Question

A nut can be opened by a wrench of length 50cm with a force of 120N. If a smaller force of 75N is applied, what will the required length of the handle of the wrench?

Answer 1

Answer 80 cm

Explaination

120 × 50 = 75× L

6000 = 75×L

L = 80 cm.

Answer 2

Answer:

80CM

Explanation:

D1=50CM = 0.5 M

F1= 120N

TORQUE REMAINS THE SAME

T = ?

1. T = F1 × D1

= 120 × 0.5

= 60 N m

2. T = F2 × D2

60 = 75 × D2

R2 = 60/75

= 15/ 12 = 4/5 = 0.8

= 0.8 m

=80 cm (°•°1cm= 1/100 m)