A man lifted a 281.5 kg load off the ground using his teeth. Suppose he can hold just three times that mass on a 30.0° slope using the same force. What is the coefficient of static friction between the load and the slope?

Answer 1

The weight of the load on the slope can be calculated by the formula:

Weight on the slope = mass * gravity * sin(angle)

where:
mass = 281.5 kg
gravity = 9.8 m/s^2
angle = 30.0°

Weight on the slope = 281.5 kg * 9.8 m/s^2 * sin(30.0°)

Weight on the slope = 1379.99 N

In order to calculate the force of static friction, we use the formula:

Force of static friction = coefficient of static friction * Normal force

where:
Normal force = weight on the slope * cos(angle)

Normal force = 1379.99 N * cos(30°)

Normal force = 1196.912 N

Since the man is able to hold three times the mass on the slope using the same force, we can calculate the maximum force the man can exert:

Maximum force = 3 * mass * gravity

Maximum force = 3 * 281.5 kg * 9.8 m/s^2

Maximum force = 8224.86 N

Since the maximum force the man can exert is equal to the force of static friction, we can set the equations equal to each other:

Force of static friction = Maximum force

coefficient of static friction * Normal force = Maximum force

coefficient of static friction = Maximum force / Normal force

coefficient of static friction = 8224.86 N / 1196.912 N

coefficient of static friction ≈ <<8224.86/1196.912=6.862>>6.862

Therefore, the coefficient of static friction between the load and the slope is approximately 6.862.