For the given torque, the minimum force exerted by the mechanic at the end of the wrench is found to be [tex]113.33\,N[/tex].
The answer is explained in the following way.
- We know that the torque is a twisting force that causes rotation. Torque is the cross product of distance vector and force vector.
- [tex]\vec{\tau} = \vec{r} \, \times \, \vec{F}[/tex]
i.e.; [tex]\tau=rFsin\theta[/tex]
- Here, we are asked to find the minimum force exerted by the mechanic at the end of the wrench.
- Here, the length of the wrench is constant. Therefore in order for the force to be minimum; for a fixed torque, [tex]sin\,\theta[/tex] must be maximum.
- So, [tex]\theta = 90 ^o \,\,\,\, \implies sin\,\theta = 1[/tex]
- Here, [tex]r = 30\,cm =0.30\,m[/tex] and
- Torque, [tex]\tau=34\,N.m[/tex]
Substituting all these values in the equation for torque, we get;
- [tex]34\,N.m=0.30\,m \times F[/tex]
- [tex]\implies F=\frac{34\,N.m}{0.30\,m} =113.33\,N[/tex]
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