When a force of 6.0 N is exerted at 30o 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?
The force acting on the end of the wrench is 1.5 N
Given
The force exerted into the wrench is given as, the angle of motion is
. The distance from the nut to the wrench end is
.
Formula Used
The formula used is that of a torque which tells us the mechanics of force which helps the object to rotate. The formula is
where
F is the force applied on the object; r is the radius of the object turning. Turning angle is given by
Explanation
The torque acting on the point A is given as
Now the torque acting on the point B is
Therefore, the force acting on the end of the wrench is 1.5 N