply a force of 400. At this point, the STATIC FRICTION FORCE SUPPLIED is EQUAL TO THE MAXIMAL STATIC FRICTIONAL FORCE. Now, if I apply a force of 500, I will accelerate because I've overcome the maximal static frictional force.

Now, what Bleargh states and is wrong, is that since in an incline the NORMAL FORCE DECREASES as THE ANGLE INCREASES, that the MAXIMAL FRICTIONAL FORCE DECREASES. I and BRT have been arguing that this isn't the case. Bleargh draws this from force max static=fnormal*mu max static.

 

[JohnnyBravo]

BRT and I tried to show that this isn't the case for an incline. In an incline the force static=mgcos(theta)*mustatic. So, we know that the fnormal=mgcos(theta). Well, since this value decreases as we increase the angle, Bleargh states that the Maximal Static Frictional Force decreases. This is wrong, since the force static at that point is NOT the maximal friction force. At all these points where it's still not sliding, it means that the force due to gravity is equal to the force due to static. As was done earlier, this gives a value of mu static=tan theta. So, the MU STATIC INCREASES as THE ANGLE INCREASES. This OVERCOMES any DECREASE in force normal to INCREASE THE STATIC FRICTIONAL FORCE SUPPLIED UNTIL IT REACHES ITS MAXIMAL VALUE.

Again you mock me, You're argument shows that you understand reading a book but you don't understand physics. What you said was irrelevant. I'm not belittling anybody. I minored in Physics and took a ton of classes. I've discussed this topic ad nauseum with students I tutor and with my professor and saw an erroneous statement.

This is not semantics, Bleargh said something that is incorrect. Bleargh wasn't even willing to admit that MU STATIC changes until after BRT stepped in with a few posts. So, don't come here and say I have no idea what I'm talking about.

We all agreed that Force Static Friction supplied increases with the angle. Then BRT and myself said this is because Mu static increases. Some posters said this wasn't true. If it wasn't, how could the Mu static supplied INCREASE while FORCE NORMAL DECREASES? I wasn't the one who pointed out Bleargh was wrong. It was BRT. I ACTUALLY AGREED WITH BLEARGH. I saw BRT explanation and went to consult my books. I then realized that I WAS WRONG. Then I tried to explain to Bleargh why he was wrong.

In closing, thanks for belittling me for consulting my books to correct a conceptual error I made.

 

[BennieBlanco]

This is really a dumb argument. The maximum force is what the force is at the moment. Theoretically it can change, SURE, but at any moment the maximum force is what it is. You may say that it can reach 400 N but it could even reach 500 N or maybe only 0 N, it depends on where the box is in time and space.

Now, what Bleargh states and is wrong, is that since in an incline the NORMAL FORCE DECREASES as THE ANGLE INCREASES, that the MAXIMAL FRICTIONAL FORCE DECREASES. I and BRT have been arguing that this isn't the case. Bleargh draws this from force max static=fnormal*mu max static.

Lol!

So your argument is this is wrong:

since in an incline the NORMAL FORCE DECREASES as THE ANGLE INCREASES, that the MAXIMAL FRICTIONAL FORCE DECREASES.

But this is right:

since in an incline the NORMAL FORCE DECREASES as THE ANGLE INCREASES, that the maximum FRICTIONAL FORCE exerted DECREASES.

WOW! Someone give this kid a nobel prize!

 

[ezsanche]

Sorry, just too funny.





JohnnyBravo, stop being so nerdy. He stated the TOTAL force that is exerted on the box is going down as the incline increases.

You say the max has one value and he stated the max is the force exerted at a moment, either is fine.

You understand this, he understands it, and even C students in physics understand this idea. Stop mincing words.

LMAO this post had me dying of laughter. Okay lets not be mean to each other.

 

[JohnnyBravo]

A 10-kg block is at rest on an inclined plane. The coefficient of static friction is 0.75. The force of static friction is 17 N. Which of the following would NOT cause the force of static friction to increase?

A. Increasing the mass of the block.
B. Increasing the angle of incline.
C. Increasing the acceleration due to gravity.
D. Increasing the coefficient of static friction.

I chose B, but the correct answer is D. I kind of understand how B would be wrong, but I'm not absolutely sure here. Originally, I thought that since F(friction) = umgcos(theta), if you increase theta, the cos(theta) will decrease and so will F(friction). But apparentely, F(friction) would not decrease--if it did, it would be the correct answer for this question.

So does this mean the coefficient of static friction will increase as cos(theta) decreases to match the force being applied on the object?

As for answer D, the only way I can see this being the correct answer if it is actually referring to the maximum coefficient of static friction it could potentially be given an applied force to counteract. In that case, the applied force will remain the same, as will the force of static friction. Can someone help with or verify my reasoning for B and D? Thanks.

You see, the OP was wondering why D is the answer and not B. The fact that the question states that INCREASING THE MAXIMAL MU wouldn't increase the force friction, implies that the FORCE FRICTION SUPPLIED isn't maximal. If the static friction supplied WERE MAXIMAL, then increasing the coefficient would increase the MAXIMAL FORCE FRICTION SUPPLIED. As it is, increasing the Max Mu, has no bearing on the 17 Newtons supplied because it isn't maximal.

Now, here's what Bleargh should say. For the given problem, if the Box were to be laid flat with the same Maximal friction, then the MAXIMAL FRICTION SUPPLIED would by 75 Newtons. Now, because the box is on an incline, The Maximal Friction Supplied IS LESS THAN WHAT IT WOULD BE IF IT WERE LYING FLAT. This is correct.

Bleargh didn't say this, he said that THE MAXIMAL STATIC FRICTION POSSIBLE ON AN INCLINE, DECREASES WITH INCREASING ANGLE WHICH ISN'T THE CASE. Once you are on an incline, then what determines the MAXIMAL STATIC FRICTION IS THE ANGLE AND THE MU VALUE. So spare me the BS and nerdy talk. If what Bleargh had stated was incorrect, then BRT who knows more than me, wouldn't have stepped in.

 

[BennieBlanco]

BRT and I tried to show that this isn't the case for an incline. In an incline the force static=mgcos(theta)*mustatic. So, we know that the fnormal=mgcos(theta). Well, since this value decreases as we increase the angle, Bleargh states that the Maximal Static Frictional Force decreases. This is wrong, since the force static at that point is NOT the maximal friction force.

Wow. Again, you say this is wrong:

the Maximal Static Frictional Force decreases. This is wrong, since the force static at that point is NOT the maximal friction force.

But this is ok:

the *OMIT* Static Frictional Force decreases.

Who cares dude. It is pretty obvious that everyone knows that the force is maximum when cos(theta) is at 0.

So, don't come here and say I have no idea what I'm talking about.

It's funny how people take 30 units or less of some subject think they are experts while others study for 8 years and come off more humble.

In closing, thanks for belittling me for consulting my books to correct a conceptual error I made.

It is how you say what you say.

You: I understand what you are saying. You are wrong.

Cool person: I think what is really going on here is ______.

 

[BennieBlanco]

LMAO this post had me dying of laughter. Okay lets not be mean to each other.

Lol.

 

[JohnnyBravo]

No you don't get it.

BRT stated why Bleargh was wrong. It's ironic. You don't get that the the static frictional force supplied doesn't always equal maximal frictional force. I'm done. I love how you completed ignored my factual example and interjected your inaccurate quotes.

 

[JohnnyBravo]

You said, "while it is true the maximum static friction decreases as you increase the angle." This is absolutely beyond any sense of the imagination wrong. You are making some very fundamental errors here, based on a lack of understanding of the mustatic term. It varies, because it is a reactionary term. See my previous post about the box on the flat surface. It will help you understand this better.



It is NOT a constant, much like the rate constant in chemical kinetics in not a constant. You have to let go of your preconceived notion that mustatic is a constant, then you'll actually get this. It is equal to tan(theta), which means it varies with angle of incline. In more general terms, it varies with applied force.

Consider pushing a box on a carpeted surface. It takes a threshold force to move the box. The harder you push, the harder it pushes back. The normal force is constant and static friction is increasing, so it has to be that mustatic is increasing.

Listen to Johnny Bravo here, because he's spot on with this.

But you're right, I'm wrong and have no idea what I'm talking about.

 

[BennieBlanco]

Bleargh didn't say this, he said that THE MAXIMAL STATIC FRICTION POSSIBLE ON AN INCLINE, DECREASES WITH INCREASING ANGLE WHICH ISN'T THE CASE.

Again, linguistics. Not physics. Lets sum up your argument in 2 sentences:

Incorrect:
THE MAXIMAL STATIC FRICTION POSSIBLE ON AN INCLINE, DECREASES WITH INCREASING ANGLE

Correct:
THE MAXIMAL STATIC FRICTION SUPPLIED ON AN INCLINE, DECREASES WITH INCREASING ANGLE


Wowsers!!! Seriously, is that your argument!? who cares!

 

[BennieBlanco]

But you're right, I'm wrong and have no idea what I'm talking about.

I didn't say you were wrong. You are right. I said you were a nerd for mincing words and making a huge deal out of it with essay posts.

 

[JohnnyBravo]

ok but here is what I am thinking since now my mind is addled by reading both sides of the arguments.

Let say we have a 50 newton object that is place on a flat table.
It takes 50 newtons to make it move from rest. (Lets say the coefficient of static friction is 1.)

It would seem to me that the max static frictional force is 50.

Now if you place the object on a ramp and increase the angle it will remain at rest. But if you increase the the ramp to 45 degrees it will be on the verge of slipping. The maxi static force is now a fraction than it was before when it was lying flat on the table.


Seems to me that the max static frictional force has changed.
So confused right now

Yes, the maximal frictional force on an incline is less than lying flat. However, that's not what bleargh said. Which is why BRT corrected him. For the record, I agreed with Bleargh. Then I read what BRT said and realized I was wrong. I then went and looked up my old books and saw that BRT was correct.

I bet Bennie thinks that if in the example that you gave that if I supply a force of 25 Newtons then that is the maximal static force. I bet he also thinks that the mu static supplied is 1 when at the time I'm pushing it's actually .5.

The point of contention is that ONCE ON AN INCLINE, the Maximal Static Friction has a MAXIMAL VALUE that DOESN'T DECREASE AS THE ANGLE INCREASES, it has ONE value that is continuously approached until the box begins to slide.

HOWEVER, and here's the kicker, that Maximal Value on the Incline DOESN'T DECREASE AS THE ANGLE INCREASES, it stays the same but is NOT EXERTED UNTIL THE MAXIMUM ANGLE IS REACHED.

Moreover, EZ, if like you said the MAXIMAL VALUE IS ATTAINED AT 45, then that is the Maximal Value because we are ONLY considering the incline. The Static friction at 45 is the MAXIMAL ON THE INCLINE, because past this point the box slides. Any value less than 45 is not the maximal. We were only considering the INCLINE, because, well, it's an incline problem. The Max stat friction at 45 on the incline will be less than while lying flat but it DOESN'T decrease with the angle as Bleargh said. MU STATIC MAX, which is what we're after, is a function of TAN(THETA) and reaches a maximal VALUE BASED ON THE MATERIAL AND NOT A DECREASING force normal.

In closing, what determines the Maximal value on the incline is the materials. To illustrate, If I have 4 identical boxes and 4 incline planes. Let's say that Incline A has a mu of .2, B, .3, C, .4, and D .5. Let's say each weighs 10 kg.

Each incline is initially laid flat on the ground, at this point the Maximal STATIC FRICTION of each is : 20, 30, 40, and 50 respectively. Now, we raise each incline slowly. What we will find is that box A will slide first say at 20, B second say at 30, C third at 40, and D 5th at 50. Why did d have the highest angle? It did because it had the highest Mu. This illustrates THAT THE MAX MU ON AN INCLINE HAS ONE VALUE AND IS DEPENDENT ON THE MU OF THE SURFACE. That's what BRT showed and it made me realize I was wrong.

 

[JohnnyBravo]

Again, linguistics. Not physics. Lets sum up your argument in 2 sentences:

Incorrect:
THE MAXIMAL STATIC FRICTION POSSIBLE ON AN INCLINE, DECREASES WITH INCREASING ANGLE

Correct:
THE MAXIMAL STATIC FRICTION SUPPLIED ON AN INCLINE, DECREASES WITH INCREASING ANGLE


Wowsers!!! Seriously, is that your argument!? who cares!

Lol, The second statement is wrong. I give up. BRT clearly explained that the Force supplied increases as the angle increases OVERCOMING the decreasing normal component.

Finally, compared to lying on the ground the maximal force on an incline is reduced. However, it doesn't decrease with an increasing angle. The angle is IRRELEVANT. It is based on Mu and has ONE value. I'm done. BRT can have a go at it if he wants.

Also, I'm no nerd. I ADMITTED i AGREED WITH BLEARGH. So, when someone points out you're wrong like BRT did. You can assume you're right, or go seek information. That makes one nerdy. This is ironic coming from the guy who wants to do 35 tests for the MCAT and has a 3 month super schedule. Newsflash, the MCAT isn't that hard. Seriously not necessary. I took 8 and it was more than enough. I'm no genius like Bleargh. If you look at the 30 plus you'll see PandaBrewMaster, and lovewalk, who prepared within a short time an to 38s. Getting a 35 -37 on the MCAT isn't to hard provided you're good at verbal. 12s on the sciences aren't hard. Getting 14s and above is but requires much more work. I averaged a 36 and and got a 37 on the real.