Chapter 8: Angular Kinetics
Problems & Exercises
8.2 Torque
- (a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850m from the hinges. What torque are you exerting relative to the hinges?
(b) Does it matter if you push at the same height as the hinges?
Solution: (a) [latex]\text{46.8 N·m}[/latex]; (b) It does not matter at what height you push. The torque depends on only the magnitude of the force applied and the perpendicular distance of the force’s application from the hinges. (Children don’t have a tougher time opening a door because they push lower than adults, they have a tougher time because they don’t push far enough from the hinges.) - When tightening a bolt, you push perpendicularly on a wrench with a force 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 the center of the bolt)?
(b) Convert this torque to footpounds. - Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible.
Solution: 23.3 N - Use the second condition for equilibrium [latex]\text{(net τ = 0)}[/latex] to calculate [latex]{F}_{\text{p}}[/latex] in the She Saw Torques example from Chapter 8.2, employing any data given or solved for in part (a) of the example.
- Repeat the She Saw Torques example from Chapter 8.2, with the center of mass of the seesaw 0.160 m to the left of the pivot (on the side of the lighter child) and assuming a mass of 12.0 kg for the seesaw. The other data given in the example remain unchanged. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium.
Solution: Given: [latex]\begin{array}{ccc}{m}_{1}& =& \text{26.0 kg,}\phantom{\rule{0.25em}{0ex}}{m}_{2}=\text{32.0 kg,}\phantom{\rule{0.25em}{0ex}}{m}_{\text{s}}=\text{12.0 kg,}\phantom{\rule{0.25em}{0ex}}\\ {r}_{1}& =& \text{1.60 m,}\phantom{\rule{0.25em}{0ex}}{r}_{\text{s}}=\text{0.160 m, find (a)}\phantom{\rule{0.25em}{0ex}}{r}_{2,\phantom{\rule{0.25em}{0ex}}}\text{(b)}\phantom{\rule{0.25em}{0ex}}{F}_{\text{p}}\end{array}[/latex]
a) Since children are balancing:
[latex]\begin{array}{}\text{net}\phantom{\rule{0.25em}{0ex}}{\tau }_{\text{cw}}=–\text{net}\phantom{\rule{0.25em}{0ex}}{\tau }_{\text{ccw}}\\ ⇒{w}_{1}{r}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}={w}_{2}{r}_{2}\\ \end{array}[/latex]
So, solving for [latex]{r}_{2}[/latex] gives:
[latex]\begin{array}{lll}{r}_{2}& =& \frac{{w}_{1}{r}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}}{{w}_{2}}=\frac{{m}_{1}{\mathit{gr}}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}}{{m}_{2}g}=\frac{{m}_{1}{r}_{1}+{m}_{\text{s}}{r}_{\text{s}}}{{m}_{2}}\\ & =& \frac{\left(\text{26.0 kg}\right)\left(\text{1.60 m}\right)+\l