A 120.0 kg crate is placed on a 15.00°

Physics

1 answer:

Citrus2011 [14]3 years ago

7 0

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

F = mgsin15°

= (120 kg)(9.8 m/s^2)sin15°

= 2820.1 N

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