The handle of a screw jack is 35 cm long and the pitch of the screw is 0.5 cm. What force must be applied at the end of the handle to lift a load of 2000N, if the efficiency of the jack is 30% [\pi = \(\frac{22}{7}\)]?
A
152.0N
B
132.0N
C
15.2N
D
440.0N
Verified Correct Answer: C.
Physical Principles & Numerical Solution
GIVEN:
Handle length r = 35cm = 0.35cm, pitch of screw = 0.5cm = 0.005cm, load W = 2000N, efficiency = 30%
V.R of screw = \(\frac{2\pi r}{p}\) = \(\frac{2 \times \frac{22}{7} \times 0.35}{0.005}\) = 440
Eff = \(\frac{\text{M.A}}{\text{V.R}}\) x 100
M.A = eff. x V. R = 0.3 x 440 = 132
Effort = \(\frac{L}{M.A}\) = \(\frac{2000}{132}\) ≈ 15.2N
Source: Myschool Past Questions Archive (Record #61901)