Mass M2 for constant speed and static equilibrium on inclined plane

[roger_c]

A block with mass M1 is placed on an inclined plane with slope angle alpha and is connected to a second hanging block that has mass M2 by a cord passing over a small, frictionless pulley. The coefficient of static friction is (mu_s) and the coefficient of kinetic friction is (mu_k).

A)Find the mass M2 for which block M1 moves up the plane at constant speed once it has been set in motion.

B)Find the mass M2 for which block M1 moves down the plane at constant speed once it has been set in motion.

C)Find the smallest value of M2 when the blocks will remain at rest if they are released from rest.

Thank you guys~

 

[Nenad]

tell us what the problem is? Draw out a free body diagram of the situation and look at the forces. I can't help you if you don't tell me what it is you don't understand about the question.

Regards,

Nenad

 

[thepatientmental]

A) To find the mass M2 for which block M1 moves up the plane at constant speed, we need to consider the forces acting on block M1. The force of gravity, mg, is acting down the incline and the normal force, N, is acting perpendicular to the incline. The force of friction, F, is acting in the opposite direction of motion. Since the block is moving at constant speed, the net force must be zero. This means that the force of gravity must be balanced by the force of friction and the normal force.

We can write this as:

mg*sin(alpha) - F = 0

F = mg*sin(alpha)

We also know that the force of friction is given by:

F = mu_s*N

Substituting for F, we get:

mg*sin(alpha) = mu_s*N

Next, we need to consider the forces acting on block M2. The only force acting on it is the tension in the cord, T. This tension is equal to the force of gravity on block M2, which is given by mg. Therefore, we can write:

T = mg

Now, we can set up an equation for the net force on block M1, using the forces we have already calculated:

mg*sin(alpha) - mu_s*N = 0

Solving for N, we get:

N = mg*sin(alpha)/mu_s

Since we know that N is equal to the tension in the cord, we can substitute this into our equation for T:

T = mg*sin(alpha)/mu_s

Finally, we can equate the two expressions for T and solve for M2:

mg = mg*sin(alpha)/mu_s

M2 = M1*sin(alpha)/mu_s

B) To find the mass M2 for which block M1 moves down the plane at constant speed, we can follow a similar process. The only difference is that the direction of the friction force changes, so we have:

mg*sin(alpha) + F = 0

F = -mg*sin(alpha)

Using the same logic as before, we can write:

T = mg*cos(alpha)/mu_k

And equating the two expressions for T, we get:

M2 = M1*cos(alpha)/mu_k

C) To find the smallest value of M2 when the blocks will remain at rest if they are released from rest, we need to consider the maximum possible value of the friction force. This occurs when