Answer:
(i) F = 25 N
(ii) F = 35.36 N
(iii) F = 43.3 N
Step-by-step 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, [tex]F_p[/tex] = 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°
[tex]F_p[/tex] = M×g×sin(θ) = 50 × sin(30°) = 25 N
The force [tex]F_p[/tex] = F = 25 N
F = 25 N
(ii) For the plane with angle of elevation = 45°
[tex]F_p[/tex] = M×g×sin(θ) = 50 × sin(45°) = 50 × (√2)/2 = 25·√2 N
[tex]F_p[/tex] = F = 25×√2 N = 35.36 N
F = 35.36 N
(iii) For the plane with angle of elevation = 60°
[tex]F_p[/tex] = M×g×sin(θ) = 50 × sin(60°) = 50 × (√3)/2 = 25·√3 N
[tex]F_p[/tex] = F = 25×√3 N = 43.3 N
F = 43.3 N.
