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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[JohnnyBravo]

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

Bleargh I see the source of confusion. You're going by this scenario:
10kg box on a board with a mu of .5. The max static friction is 50.

Now, as I lift the board, force static is increasing to offset the increasing force due to gravity. Now, why is it increasing, because mu supplied is overcoming the decreasing force normal component. The force static will INCREASE TO A MAXIMAL VALUE say at 50 degrees. After this, say 51 it slides.

You are right that this maximal value on the incline is less than lying done due to a reduced force normal. You are wrong in stating that THE MAXIMAL FRICTIONAL FORCE DECREASES WITH INCREASING ANGLE. The Maximal frictional force on an incline is dependent on the MU VALUE AND AND HAS ONE CONSTANT VALUE THAT IS REACHED AT THE MAXIMAL ANGLE. This is what BRT was saying. We didn't realize you were talking about from lying flat.

If there are 4 boards with different MAX MU STATICS and we put four identical boxes on each one. The boxes will begin to slide at different angles. The point at which they slide is the MAXIMAL STATIC FRICTION ON THE INCLINE. This value doesn't vary, it IS INTRINSIC DUE TO THE MAXIMAL MU WHILE LYING FLAT ON THE GROUND. Now, because the Maximum Mu determines the MAXIMUM POSSIBLE STATIC FORCE POSSIBLE, and STATIC FRICTION SUPPLIED INCREASES WITH INCREASING ANGLE, each block slides at a different angle with the largest Mu value incline having the largest angle.

 

[RogueUnicorn]

i haven't read all of this because it's just too long for now, but johnnybravo, i seriously think, like i've stated waaaaaay long ago, we're not in disagreement, but instead we're talking semantics of what you define as maximal static friction. to you, it's the critical point at which supplied=max which has to be constant. to me, it's just max.

edit: guys i can't believe *i* am the one saying this, but let's keep the flaming on the down low. mmkay?

 

[ezsanche]

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.

Okay I get with your saying but there is the one thing that trips me up.

You are saying that as long that a object is on an incline the maximal static frictional force will remain constant. That the force of friction is a reactionary force that takes up the value of the applied force as long as the maximal static frictional force is not reached.

And that this is true only if we are talking about an object on an incline.

Well, let says that in my previous example the object was not lying flat on a table. It is placed on an incline of 1 degree. If we were to use the equation mu X Mg X Cos 1 we will calculate a value that is close to 49.98 newtons. However you contend that this not true because it is not on the verge of slipping, so this is not the value of maximal static frictional force.

BTW, the frictional force that the object is experiencing is at a value close to zero.

Now here is another scenario, imagine that on the same incline of 1 degree, there is a person at the higher end of the ramp pushing down on the object.
Let say he pushes with a force of 25 newtons. The frictional force that the object is experiencing is 25 newton plus the force of gravity which is small in comparision to what the man is applying. Now lets say that he pushes even harder at a force of 49.98 newton. At this time the object is at the verge of slipping. The maximal frictional force is now 49.98

Now in the last scenario there is no man applying a force. We put the object on an incline of 45 degrees. At this time, the object is on the verge of slipping. If we were to use the equation, mu X Mg X Cos 45, we
will calculate the Max static frictional force to be around 35.35 Newton.

The maximal frictional force is different in these two scenario.

(mu= 1 Mg= 50 )

 

[Tokspor]

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

I have a question about BerkReviewTeach's post. Was ezsanche correct in using the formula F(friction) = mgsin(theta), because of the reasoning that F(friction), as long as we're not talking about its maximum value, is a reactionary force to the force of gravity? If so, then it does make sense to me to set F(friction) equal to it.

So once the static friction reaches its maximum value, then we can use the equation F(friction) = mu(maximum)mgcos(theta), right?

But am I also correct in reading here that the coefficient of static friction is not constant? If that's the case, for any given angle, would it make sense to use F(friction) = mu(given angle)mgcos(theta) = mgsin(theta)? And an increasing "mu(given angle)" offsets a decreasing "cos(theta)" value?

 

[JohnnyBravo]

Okay I get with your saying but there is the one thing that trips me up.

You are saying that as long that a object is on an incline the maximal static frictional force will remain constant. That the force of friction is a reactionary force that takes up the value of the applied force as long as the maximal static frictional force is not reached.

And that this is true only if we are talking about an object on an incline.

Well, let says that in my previous example the object was not lying flat on a table. It is placed on an incline of 1 degree. If we were to use the equation mu X Mg X Cos 1 we will calculate a value that is close to 49.98 newtons. However you contend that this not true because it is not on the verge of slipping, so this is not the value of maximal static frictional force.

BTW, the frictional force that the object is experiencing is at a value close to zero.

Now here is another scenario, imagine that on the same incline of 1 degree, there is a person at the higher end of the ramp pushing down on the object.
Let say he pushes with a force of 25 newtons. The frictional force that the object is experiencing is 25 newton plus the force of gravity which is small in comparision to what the man is applying. Now lets say that he pushes even harder at a force of 49.98 newton. At this time the object is at the verge of slipping. The maximal frictional force is now 49.98

Now in the last scenario there is no man applying a force. We put the object on an incline of 45 degrees. At this time, the object is on the verge of slipping. If we were to use the equation, mu X Mg X Cos 45, we
will calculate the Max static frictional force to be around 35.35 Newton.

The maximal frictional force is different in these two scenario.

(mu= 1 Mg= 50 )

Exactly. EZ, I guess I was referencing something else. The Maximum POTENTIAL STATIC FRICTION DECREASES WITH INCREASING ANGLE BECAUSE OF THE DECREASING NORMAL FORCE. In that regard, BLEARGH is correct and this is what I though originally.

However, like you showed in the 49.98 versus the 35.35, that the MAXIMAL Mu STATIC COEFFICIENT INCREASES WITH INCREASING ANGLE. This is what causes the box not to slip as the FORCE GRAVITY INCREASES. At the 35.35, the Mu Static is equal to ONE just like ON LEVEL BOARD. At all the other angles it is less than one.

Now, what is of IMPORTANCE in your example is that ALTHOUGH the MAXIMAL STATIC FORCE VALUE ON AN INCLINE is at say one degree, that ASSUMING we let it go until it slides, that the MAXIMAL STATIC FRICTIONAL FORCE SUPPLIED is actually 35.35.

Why? At one degree it won't slide. At two it still won't, not until we RAISE IT TO 45 WILL IT BEGIN TO SLIP because at that point the FORCE STATIC PROVIDED IS MAXIMAL. This is what BRT pointed out. I WAS CONFUSED by this and have no shame admitting that. So, it made me reconsider the problem.

On inclines, what provides the static friction is the Fnormal and mu. If you PUSH, like you did, then that changes everything. Good discussion all.

 

[BennieBlanco]

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.

Getting a 35 -37 on the MCAT isn't to hard provided you're good at verbal.

Johnny Bravo-Type response:

This should say "TOO hard", not "to hard". I guess you aren't good at verbal.

You are wrong. Everything is wrong. Wrong.

**insert long useless 5 paragraph essay on proper use of the word "too" and "to", while consistently saying ___ is wrong. Wrong.**

[/BRAVO RESPONSE]

This isn't a graded assignment, people are less scrupulous on internet forums! WHAT?!

Normal person response:
Oh wait, I just realized you know it was supposed to be "too" and probably just made a semantic error. I guess I don't need to bore everyone with a long essay about something everyone already knows! Maybe I don't even post about it. hmmmmm.

Wow, that is so easy.

p.s. I find it flattering that you are following my study schedule, I have posted it maybe once or twice in the past 60 days. Do you also subscribe to my blog? Thx for adding to the readership! Do you mind if I use your term "3 month super schedule" for a later thread? thx again little buddy.

My prep will be more extensive because #1, I've been a pre-med for 11 months unlike many who have less extensive prep routines and #2 half of my pre-reqs were done in my first degree 4 years ago. Everyone is unique Mr. Bravo

 

[JohnnyBravo]

I have a question about BerkReviewTeach's post. Was ezsanche correct in using the formula F(friction) = mgsin(theta), because of the reasoning that F(friction), as long as we're not talking about its maximum value, is a reactionary force to the force of gravity? If so, then it does make sense to me to set F(friction) equal to it.

Yes, assuming that it doesn't move, the static frictional force must equal the force due to gravity .

So once the static friction reaches its maximum value, then we can use the equation F(friction) = mu(maximum)mgcos(theta), right?

Actually, you can use that AT ANY POINT IN WHICH THE BOX ISN'T MOVING. I will use EZ's example. The box weighs 10kgs Let's say that the Mu of an incline is 1. On level Ground THE MAXIMAL POTENTIAL STATIC FRICTION IS 10KG*10*1=100. Once on an incline it will DECREASE TO A LOWER VALUE. Note at this point that MU SUPPLIED IS 1. Let's say we have a box and an incline and we raise it from 0. Let's say we raise to 30. At this point the FORCE STATIC SUPPLIED = FORCE GRAVITY.

So, force static supplied=Mu(supplied)mgcos(theta)=mgsine(theta)
As, BRT showed, this comes to Mu supplied=tan(theta). At 30 degrees this value is .57, lower than 1! The FORCE STATIC SUPPLIED is 50, lower than the 100! Now, as EZ said, if you were to push on it the POTENTIAL STATIC SUPPLIED AT THIS ANGLE WOULD BE 86.7 NEWTONS WHICH IS LOWER THAN 100 BUT HIGHER THAN 50. However, BRT AND I ASSUMED NO PUSHING. This is IMPORTANT BECAUSE IT HIGHLIGHTS WHAT BLEARGH SAID.

But am I also correct in reading here that the coefficient of static friction is not constant? If that's the case, for any given angle, would it make sense to use F(friction) = mu(given angle)mgcos(theta) = mgsin(theta)? And an increasing "mu(given angle)" offsets a decreasing "cos(theta)" value?

THE COEFFICIENT SUPPLIED IS NOT CONSTANT. THE MAXIMAL IS CONSTANT, IT'S 1 IN THE EXAMPLE I GAVE. THE SUPPLIED VALUE INCREASES TO ONE AT THE MAXIMAL POINT.

Actually. This is what set off the argument. Bleargh was right in SAYING THAT AS THE ANGLE DECREASES, THE MAXIMAL STATIC FRICTION DECREASES BECAUSE FORCE NORMAL DECREASES, DESPITE THE FACT THAT AS YOU RAISE THE ANGLE THE FORCE SUPPLIED INCREASES UNTIL IT REACHES A MAXIMAL VALUE. THE EMPHASIS ON MAXIMAL, THIS ISN'T AS LARGE AS POSSIBLE DUE TO THE REDUCED NORMAL FORCE.

Now, let's say the max angle is 45 before it slips. Well, here Mgsin(45)= 70.7 Newtons. So MU SUPPLIED = TAN(45)=1!!. Now, although this 70.7 is less than 100, note the MU SUPPLIED= MU MAXIMAL!!! ALTHOUGH, FORCE STATIC SUPPLIED IS LESS THAN 100. If you were to push here, than the MAXIMAL FORCE STATIC AT THIS ANGLE IS 70.7 WHICH IS THE SAME AS FORCE SUPPLIED. So, ASSUMING NO PUSHING, this is the ACTUAL MAX STATIC YOU WOULD SEE, BUT NOT THE THEORETICAL MAX VALUE OF 100.

In closing, BLEARGH stated the following: as the angle increases, the maximum potential static friction decreases due to a decreasing normal force What BRT pointed out, As the angle increases, the Mu Supplied increases to the Maximal MU value, this is what causes the Mu Supplied to increase despite the DECREASING NORMAL FORCE COMPONENT. If you assume like in the ORIGINAL PROBLEM, that you RAISE the incline level WITHOUT PUSHING, then the MAXIMAL FORCE SUPPLIED INCREASES WITH INCREASING ANGLE. THIS MAXIMAL FORCE SUPPLIED IS NOT THE POTENTIAL MAXIMAL SUPPLIED( 70.7 VERSUS 100)

BLEARGH, I'm sorry if I rubbed you wthe wrong way. I actually admire you and always check your posts as I learn something new. I just wrote out my thoughts because it's easier for me that way. I'll start teaching for Kaplan this summer and I like reading these forums to remain fresh. Bennie, don't interject into a post. This was me and Bleargh discussing. I already told BLEARGH I ADMIRED HIM, and was just going back and forth to clear something up.

So, BLEARGH AND I WERE CORRECT IN OUR ORIGINAL ASSUMPTION. BRT pointed out a NUANCE that I hadn't picked up on, and it made me assume what BLEARGH SAID AND I AGREED WITH WAS INCORRECT WHEN IT ACTUALLY PROVES OUR POINT. Great question OP.

 

[JohnnyBravo]

Johnny Bravo-Type response:

This should say "TOO hard", not "to hard". I guess you aren't good at verbal.

You are wrong. Everything is wrong. Wrong.

**insert long useless 5 paragraph essay on proper use of the word "too" and "to", while consistently saying ___ is wrong. Wrong.**

[/BRAVO RESPONSE]

This isn't a graded assignment, people are less scrupulous on internet forums! WHAT?!

Normal person response:
Oh wait, I just realized you know it was supposed to be "too" and probably just made a semantic error. I guess I don't need to bore everyone with a long essay about something everyone already knows! Maybe I don't even post about it. hmmmmm.

Wow, that is so easy.

p.s. I find it flattering that you are following my study schedule, I have posted it maybe once or twice in the past 60 days. Do you also subscribe to my blog? Thx for adding to the readership! Do you mind if I use your term "3 month super schedule" for a later thread? thx again little buddy.

I wrote that at 3 Am or there abouts. I did only get a 9 but I'll take my 14's in science all day with my 3 weeks of cramming.

 

[BennieBlanco]

I wrote that at 3 Am or there abouts. I did only get a 9 but I'll take my 14's in science all day with my 3 weeks of cramming.

Dude.

I'm not criticizing your verbal skills. I'm just pointing out that it is silly to do things like that.

I don't care if you got a 6 or a 15. People aren't their scores, I judge people by how they treat others not by how good they are at things.

 

[RogueUnicorn]

seriously guys. enough flaming. we're all friends here.

 

[wanderer]

What just happened here?

Stop arguing about nothing. The point is to help others on this forum, not confuse them.

 

[BennieBlanco]

What just happened here?

Stop arguing about nothing. The point is to help others on this forum, not confuse them.

seriously guys. enough flaming. we're all friends here.

I'm finished.

Please subscribe to Johny's blog at, www.noseriouslydudeyourewrong.com

 

[BennieBlanco]

edit...

unproductive comment.

 

[JohnnyBravo]

Nice try. I have a great memory. I read the January thread back in October because I have a friend who posts there and is taking then. I love how you said you would take 35 tests and bragged about your 14 hour sales day. Funny, how your rubbed a ton of kids in that thread the wrong way. In fact, it got to the point where you stated you ''would no longer post there and would only post to form a study group to form notes." Well, you're still posting I presume. Diminishing returns buddy.

 

[JohnnyBravo]

holy crap.

No one reads these posts dude.

Get a life. Tokspor had a question and I answered it. Just like TBR books, it goes into greater depth than necessary. If I were in live conversation I could simplify it. However, I feel it much better to give a thorough explanation step by step. Stop commenting.

 

[BennieBlanco]

...

edit, unproductive comment.

 

[JohnnyBravo]

. . .



seriously. 3 posts about the fact that frictional force has a maximum value?

I just wrote a paper that is shorter than these three posts and I used 5 research papers as a source.

It is funny that you keep making attacks on me from posts from months ago, while I'm talking about what you just wrote. I didn't realize you cared so much

You do realize you started this right? I admitted I was wrong in a few post before you came in here. I'm done. If you had taken the time to read my earlier posts which weren't long, you would have seen that BRT pointed out a nuance that was overlooked by myself AND BLEARGH. No, I read a post a few months ago and think it's fitting that you rubbed me the wrong way but realized you have a habit of doing this...... So, I'm not offended in the least. Good day.

 

[BennieBlanco]

...

edit, not productive comment.

 

[JohnnyBravo]

. . .



seriously. 3 posts about the fact that frictional force has a maximum value?

I just wrote a paper that is shorter than these three posts and I used 5 research papers as a source.

It is funny that you keep making attacks on me from posts from months ago, while I'm talking about what you just wrote. I didn't realize you cared so much

It wasn't little. Very few people realize that as the angle of inclination increases, the mu static supplied increases to a MAXIMAL VALUE. BRT pointed this out.

Finally, what's ironic is you stated a false statement which is EXACTLY OPPOSITE TO WHAT BRT STATED. Then you mock me on SEMANTICS. This wasn't on semantics. So, I find it funny that you rebuke me for saying so little in a roundabout fashion.

I won't respond to you anymore. What BRT argued and was correct, that MU SUPPLIED INCREASES AS THE ANGLE INCREASES. THIS OVERCOMES THE INCREASING GRAVITATIONAL FORCE. Now, the angle at which the block starts to slide, is the MAXIMAL FORCE STATIC SUPPLIED WHICH INCREASES AS THE ANGLE INCREASES TO THAT MAXIMAL ANGLE. Now, this maximal force exerted is LESS than the POTENTIAL MAXIMUM FORCE ON LEVEL GROUND. At this point, MU SUPPLIED = MU MAXIMAL even though FORCE STATIC SUPPLIED is LESS THAN MAXIMAL POTENTIAL STATIC FRICTION due to a REDUCED NORMAL FORCE.

Vihsadas and Bluemonkey used to dominate the help forum much like BLEARGH. When a question such as this would arise, they would delve into detail like I did to give a true conceptual understanding. That's what I tried to do. I don't think anyone took my wording the wrong way except for you. You're the one who interjected worthless drivel about gravity, in a belittling manner. It served no purpose to our discussion. This is my final post here. Good luck on the MCAT and my belittling ways will no longer bother anyone.

 

[BennieBlanco]

...

I have decided to withdraw my argument.

I wish good luck to Bravo and I have misjudged this situation.

I have a hot head sometimes. Oops.

apologies.

 

[JohnnyBravo]

...

I have decided to withdraw my argument.

I wish good luck to Bravo and I have misjudged this situation.

I have a hot head sometimes. Oops.

apologies.

Me too. I was a jerk too. From now on, I will write my thoughts on MS word, simplify them, and then post them here. That would have prevent all this.

 

[RogueUnicorn]

um ok. so what's the conclusion here?

 

[JohnnyBravo]

um ok. so what's the conclusion here?

Bleargh, you were CORRECT.Your statement of the following: As the angle of inclination INCREASES, the MAXIMUM POTENTIAL STATIC FRICTION SUPPLIED DECREASES BECAUSE FORCE NORMAL DECREASES. HOWEVER, the MU SUPPLIED INCREASES TO THE MU MAXIMUM AS THE ANGLE INCREASES, this is what keeps the block from slipping.

Short example. A Mu of 1 and a box of 10 kg. The MAX STATIC POTENTIAL FRICTION=100 Newtons before it moves.

On an incline, at 45 degrees before it slips, the MAX STATIC POTENTIAL FALLS TO 70.7 NEWTONS, LOWER THAN THE 100 AS BLEARGH STATED due to LOWER NORMAL FORCE. However, the Mu STATIC SUPPLIED IS EQUAL TO ONE JUST LIKE IN THE MAX STATIC FRICTION 100, BUT THE FORCE IS LESS DUE TO NORMAL FORCE LIKE BLEARGH STATED.

BRT pointed out that Mu supplied increases and this made me think what you said and I though to be true, was wrong. However, they actually complement each other

Finally, Bleargh's statement makes sense: 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.

 

[RogueUnicorn]

ok this is what i was thinking all along. very nice. good to see you two made up

 

[wanderer]

Uh, mu is a constant, it cannot increase for a given interface (or at least that is what is assumed). Fs is what changes when the angle is changed.

 

[RogueUnicorn]

Uh, mu is a constant, it cannot increase for a given interface (or at least that is what is assumed). Fs is what changes when the angle is changed.

technically speaking, mu changes from 0 up to the "constant" that we are given.. if i understand it right. because F(friction) = mu * F(normal), and F(normal) is not changing if you pushed a stationary box, the only thing that can "change" is the coefficient... but that is a very befuddling way of looking at it... ignore me ye studiers of MCAT. it's TMI.

 

[wanderer]

Well I thought that Fs = mu * Fn is only valid for the maximum value of Fs. In other words mu is a constant and Fs always equal to Fopposing until mu*Fn. If an object is not about to slip, you wouldn't really describe it in terms of mu and Fn.

 

[RogueUnicorn]

Well I thought that Fs = mu * Fn is only valid for the maximum value of Fs. In other words mu is a constant and Fs always equal to Fopposing until mu*Fn. If an object is not about to slip, you wouldn't really describe it in terms of mu and Fn.

you're right, you wouldn't. but that's why i put "technically" up there. i could be wrong.

 

[watchntv]

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.

I'm studying for 1/30 mcat and I'm trying to figure out why the answer is D and not B, I read a bunch of the responses, was there a final consensus on why the answer is D?

 

[JohnnyBravo]

I'm studying for 1/30 mcat and I'm trying to figure out why the answer is D and not B, I read a bunch of the responses, was there a final consensus on why the answer is D?

I'm studying for 1/30 mcat and I'm trying to figure out why the answer is D and not B, I read a bunch of the responses, was there a final consensus on why the answer is D?

Now, to the angle situation. Let's say that we a Mu of .75 like in the problem. The force static supplied is 17. We have a 10 Kg block. The reason B is also right, is BECAUSE SINCE FNORMAL DECREASES WITH INCREASING NORMAL, FORCE SUPPLIED DECREASES. This is why as you raise a block it EVENTUALLY FALLS DOWN. As the angle INCREASES, the Force static supplied DECREASES.

We know that this 17=force gravity=mgsine(theta). so .17=sin(theta) sine inverse of (17) which gives an angle of 9.78 degrees. So, as BRT showed, tan(theta)=mu supplied so tan(9.78)=.17 is what is supplied EVEN THOUGH THE MAXIMUM VALUE IS .75. So, Increase MU Max (.75, won't do anything).

What caused the distention above is that Mu SUPPLIED INCREASES. If the ANGLE STAYS THE SAME AND WE INCREASE MAX MU, MU SUPPLIED AND FORCE NORMAL REMAIN THE SAME. That is, it will go from .17 to .75 but the FORCE STATIC SUPPLIED WHEN MU =.75 WILL BE LOWER THAN MU=.17 BECAUSE OF A REDUCED NORMAL FORCE.

a. MASS INCREASES FORCE NORMAL
B. DECREASE FORCE NORMAL
c. Gravity increases and so does force normal
d. Nothing changes

So, you're right. B is ALSO CORRECT. Great Catch. However, note that the BEST ANSWER IS D. Why, the mcat tests basic concepts, and wants you to realize that the FORCE STATIC SUPPLIED ISN'T ALWAYS MAXIMAL. Answer B is a less emphasized point.

 

[watchntv]

Now, to the angle situation. Let's say that we a Mu of .75 like in the problem. The force static supplied is 17. We have a 10 Kg block. The reason B is also right, is BECAUSE SINCE FNORMAL DECREASES WITH INCREASING NORMAL, FORCE SUPPLIED DECREASES. This is why as you raise a block it EVENTUALLY FALLS DOWN. As the angle INCREASES, the Force static supplied DECREASES.

We know that this 17=force gravity=mgsine(theta). so .17=sin(theta) sine inverse of (17) which gives an angle of 9.78 degrees. So, as BRT showed, tan(theta)=mu supplied so tan(9.78)=.17 is what is supplied EVEN THOUGH THE MAXIMUM VALUE IS .75. So, Increase MU Max (.75, won't do anything).

What caused the distention above is that Mu SUPPLIED INCREASES. If the ANGLE STAYS THE SAME AND WE INCREASE MAX MU, MU SUPPLIED AND FORCE NORMAL REMAIN THE SAME. That is, it will go from .17 to .75 but the FORCE STATIC SUPPLIED WHEN MU =.75 WILL BE LOWER THAN MU=.17 BECAUSE OF A REDUCED NORMAL FORCE.

a. MASS INCREASES FORCE NORMAL
B. DECREASE FORCE NORMAL
c. Gravity increases and so does force normal
d. Nothing changes

So, you're right. B is ALSO CORRECT. Great Catch. However, note that the BEST ANSWER IS D. Why, the mcat tests basic concepts, and wants you to realize that the FORCE STATIC SUPPLIED ISN'T ALWAYS MAXIMAL. Answer B is a less emphasized point.

I am lost in what you answered me
here's the basic way I went about this'

problem(restated so we know what's going on)
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?
I am looking for what will cause the (Fn)(Us)=Fs(max) =(Force normal)(coeffienct of static friction)=Max force of static friction

A. Increasing the mass of the block.no this will directly increase static friction to increase
B. Increasing the angle of incline.--raising the angle causes the static friction to decrease, as shown by raise it too high and the block slides this fits what the question wants
C. Increasing the acceleration due to gravity.this will increase static friction as shown when I break down the equations
fsmax = Us * N
N= mg
D. Increasing the coefficient of static friction.if I changed the place the block is on from ice to cardboard, I am increasing the coeffienicent of friction which obviously increases force of static friction , which I dont want to do, so this isnt the answer

you have D as "nothing changes", that wasnt an answer choice,
again, why isnt this D?
MCAT tests basic concepts, so this answer explanation should be really simple.

 

[ezsanche]

I am lost in what you answered me
here's the basic way I went about this'

problem(restated so we know what's going on)
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?
I am looking for what will cause the (Fn)(Us)=Fs(max) =(Force normal)(coeffienct of static friction)=Max force of static friction

A. Increasing the mass of the block.no this will directly increase static friction to increase
B. Increasing the angle of incline.--raising the angle causes the static friction to decrease, as shown by raise it too high and the block slides this fits what the question wants
C. Increasing the acceleration due to gravity.this will increase static friction as shown when I break down the equations
fsmax = Us * N
N= mg
D. Increasing the coefficient of static friction.if I changed the place the block is on from ice to cardboard, I am increasing the coeffienicent of friction which obviously increases force of static friction , which I dont want to do, so this isnt the answer

you have D as "nothing changes", that wasnt an answer choice,
again, why isnt this D?
MCAT tests basic concepts, so this answer explanation should be really simple. [/QUOTE]

It is simple. The equation that you should be using is

F(friction) = mg sin(theta) AND NOTT (Fn)(Us)=Fs(max) (And why is that?)

Now look at each answer choice and see how the force of friction changes in the equation as you increase each of the variables listed in the problem.

 

[watchntv]

It is simple. The equation that you should be using is

F(friction) = mg sin(theta) AND NOTT (Fn)(Us)=Fs(max) (And why is that?)

Now look at each answer choice and see how the force of friction changes in the equation as you increase each of the variables listed in the problem.

yeah, I am not getting your point...I asked a question and you give me a question(s)?
not very helpful

 

[ezsanche]

yeah, I am not getting your point...I asked a question and you give me a question(s)?
not very helpful

Sorry, I was trying to make you think about the problem critically, but I guess I was asking too much from you. This has already been discussed in detail. Read the previous posts for the answer and explanation.

 

[watchntv]

Sorry, I was trying to make you think about the problem critically, but I guess I was asking too much from you. This has already been discussed in detail. Read the previous posts for the answer and explanation.

I accept you apology

 

[Seraph 84]

I am surprised this thread is still alive, I had a few good laughs.

For the block on the incline:

F(FR) - mgsin(phi) = 0 (the block is at rest so it must not be accelerating)

F(FR) = mgsin(phi)

Looks like the mass, gravity, and the angle will all affect the frictional force on the block.

The coefficient is only to be accounted for when analyzing the maximum frictional force that can be supplied from the two interacting surfaces.

 

[puravida85]

I am glad I read this discussion, it has helped me review inclines and friction. I don't know why people are so confused on why the answer is D. First off, by process of elimination, A,B,&C are wrong. So it leaves you to either guess D or think about it for a second. My way of seeing this is that if you increase the static coefficient, what good does it do? Yes, you can plug it into the F = u(s) * Fn and get a larger F but that's probably what Kaplan was hoping you'd think.

Instead think about the nature of the coefficient of static friction. It is a component of the frictional force. The frictional force is needed to oppose whatever force wants it to slide, if you will. Like the words of many here, it is a reactive force. Therefore simply just increasing its value does not increase the frictional force (F). It makes no sense and it is easy to pick out D. If this same nonsense logic were applied to Newton's second law F = m *a, than we'd have a field day of trying to solve problems trying to just plug in some random F to solve for m or a. The physics formulas are elegant in design (as with the MCAT) and weren't designed for purely plugging and chugging (although I've admit to it helping pass certain physics classes).

 

[puravida85]

I am surprised this thread is still alive, I had a few good laughs.

For the block on the incline:

F(FR) - mgsin(phi) = 0 (the block is at rest so it must not be accelerating)

F(FR) = mgsin(phi)

Looks like the mass, gravity, and the angle will all affect the frictional force on the block.

The coefficient is only to be accounted for when analyzing the maximum frictional force that can be supplied from the two interacting surfaces.

Why the mgsin(phi) ? Phi and Theta are two different greek letters. One has a vertical line across the O (phi) and the other has a horizontal line across the O (Theta). In physics, we use theta to denote angles. Sorry, just had to make the correction as worthless as it may seem.

 

[Seraph 84]

Why the mgsin(phi) ? Phi and Theta are two different greek letters. One has a vertical line across the O (phi) and the other has a horizontal line across the O (Theta). In physics, we use theta to denote angles. Sorry, just had to make the correction as worthless as it may seem.

No, in physics, phi is used to represent an angle. Theta, and phi are the ones most commonly used. I usually use phi.

 

[puravida85]

No, in physics, phi is used to represent an angle. Theta, and phi are the ones most commonly used. I usually use phi.

Thanks, I never knew that. Blame it on my textbooks and professors.