Mass, friction and equilibrium in a system

[furor celtica]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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.

 

[da_nang]

Assuming the string is massless and frictionless, you can think of the two extreme cases.
Case 1: mass m is sliding down. To prevent it from sliding down, the friction together with the tension must push/pull it up.
Case 2: mass m is sliding up. To prevent it from sliding up, the friction and gravitational force must push it down.

 

[furor celtica]

ok so i get
in the first case: 10M > 10m(sinA)µ
and in the second case: 10M x µ < 10m(sinA)
i know these are incorrect but still, show me where and how.

 

[da_nang]

You're missing the other component for the gravity force. Namely, G_x is the component of the gravitational force parallell to the track acting on mass m.

I suggest you also use a free body diagram and determine what components to use and what their directions are. For example, in the first case, the forces you want to use comes from Newton's second law of motion: T + F = G_x (extreme case), where T is the tension, F friction. You want to stop it from sliding down, thus T + F ≥ G_x, since G_x is what's pulling m down.