Question

A body resting on a horizontal plane required a pull of 200N inclined at 40° to the plane to initiate the motion . it was also found that a push . determine weight of the body and coefficient of friction

Answer 1

Answer:

0.1714

Step-by-step explanation:

Weight of the body in newtons = W (Given)

Normal reaction = Rn (Given)

∅ = 30° (Given)

Coefficient of friction = u (Given)

Force of friction = F (Given)

Resolving the forces horizontally,

F = 220 cos 30º = 220 × 0.866 = 190.5 N

F = 180 cos 30º = 180 × 0.866 = 156 N

Now,

Resolving the forces vertically,

RN = W – 180 sin 30º = W – 180 × 0.5 =(W– 90) N.

where F = μ. Rn or 156 = μ (W – 90)

= RN =W+220 sin 30º = W + 220 × 0.5 = (W + 110) N

where F = μ. RN or 190.5 = μ (W + 110)

W = 1000 N, and μ = 0.1714

Thus, the weight of the body and the coefficient of friction is 0.1714

Answer 2

Answer:

0.1714

Step-by-step explanation:

Weight of the body in newtons = W (Given)

Normal reaction = Rn (Given)

∅ = 30° (Given)

Coefficient of friction = u (Given)

Force of friction = F (Given)

Resolving the forces horizontally,

F = 220 cos 30º = 220 × 0.866 = 190.5 N

F = 180 cos 30º = 180 × 0.866 = 156 N

Now,

Resolving the forces vertically,

RN = W – 180 sin 30º = W – 180 × 0.5 =(W– 90) N.

where  F = μ. Rn or 156 = μ (W – 90)

= RN =W+220 sin 30º = W + 220 × 0.5 = (W + 110) N

where F = μ. RN or 190.5 = μ (W + 110)

W = 1000 N, and μ = 0.1714

Thus, the weight of the body and the coefficient of friction is 0.1714