Question

A nut is opened by a wrench of length 10 cm. If the least force required is 5.0 N, find the moment of force needed to turn the nut.

Answer 1

Step 1: Given that:

Length($l$) of wrench = $10cm$ = $0.1m$

Least force($F$) = $5N$

Step 2: Calculation of the moment of force:

The moment of force(τ) is given as:

$\text{τ}=F\times l$

Where; F is the force applied and l is the perpendicular distance of the point of rotation from the point of action of force.

Putting the values, we get;

$\text{τ}=5N\times 0.1m$

$\text{τ}=0.5N-m$

Thus,

The moment of force is $0.5N-m$