Question

When tightening a bolt, you push perpendicularly on a wrench with aforce of 165 N at a distance of 0.140 m from the center of the bolt. (a)How much torque are you exerting in newton × meters (relative to thecenter of the bolt)? (b) Convert this torque to foot-pounds.

Short answer

(a)The torque is$23.1\mathrm{N}\cdot \mathrm{m}$.

(b)The torque is$17.4\mathrm{f}\mathrm{o}\mathrm{o}\mathrm{t}\cdot \mathrm{p}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{l}$.

Step-by-step solution

Step 1:Given data and torque calculation

• The force is$F=165\mathrm{N}$.
• The perpendicular distance from the hinge is$r=0.140\mathrm{m}$.

The torque is,

$\tau =r\times F$

Here$r$is the perpendicular distance and substituting the values we get,

The torque

$\begin{array}{c}\tau =0.140\times 165\\ =23.1\mathrm{N}\cdot \mathrm{m}\end{array}$

Step 2:Calculation of torque in the foot-poundal

(b)

The torque is$23.1\mathrm{N}\cdot \mathrm{m}$.

We know,

$1\mathrm{m}=3.28\mathrm{f}\mathrm{e}\mathrm{e}\mathrm{t}$

$1\mathrm{N}=0.225\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}$

So,

$\begin{array}{c}\tau =23.1\mathrm{N}\cdot \mathrm{m}\\ =23.1\times 0.225\times 3.28\\ =17.04\mathrm{f}\mathrm{o}\mathrm{o}\mathrm{t}\cdot \mathrm{p}\mathrm{o}\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{a}\mathrm{l}\end{array}$