#### Sample problem

A 2.0 kg block is placed on a 30 degree inclined plane. The block is connected to a mass m by a string that passes over a pulley at the top of the incline. Mass m hangs vertically from the string. The coefficients of friction for the block and the plane are: ms = 0.4 and mk = 0.3.

If m is 0.2 kg and the system is released from rest, what is the acceleration?

The place to start, as usual, is with the free-body diagrams of the block and the mass. This is a little problematic because we don't know whether the system is accelerating, so it's not immediately clear how to handle the force of friction.

Will the system accelerate? If so, which way will it accelerate?

You can take an educated guess, work things out, and see if your final result is consistent with your guess. Another option is to do some calculations to figure out what's going on - we'll try that.

Align the coordinate system with the incline.
+x = down the ramp and +y = perpendicular to the ramp.

The force of gravity on the block, with mass M, needs to be split into components.
Mg sin(θ) = 9.8 N down the slope.
Mg cos(θ) = 17.0 N into the slope.

Apply Newton's second law in the y-direction to determine the normal force.

ΣFy = M ay = 0

N - Mg cos(θ) = 0

N = Mg cos(θ) = 16.97 N

Use this to determine the maximum value of the force of static friction.

fs max = μs N = 0.4 x 16.97 = 6.8 N.

Can the combined effect of the force of static friction (6.8 N) and the weight of mass m (1.96 N) balance the component of the force of gravity trying to pull the block down the slope (9.8 N)?

No. So, the block must accelerate down the slope. The free-body diagram should show the force of kinetic friction acting up the slope.

Apply Newton's second law to the block in the x-direction:

ΣFx = M ax

Mg sin(θ) - fk - T = M ax

Apply Newton's second law to the mass m to get an expression for the tension. Note that if the positive direction for the block is down the slope, we must take positive up for the mass.

ΣF = ma

T - mg = ma

T = ma + mg

Substitute this into the expression we got above. The block and mass have the same acceleration, so just call that a everywhere:

Mg sin(θ) - μk N - ma - mg = Ma

Mg sin(θ) - μk Mg cos(θ) - mg = Ma + ma
a =
 Mg sin(θ) - μk Mg cos(θ) - mg m + M

Think about whether this makes sense. If so, plug in the values to find that:
a =
 9.8 - (0.3 x 16.97) - 1.96 2.2
= 1.25 m/s2

Bonus question: determine the values of A, B, and C that make the following statements true in this problem:

• If m < A, the block accelerates down the slope.
• If A < m < B, the block is stationary and the force of friction acts up the slope.
• If m = B, the block is stationary and the force of friction is zero.
• If B < m < C, the block is stationary and the force of friction acts down the slope.
• If m > C the block accelerates up the slope.

Answers: A = 0.31 kg, B = 1.0 kg, and C = 1.7 kg.