Question

A screw jack of pitch 5 mm is used to lift the tyre of a vehicle of load 200 kg with the help of a handle of length 0.5 m. Neglecting the friction force between screw and nut of the jack, the least force required to raise the load is

• A. 1.2 N
• B. 2.2N
• C. 3.2N
• D. 4.2N

Answer 1

Answer: (C)

3.2N

Answer 2

Work done in one rotation = work done against gravity

i.e., F × 2π × $\frac { 1 } { 2 }$ = 200 × 10 × 5 × 10^-3

(F × 2πl = (mg) × pitch)

F = $\frac { 10 } { 3.14 }$ = 3.18 N