Question

Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont know how to taxle this problem especially because of the fact that the problem is talking about pushing and not tension.

it writes
"A 1,480-N crate is being pushed across a level floor at a constant speed by a force F of 370 N at an angle of 20.0 degrees below the horizontal, as shown in the figure a below. The floor has some amount of friction. "

The questions ask me to make a "free-body diagram"
- write an equation each for the x & y components of the resultant force

-determine the normal force and lastly what exactly is the coefficient of kinetic friction between the crate and the floor

Answer 1

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn ⇒ µ = f/n ≈ 0.216