Question

A block of wood weighing 18 pounds rests on an inclined plane that makes an angle of 60 with the horizontal. How much of the friction force is needed to prevent the block from sliding on the plane?

Round answer to nearest whole number. Do not use a space or decimal in your answer or it will be marked wrong

Answer 1

Force of friction will be 69 N

Explanation:

Given a block of wood weighing 18 pounds rests on an inclined plane that makes an angle of 60 with the horizontal. we have to find the force of friction needed to prevent the block from sliding on the plane.

As the plane makes an inclination, vertical force will act on the wood to slide. Therefore, the friction force must be equal to vertical force so that the wood not to slide.

`F = mgsin\theta` where F is friction force, m is weight of an object and
`\theta` is inclination angle made from the horizontal.

As 1 pound=0.453592 kg

∴ 18 lbs = 8.165 kg

`F = (m)(g)(sin\theta)`

= 8.165 x 9.8 x sin 60°

= 69.29 N ≈ 69 N

Answer 2

The easiest way to figure out the friction force in the problem is to get the force acting on the vertical. As the plane makes an inclination, vertical force will act on the wood to slide. Therefore, the friction force must be equal to the vertical force in order for the wood not to slide. We will use F = Wx sin ∅.

F = friction force
W = weight of an object
∅ = inclination angle made from the horizontal

18 lbs = 8.165 kg
F = (m)(a)(sin∅)
F = 8.165 x 9.8 x sin 60
F = 69.29 N ≈ 69 N (answer)