- #1

- 6

- 0

## Homework Statement

The worker pushes downward and to the left on the box at an angle of 35°. The box has a mass of 135 kg. The box moves horizontally across the floor to the left at a constant velocity. The force applied by the man is 2030 N∠215°. What is the value of the friction force?

## Homework Equations

Fgravity · sin35° = Fgravityx

Ffriction + Fgravityx = m · a

f = μ · Fs

## The Attempt at a Solution

sin35° =

Fgravity · sin35° = Fgravityx

m · g · sin35° = Fgravityx

(135 kg)(9.81) sin35° = Fgravityx

759.6 N = Fgravityx

Ffriction + (759.6 N) = (135 kg)(0 m/s2)

Ffriction - 759.6 N = 0 N

Ffriction = 759.6 N

I know that this isn't correct, because my study guide is multiple choice. My options are:

a. 1164 N

b. 1324 N

c. 1663 N

d. 2489 N

I'm really not sure what I'm missing or if I am going the wrong way with finding the value of the friction force. My guess is that c. 1663 N is correct, but that is only a guess. I'd like to be able to prove which answer it is. Thank you for any help!