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 losen 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

A

Explanation

A force 6 N acting at angle of `30^@` is just able to loosen the wrench at a distance 8 cm from it.
therefore total torque acting at A about the point O.

=6sin30^@xx(8/100)`
threfore total torque required at b about the point O.
`=Fxx16/100`
`rarr Fxx16/100=6sin30^@xx8/100
`rarr F=(8xx3)/16=1.5N`