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Old Sep 28th 2013, 05:13 PM   #1
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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<<m2 are connected by a massless inextensible string and a massless pulley as shown in the figure below. The pulley is rigidly connected to the top of a wedge with angle θ . The coefficient of friction between the blocks is μ . The surface between the lower block and the wedge is frictionless. See this image.

What are the magnitudes of the acceleration of the two blocks? Express your answer in terms of g, μ, m1, m2, and θ.

Answer: g*( (m_2 - m_1)*sin(theta)- 2*mu*m_1*cos(theta) )/(m_2+m_1)

Now I understand everything about this except the forces of gravity involved on the second mass. I would have thought that the total forces acting on this mass would be the force of gravity down the plane, the force of friction up the plane, and the force of tension up the plane. And I would think that the force of gravity down the plane would be (m_1+m_2)g*sin(theta). Why do we use only m_2*g*sin(theta)? Doesn't m1 exert a downward force on m2? Doesn't that increase the aggregate force on m2 going down the plane?

3) A block of mass mB = 19 kg is on top of a long slab of mass mS = 7 kg, and the slab is on top of a horizontal table as shown. A horizontal force of magnitude F = 564 N is applied on the block. As a result the block moves relative to the slab and the slab moves relative to the table. There is friction between all surfaces. The coefficient of kinetic friction between the block and the slab is μ1 = 0.8, and the coefficient of kinetic friction between the slab and the table is μ2 = 0.1. take g to be 9.81 m/s2, and enter your answer to 3 significant figures. See this image.

(a) What is the magnitude of the block's acceleration?

Answer: (564-.8*19*9.81-.1*(7)*9.81)/19

Again, I would have thought that the force on the block would have been F - Friction, i.e. 564-.8*19*9.81. I don't really understand how the friction that the other object experiences contributes to the force experienced by the block.

(b) What is the magnitude of the slab's acceleration?

Answer: (.8*19*9.81-0.1*(7+19)*9.81)/7

Now this answer makes perfect sense to me, but it seems to be in contradiction with the method used in a previous answer! Here, we actually account for the force that the block exerts downward on the slab! In the second problem, this was neglected.

Specific explanations about these problems, or general explanations about how friction works between two moving bodies, would be appreciated, thank you.
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Old Sep 28th 2013, 08:40 PM   #2
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Originally Posted by Addem View Post
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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Old Sep 29th 2013, 12:14 PM   #3
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Originally Posted by topsquark View Post
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.
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Old Sep 29th 2013, 06:20 PM   #4
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Originally Posted by Addem View Post
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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Old Sep 29th 2013, 07:55 PM   #5
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Originally Posted by Addem View Post
2) Two blocks with masses m1 and m2 such that m1<<m2 are connected by a massless inextensible string and a massless pulley as shown in the figure below. The pulley is rigidly connected to the top of a wedge with angle θ . The coefficient of friction between the blocks is μ . The surface between the lower block and the wedge is frictionless. See this image.

What are the magnitudes of the acceleration of the two blocks? Express your answer in terms of g, μ, m1, m2, and θ.

Answer: g*( (m_2 - m_1)*sin(theta)- 2*mu*m_1*cos(theta) )/(m_2+m_1)

Now I understand everything about this except the forces of gravity involved on the second mass. I would have thought that the total forces acting on this mass would be the force of gravity down the plane, the force of friction up the plane, and the force of tension up the plane. And I would think that the force of gravity down the plane would be (m_1+m_2)g*sin(theta). Why do we use only m_2*g*sin(theta)? Doesn't m1 exert a downward force on m2? Doesn't that increase the aggregate force on m2 going down the plane?
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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Old Dec 22nd 2013, 01:15 AM   #6
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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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