Question

When a force of 6.0 N is exerted at 30° to a wrench at a distance of 8 cm from the nut, it is just able to loosen the nut. What force F would be sufficient to loosen it if it acts perpendicularly to the wrench at 16 cm from the nut.?

Answer 1

Given in the question :- Refer to the attachment

Force of 6N is acting on a body with the angle 30°, Now we have to resolve the force of 6 N along the wrench and perpendicular to it.

$P = 6. cos 30^$

Now, Total torque working on point A = 0

Hence The force component which is perpendicular to the wrench

then

$P' =6 sin30^0$

P' =3.0 N

The torque ( nut ) = 3.0 × 8/100

=0.24 N-m ----(i)

The torque of force F (nut)

= F× 16/100 N-m ------(ii)

Equating the two we get,

$F * 16/100 =0.24$

F = 24/16

F =3/2

F=1.5 N

Hope it Helps :-)