Physics Help Forum mass, friction and equilibrium in a system

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 May 10th 2011, 10:34 PM #1 Member   Join Date: Mar 2010 Posts: 40 mass, friction and equilibrium in a system A particle of mass m is placed on a rough track which goes up at an angle A to the horizontal, where sinA=0.6 and cosA=0.8. The coefficient of friction is 0.5. A string is attached to the particle, and a particle of mass M is attached to the other end of the string. The string runs up the track, passes over a smooth bar at the top of the track, and then hangs vertically. Find the interval of values of M for which the system can rest in equilibrium. I thought this would be easy but I think I started with the wrong idea and it messed me up. So the particle with mass M has a vertical downward force of 10M. Thats going to be constant. However the force exerted by the first particle is going to depend on the direction of motion, right? I.e. against which force the friction will work. So first of all I took 10M>(10m)(sinA)(µ) which makes M>0.3m which I already know is wrong. And then for the other possibility I'm not even sure which equations to use, and I'm getting all mixed up with the tension in the string...any help appreciated.
May 11th 2011, 05:27 AM   #2

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 Originally Posted by furor celtica A particle of mass m is placed on a rough track which goes up at an angle A to the horizontal, where sinA=0.6 and cosA=0.8. The coefficient of friction is 0.5. A string is attached to the particle, and a particle of mass M is attached to the other end of the string. The string runs up the track, passes over a smooth bar at the top of the track, and then hangs vertically. Find the interval of values of M for which the system can rest in equilibrium. I thought this would be easy but I think I started with the wrong idea and it messed me up. So the particle with mass M has a vertical downward force of 10M. Thats going to be constant. However the force exerted by the first particle is going to depend on the direction of motion, right? I.e. against which force the friction will work. So first of all I took 10M>(10m)(sinA)(µ) which makes M>0.3m which I already know is wrong. And then for the other possibility I'm not even sure which equations to use, and I'm getting all mixed up with the tension in the string...any help appreciated.
We can work this out in more detail if you like, but the key is that the static friction force opposes the direction of the "intended" motion. That is to say it will oppose the direction the system wants to go. So we need to decide which way the block on the incline is going to want to move. So calculate the acceleration (assuming a given m and M) that the system will have without friction being involved. That will tell you which way the system is going to try to accelerate in.

For example, if M/m < 0.8 and in the absence of friction the block on the incline will slide down the incline. (Prove this.) So the static friction force must face up the slope. This gives a static friction force of f = (0.8m - M)g. The static friction force will range from 0 N to (mu)N. This condition will give you a range of values for M. Then do the problem again for M/m > 0.8.

-Dan
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 May 11th 2011, 07:33 AM #3 Member   Join Date: Mar 2010 Posts: 40 ok so i get for the first particle not to slip downwards: 10M > 10m(sinA)µ and for it not to be pulled upwards: 10M x µ < 10m(sinA) i know these are incorrect but still, show me where and how.
 May 11th 2011, 08:55 AM #4 Physics Team     Join Date: Jun 2010 Location: Mauritius Posts: 609 Well, I would do it like that. 1. Assume that m is larger and instead is pulling M up, and the force of friction is just opposing the downfall of m. 2. Assume that m is smaller and M pulls m up the ramp, with friction just preventing m to move up. First case: On M, we have: T - Mg = 0 On m, mg sinA = µ mg cosA + T You can work on the highest value of m. (Let that be x) Case 2: T - Mg = 0 mg sinA + µ mg cosA = T You get the lower boundary. (Let that be y) Then, the interval is: y <= m <= x __________________ Jerry (Got my results!) It is easier to protect your feet with slippers than to cover the earth with carpet. No one can go back and change a bad beginning; but anyone can start now and create a successful ending. If a problem can be solved, no need to worry about it. If it cannot be solved what is the use of worrying?

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