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# A crate resting on a rough hor

A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is 0.40. To start the crate moving with the weakest possible applied force, in what direction should the force be applied?

A. Horizontal

B. 24度 below the horizontal

C. 22度 above the horizontal

D. 24度 above the horizontal

E. 66度 below the horizontal

### 1 Answer

- 六呎將軍Lv 71 decade agoFavorite Answer
When the force F makes an angle θ with the horizontal, we have:

Horizontal component of the force = F cos θ

Normal reaction on the crate = mg - F sin θ (m = mass of crate)

So static friction = k(mg - F sin θ) = 0.4mg - 0.4F sin θ

Net force on the crate = F cos θ - (0.4mg - 0.4F sin θ)

= F cos θ + 0.4F sin θ - 0.4 mg

So when F cos θ + 0.4F sin θ is maximized, the acceleration will be the greatest:

F cos θ + 0.4F sin θ = F(cos θ + 0.4 sin θ)

= 1.07F (sin 68 cos θ + cos 68 sin θ)

= 1.07F sin (68 + θ)

So when θ = 22, the net force is max.

Ans = C

Source(s): Myself