Question

Marvin exerts 40.0 N on a chain attached to a crate. The chain remains 20 degrees above the horizontal as the crate slides at constant velocity on a horizontal surface.

a) How much of Marvin's force is horizontal?
b? How much of Marvin's force is vertical?
c) How much friction acts on the crate?
d) By how much does Marvin's pull affect the normal force of the crate?
e) Would the friction force be greater or less if the chain were horizontal?

Answer 1

a) The horizontal component of Marvin's force can be calculated as follows:

Horizontal component = Force x Cosine of the angle

Horizontal component = 40.0 N x Cos(20°)

Horizontal component = 37.03 N (rounded to two decimal places)

Therefore, 37.03 N of Marvin's force is horizontal.

b) The vertical component of Marvin's force can be calculated as follows:

Vertical component = Force x Sine of the angle

Vertical component = 40.0 N x Sin(20°)

Vertical component = 13.62 N (rounded to two decimal places)

Therefore, 13.62 N of Marvin's force is vertical.

c) Since the crate is moving at a constant velocity, we know that the net force on the crate is zero. Therefore, the force of friction acting on the crate must be equal in magnitude and opposite in direction to the horizontal component of Marvin's force.

Friction force = 37.03 N

d) Marvin's pull does not affect the normal force on the crate. The normal force is the force exerted by the surface on the crate perpendicular to the surface, and it is equal in magnitude to the weight of the crate. Marvin's force is exerted parallel to the surface, so it does not affect the normal force.

e) If the chain were horizontal, there would be no vertical component of Marvin's force. Therefore, the force of friction would need to be supplied entirely by the normal force, which would increase. As a result, the friction force would be greater if the chain were horizontal.