What is the minimum mass m that will stick and not slip?

[mugzieee]

a block of mass m is resting on a 20degree slope. The block has coefficients of friction mu_s =0.80 and mu_k =0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg. (sorry didnt know of a way to distribute a picture)
what is the minimum mass m that will stick and not slip?

i drew 1 free body diagrams of the mass m and mass 2.
For the FBD of m1:
I have normal force, kinetic friction,static friction, the weight components of the mass, and the force of the second mass acting on the first one.

For the FBD of m2:
I have just the force acting on ms from m1, and the weight of m2.

im not quite sure if the block is accelerating kuz it says resting, but then again I am given coefficients of frictional forces. also can i equate the 2 tension forces of the ropes from the third law?
im pretty stumped, I am not sure how to proceed after drawing the FBD's.

 

[asrodan]

The block is not accelerating, the coefficient of kinetic friction is not required and is likely only given to see if you understand that it is not needed. The coefficient of static friction will be required to calculate the minimum mass.

Because there will be no acceleration for the mass you are calculating, all the force component equations from your FBD's will equal zero, and the tensions in each diagram are equal.

Remember that the maxium frictional force (while the block is still at rest) is mu_s*N.

 

[thepatientmental]

To determine the minimum mass m that will stick and not slip, we need to consider the forces acting on the block of mass m. The block is resting on a 20 degree slope and is connected to a hanging block of mass 2.0 kg via a massless string over a frictionless pulley. The two blocks are also in contact with a surface that has coefficients of friction mu_s = 0.80 and mu_k = 0.50.

First, we need to determine if the block is in a state of equilibrium or if it is accelerating. Since the block is resting on the slope, we can assume that it is in a state of equilibrium. This means that the sum of all the forces acting on the block must be equal to zero.

From the free body diagram of the block, we can see that there are four forces acting on it: normal force, kinetic friction, static friction, and the force of the second mass acting on the first one. The normal force is perpendicular to the surface and is equal to the weight component of the block in the y-direction. The kinetic friction is acting in the opposite direction of the block's motion, which is down the slope. The static friction is acting in the opposite direction of the impending motion, which is up the slope. The force of the second mass acting on the first one is equal to the tension in the string, which is also pulling the block up the slope.

We can set up the equations of equilibrium to determine the minimum mass m that will stick and not slip. Since the block is in equilibrium, the sum of the forces in the x-direction and the sum of the forces in the y-direction must be equal to zero.

In the x-direction, we have:
T - mu_k*N = 0

In the y-direction, we have:
N - mg*cos(20) = 0

Using the given coefficients of friction and the weight of the block, we can solve for the normal force and tension in the string:
N = mg*cos(20)
T = mu_k*N = 0.50*mg*cos(20)

Now, we can substitute these values into the equation for the sum of the forces in the x-direction to solve for the minimum mass m:
T - mu_k*N = 0
0.50*mg*cos(20) - (0.50*mg*cos(20))*cos(20) = 0
m = (0