A 120.0 kg crate is placed on a 15.00° frictionless slope (covered with ice). How
much force must be applied along the
(upward) slope to keep the crate from
sliding down?
[?] N

F = 2820.1 N

Explanation:

Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as

Fnet = ma = 0 (a = 0 no sliding)

= F - mgsin15°

= 0

or

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

15.3 cos 65 degrees trig