Question

Please help me out
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Answer 1

(i) F = 25 N

(ii) F = 35.36 N

(iii) F = 43.3 N

Explanation:

We note that the component of force acting along the incline surface is given by the relation

Force, acting parallel to the incline plane,
`F_p` = M×g×sin(θ)

Given that the weight of the body is given as 50 N, we have;

M×g = 50 N

Therefore we have;

(i) For the plane with angle of elevation = 30°

`F_p` = M×g×sin(θ) = 50 × sin(30°) = 25 N

The force
`F_p` = F = 25 N

F = 25 N

(ii) For the plane with angle of elevation = 45°

`F_p` = M×g×sin(θ) = 50 × sin(45°) = 50 × (√2)/2 = 25·√2 N

`F_p` = F = 25×√2 N = 35.36 N

F = 35.36 N

(iii) For the plane with angle of elevation = 60°

`F_p` = M×g×sin(θ) = 50 × sin(60°) = 50 × (√3)/2 = 25·√3 N

`F_p` = F = 25×√3 N = 43.3 N

F = 43.3 N.