Physics Help Forum Friction - very confused about friction

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 Sep 28th 2013, 06:13 PM #1 Junior Member   Join Date: Jul 2013 Posts: 5 Friction - very confused about friction Here are three questions in which I know the answers (they're given after I state the problem) but I don't understand the logic behind them. 1) Consider two blocks that are resting one on top of the other. The lower block has mass m2 = 4.6 kg and is resting on a frictionless table. The upper block has mass m1 = 2.7 kg. Suppose the coefficient of static friction between the two blocks is given by μs = 0.1. Part a) A force of magnitude F is applied as shown in the left figure above (see this image). What is the maximum force for which the upper block can be pushed horizontally so that the two blocks move together without slipping? Answer: (2.7+4.6)*(.1*9.8*2.7/4.6) Now I would have thought that the answer would simply be: Find the force of friction of the second mass on the first, and this must be equal to the force applied to break it loose. Obviously, according to this answer, that is not correct--but why? 2) Two blocks with masses m1 and m2 such that m1<
Sep 28th 2013, 09:40 PM   #2

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 Originally Posted by Addem 1) Consider two blocks that are resting one on top of the other. The lower block has mass m2 = 4.6 kg and is resting on a frictionless table. The upper block has mass m1 = 2.7 kg. Suppose the coefficient of static friction between the two blocks is given by μs = 0.1. Part a) A force of magnitude F is applied as shown in the left figure above (see this image). What is the maximum force for which the upper block can be pushed horizontally so that the two blocks move together without slipping? Answer: (2.7+4.6)*(.1*9.8*2.7/4.6) Now I would have thought that the answer would simply be: Find the force of friction of the second mass on the first, and this must be equal to the force applied to break it loose. Obviously, according to this answer, that is not correct--but why?
One of us made a mistake. I get the same answer, but with the m1 and m2 switched.:
(2.7 + 4.6)(.1*9.8*4.6/2.7)

To answer your question, your friction comment is correct. But in order to find the friction force on m2 you have to find an equation for it for m1 (they are 3rd Law pairs.) And remember that the whole system is accelerating, so you can't put a = 0.

Before I will even think of helping you more with this answer, I'm going to need to see your work. This one's too big to just spot-check.

-Dan
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Sep 29th 2013, 01:14 PM   #3
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 Originally Posted by topsquark One of us made a mistake. I get the same answer, but with the m1 and m2 switched.: (2.7 + 4.6)(.1*9.8*4.6/2.7) To answer your question, your friction comment is correct. But in order to find the friction force on m2 you have to find an equation for it for m1 (they are 3rd Law pairs.) And remember that the whole system is accelerating, so you can't put a = 0. Before I will even think of helping you more with this answer, I'm going to need to see your work. This one's too big to just spot-check. -Dan
Hello Dan, thank you for your help! I really appreciate it.

As for the mistake, I am perfectly happy to claim it for my own. I did submit it to a grader (this problem is associated with an online MIT course) and the grader accepted the answer as correct, but I'm also perfectly willing to accept that the person who programmed the grader made a mistake. I'm the last person qualified to adjudicate.

In any case, about your answer, I hope I can ask another question about this. Wouldn't the force of friction applied by m2 on m1 just be given by the proportion of the normal force? And wouldn't the normal force have magnitude matching the force of gravity? So the normal force has magnitude m1*g and the frictional force therefore be mu*m1*g, or am I missing something?

You asked to see my work--did you mean for the first question, or for the others? I'll assume you meant for the second question, since I think I've shown you my basic logic for the first question in the paragraph above.

In the second question, I reasoned (apparently incorrectly) this way:

I choose the positive direction parallel to the plane to be up-and-right.

The net of the forces on m1 is m1*a, which is also [gravity's force down the plane]+[friction's force down the plane]+[tension's force up the plane] = -m1*g*sin(theta) + mu*m1*g*cos(theta) + T.

The net of the forces on m2 is -m2*a = [g down] + [f up] + [T up] = -(m1+m2)*g*sin(theta) + mu*m1*g*cos(theta) + T.

I think my logic surrounding m1 is correct and clear, but correct me if I'm wrong, so I'll just focus on the forces on m2. In particular, my reasoning behind [g down] seems to be what's wrong. And in more particular, I would have thought that m1 would exert a downward force on m2 and therefore further contribute to how much gravitational force would push m2 down the plane. Apparently not, but why? It must push m2 down somewhat, right? I mean, if m2 were absent, then m1 would fall down a little even under the force of the tension, right? So anyway, exactly why [g down] =/= -(m1+m2)*g*sin(theta) is so, seems to be what's confusing me here. When I change that to just -m2*g*sin(theta), it yields the right answer.

Sep 29th 2013, 07:20 PM   #4

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 Originally Posted by Addem In any case, about your answer, I hope I can ask another question about this. Wouldn't the force of friction applied by m2 on m1 just be given by the proportion of the normal force? And wouldn't the normal force have magnitude matching the force of gravity? So the normal force has magnitude m1*g and the frictional force therefore be mu*m1*g, or am I missing something?
My formula was correct but I switched the numerical values for m1 and m2. So we agree.

As to the friction force on m1 by the normal force on it this is correct. But one thing to be careful about: the normal force is as big as it needs to be to do the job. The normal force is usually mg, but don't count on it. For example, there are not one but two normal forces on m2...The normal on 2 by 1 and the normal on 2 by the floor. So which mass do you use? It's always better to leave N as an unknown and compute it using Newton's 2nd.

I haven't looked at the other two problems yet.

-Dan
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Sep 29th 2013, 08:55 PM   #5

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 Originally Posted by Addem 2) Two blocks with masses m1 and m2 such that m1<
Well, we agree on the formula this time.

Recall that when we apply Newton's laws the acceleration term involves the mass of the object in the FBD. You can't really do this problem using both masses as one since they move in different directions. So the m1 and m2 terms cannot combine to measure the friction between the surfaces. Now if you are referring to the normal force on m2 from the incline, then yes this is the case.

-Dan
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 Dec 22nd 2013, 02:15 AM #6 Junior Member   Join Date: Dec 2013 Posts: 1 Hey! \ sorry for bumping into this thread uninvited. I'm just curious. Is the answer for 3.a) correct? I also don't understand why the friction of the floor to the slab would impact on the block's acceleration. thanks!

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