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% [ฯ = \(\frac{22}{7}\)]?
- 152.0N
- 132.0N
-
15.2N
- 440.0N
The correct answer is: C
Explanation
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 = \(\frac{L}{M.A}\) = \(\frac{2000}{132}\) โ 15.2N