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…

Analyze the question: A screw jack has a handle (effort arm) of radius r = 35 cm and a screw pitch of 0.5 cm. We need the effort F that lifts a load of 2000 N when the efficiency is 30%.

Concept:

• Velocity ratio of a screw jack: VR = 2πr / pitch
• Mechanical advantage: MA = Load / Effort
• Efficiency = MA / VR × 100%

Solution:

1. Velocity ratio: VR = 2πr / p = (2 × 22/7 × 35) / 0.5 = 220 / 0.5 = 440

2. Use the efficiency: Efficiency = (MA / VR) × 100 30 = (MA / 440) × 100 MA = 0.30 × 440 = 132

3. Find the effort: MA = Load / Effort Effort = 2000 / 132 ≈ 15.15 N ≈ 15.2 N

Checking the options: 132 is the mechanical advantage, not the force, so B is a trap. The effort is about 15.2 N, which matches option C.

Answer: C (15.2 N)

Tip: For screw jack questions, always compute VR = 2πr/pitch first, then MA = efficiency × VR, then Effort = Load/MA. Don't stop at MA, because a number like 132 appears as a distractor option.