Mass on incline will it slip or stick?

[charan1]

Homework Statement

Figure shows a block of mass m resting on a 20* slope. The block has coefficients of friction uS=0.80 and uK=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.

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

Part B:
If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

Homework Equations

Fnet=ma
us=F/N

The Attempt at a Solution

I really have no idea how to do Part B, but I tried part A and was real confident with my answer but it turned out wrong. This is what I did.

(2kg)(9.8)=19.6N for the hanging mass

assuming that the Hanging mass = to the tension force on the block with mass m I then did this...

19.6N=m(9.8)cos20*(.8)
m=2.66kg and this is wrong

Were did I go wrong?

Thanks

 

[tiny-tim]

Welcome to PF!

Figure shows a block of mass m resting on a 20* slope. The block has coefficients of friction uS=0.80 and uK=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.

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

Part B:
If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?
…
(2kg)(9.8)=19.6N for the hanging mass
…
19.6N=m(9.8)cos20*(.8)
m=2.66kg and this is wrong

Hi charan1! Welcome to PF!

(Useful tip: don't bother to multiply everything by 9.8 … all the terms have 9.8, so just call it g, and cancel it at the end. )

(oh … and have a mu: µ )

I take it you were applying Newton's second law in the direction of the slope.

That means that you must include all the forces … you left out the force due to gravity.

(For part B, use µ_K instead of µ_S)

 

[baba89]

for sharing your attempt at a solution and asking for help with where you went wrong. It shows that you are willing to learn and improve.

To answer Part A, we need to consider the forces acting on the block on the incline. These are the weight of the block (mg), the normal force (N), and the force of friction (Ff). Since the block is on the verge of slipping, the frictional force will be at its maximum value, which is given by uS*N. Therefore, we can set up an equation for the equilibrium of forces:

mg = N + uS*N

Solving for N, we get:

N = mg/(1+uS)

Since the tension force in the string is equal to the weight of the hanging mass, we can also set up an equation for the equilibrium of forces for the hanging mass:

mg = T

Substituting the expression for N from the first equation into the second equation, we get:

mg = mg/(1+uS) + uK*mg

Solving for m, we get:

m = 1/(1+uS+uK) = 1/(1+0.8+0.5) = 1/2.3 = 0.435 kg

Therefore, the minimum mass that will stick and not slip is 0.435 kg.

For Part B, we can use the equation Fnet = ma, where Fnet is the net force acting on the block, m is the mass of the block, and a is the acceleration of the block. The net force is given by:

Fnet = T - mg*sin20* - uK*N

Substituting the expression for N from the first equation in Part A, we get:

Fnet = T - mg*sin20* - uK*mg/(1+uS)

Substituting the value of m from Part A, we get:

Fnet = 19.6 N - (0.435 kg)(9.8 m/s^2)*sin20* - 0.5*(0.435 kg)(9.8 m/s^2)/(1+0.8)

Solving for Fnet, we get:

Fnet = 9.65 N

Substituting this value into the equation Fnet = ma, we get:

9.65 N = (0.435 kg)*a

Solving for a, we