The bolts on the cylinder head of an engine require tightening to a torque of 95 m N. If a wrench is 28 cm long, what force perpendicular to the wrench must the mechanic exert at its end? If the six-sided bolt head is 15 mm across (Fig. 8–44), estimate the force applied near each of the six points by a wrench.

The magnitude of the torque is \(\tau = 95\;{\rm{m}}\;{\rm{N}}\).

The length of the wrench is\(l = 28\;{\rm{cm}} = 0.28\;{\rm{m}}\).

The diameter of the bolt is \(d = 15\;{\rm{mm}} = 0.015\;{\rm{m}}\).

Torque is defined as the measure of the force that causes an object to rotate about an axis. It is also called the rotational equivalent of linear force.

Mathematically, torque is represented as follows:

\(\begin{align}\vec \tau &= \vec r \times \vec F\\\left| \tau \right| &= rF\sin \theta \end{align}\) … (i)

Here, F is the applied force, and r is the perpendicular distance between the point about which the torque is calculated and the point of application of force.

From equation (i), the magnitude of the torque is given by the following equation:

\(\tau = rF\sin \theta \).

Hence, the force applied is perpendicular\(\left( {\theta = {{90}^{\rm{o}}}} \right)\)to the wrench. The force will be given as follows:

\(\begin{align}\tau &= rF\sin {90^{\rm{o}}}\\F &= \frac{\tau }{r}\\ &= \frac{{95\;{\rm{m}}\;{\rm{N}}}}{{0.28\;{\rm{m}}}}\\ \approx 340\;{\rm{N}}\end{align}\)

Thus, the force applied to the wrench is 340 N.

The net torque is still 95 m N and is produced by six forces, one at each of the six points on the bolt by the wrench. Suppose these forces are perpendicular to the lever arms.

The net torque will be given as follows:

\({\tau _{{\rm{net}}}} = 6\left( {{r_{{\rm{point}}}}{F_{{\rm{point}}}}} \right)\)

Here,\({F_{{\rm{point}}}}\)is the force at each point, and\({r_{{\rm{point}}}}\)is the radius of the bolt.\( \Rightarrow {r_{{\rm{point}}}} = \frac{{0.015\;{\rm{m}}}}{2} = 0.0075\;{\rm{m}}\)

Therefore, the force applied near each point will be given by the following:

\(\begin{align}{F_{{\rm{point}}}} &= \frac{{{\tau _{{\rm{net}}}}}}{{6{r_{{\rm{point}}}}}}\\ &= \frac{{95\;{\rm{m}}\;{\rm{N}}}}{{6\left( {0.0075\;{\rm{m}}} \right)}}\\ \approx 2100\;{\rm{N}}\end{align}\)