Torque — Question 4

Question 4

A limited handle angle sets a minimum applied force

A wrench extends $0.25\,\mathrm m$ from a bolt. The force can make an angle $\phi$ between $30^\circ$ and $60^\circ$ with the handle, in the sense that gives positive torque. Required torque is $12\,\mathrm{N\,m}$.

1. Find the smallest possible force and its angle.

2. Would a $50\,\mathrm N$ force be sufficient anywhere in the allowed range?

Question 4 - Solution

A limited handle angle sets a minimum applied force

(a) Derivation and calculation

For a specified positive moment, $$F(\phi)=\frac{\tau}{r\sin\phi}=\frac{12}{0.25\sin\phi}.$$ On $30^\circ\le\phi\le60^\circ$, sine increases, so $$\boxed{\phi_{\mathrm{opt}}=60^\circ},\qquad \boxed{F_{\min}=\frac{12}{0.25(\sqrt3/2)}=32\sqrt3=55.43\ \mathrm N}.$$

(b) Result and interpretation

The greatest moment available from 50 N is $$\boxed{\tau_{\max}=0.25(50)\sin60^\circ=10.83\ \mathrm{N\,m}}<12\ \mathrm{N\,m}.$$ No permitted angle is sufficient. A perpendicular pull is outside the allowed interval.

Physical check

The constrained force exceeds the unconstrained minimum $12/0.25=48$ N.