Question

A 10-kg block is suspended in air by a single string that passes through a pulley and is attached on the other side to a 35-kg block that is on the verge of sliding on the ground. What is the coefficient of static friction between the larger block and the ground? (Note: The magnitude of the vertical force the small block exerts on the large block is 87 N.)

Answer 1

Answer:

0.286

Explanation:

Let g = 10m/s2, and assume the pulley is frictionless, the weight by the 10kg block would exert a force on the 35kg block on the ground

Weight of the 10 kg-block = mg = 10 * 10 = 100 N

This force is balanced by static friction force from the ground due to the normal force of the 35kg block. So the friction force would also be 100N

The weight and normal force of the 35kg block is N = Mg = 35*10 = 350 N

The coefficient of the friction force if friction force divided by normal force:

$\mu = \frac{F_f}{N} = \frac{100}{350} = 0.286$