Consider the figure below. (Let w1 = 150 N and w2 = 38.0 N.)

Question

(a) What is the minimum force of friction required to hold the system of the figure in equilibrium?
N

(b) What coefficient of static friction between the 150-N block and the table ensures equilibrium?


(c) If the coefficient of kinetic friction between the 150-N block and the table is 0.127, what hanging weight should replace the 38.0-N weight to allow the system to move at a constant speed once it is set in motion?
N

Answer 1

To solve these problems, we first need to analyze the forces acting on both blocks. Let's call the 150-N block A and the 38.0-N weight B.

(a) The minimum force of friction required to hold the system in equilibrium is the force that balances the gravitational force acting on block B. So,

Friction force = w2 = 38.0 N

(b) The coefficient of static friction between the 150-N block and the table ensures equilibrium can be found by the formula:

Friction force = static friction coefficient * Normal force

Since the system is in equilibrium, the normal force on block A is equal to its weight, w1.

Static friction coefficient = Friction force / Normal force
Static friction coefficient = 38.0 N / 150 N = 0.2533

(c) To find the hanging weight that allows the system to move at a constant speed with a given coefficient of kinetic friction, we first find the kinetic friction force acting on the 150-N block:

Kinetic friction force = kinetic friction coefficient * Normal force
Kinetic friction force = 0.127 * 150 N = 19.05 N

Now, we need to find the weight that provides this amount of force as tension in the string. Since the system is moving at a constant speed, the tension in the string equals the kinetic friction force:

Tension = 19.05 N

This tension is also equal to the gravitational force acting on the hanging weight:

Hanging weight = Tension = 19.05 N