When a force of 6.0 N is exerted at 30° to a wrench at a distance of

1.	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» *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 :-)*

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»

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 :-)

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