a) $\text{46.8 N\xb7m}$

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.)

Given:

$\begin{array}{ccc}{m}_{1}& =& \text{26.0 kg,}\phantom{\rule{0.25em}{0ex}}{m}_{2}=\text{32.0 kg,}\phantom{\rule{0.25em}{0ex}}{m}_{\text{s}}=\text{12.0 kg,}\phantom{\rule{0.25em}{0ex}}\\ {r}_{1}& =& \text{1.60 m,}\phantom{\rule{0.25em}{0ex}}{r}_{\text{s}}=\text{0.160 m, find (a)}\phantom{\rule{0.25em}{0ex}}{r}_{2\mathrm{,}\phantom{\rule{0.25em}{0ex}}}\text{(b)}\phantom{\rule{0.25em}{0ex}}{F}_{\text{p}}\end{array}$

a) Since children are balancing:

$\begin{array}{}\text{net}\phantom{\rule{0.25em}{0ex}}{\tau}_{\text{cw}}=\u2013\text{net}\phantom{\rule{0.25em}{0ex}}{\tau}_{\text{ccw}}\\ \Rightarrow {w}_{1}{r}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}={w}_{2}{r}_{2}\\ \end{array}$

So, solving for ${r}_{2}$ gives:

$\begin{array}{lll}{r}_{2}& =& \frac{{w}_{1}{r}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}}{{w}_{2}}=\frac{{m}_{1}{\mathit{gr}}_{1}+{m}_{\text{s}}{\mathit{gr}}_{\text{s}}}{{m}_{2}g}=\frac{{m}_{1}{r}_{1}+{m}_{\text{s}}{r}_{\text{s}}}{{m}_{2}}\\ & =& \frac{(\text{26.0 kg})(\text{1.60 m})+(\text{12.0 kg})(\text{0.160 m})}{\text{32.0 kg}}\\ & =& \text{1.36 m}\end{array}$

b) Since the children are not moving:

$\begin{array}{}\text{net}\phantom{\rule{0.25em}{0ex}}F=0={F}_{\text{p}}\u2013{w}_{1}\u2013{w}_{2}\u2013{w}_{\text{s}}\\ \Rightarrow {F}_{\text{p}}={w}_{1}+{w}_{2}+{w}_{\mathrm{s}}\end{array}$

So that

$\begin{array}{lll}{F}_{\text{p}}& =& (\text{26.0 kg}+\text{32.0 kg}+\text{12.0 kg})(\text{9.80}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2})\\ & =& \text{686 N}\end{array}$

a) $\text{2.55}{\mathrm{\times 10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N, 16.3\xba to the left of vertical (i.e., toward the wall)}$

b) 0.292

a) 0.167, or about one-sixth of the weight is supported by the opposite shore.

b) $F=2\text{.}0\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N}$, straight up.

$\begin{array}{lll}{F}_{\text{B}}& =& \text{470 N;}\phantom{\rule{0.25em}{0ex}}{r}_{1}=\text{4.00 cm;}\phantom{\rule{0.25em}{0ex}}{w}_{\text{a}}=\text{2.50 kg;}\phantom{\rule{0.25em}{0ex}}{r}_{2}=\text{16.0 cm;}{w}_{\text{b}}=\text{4.00 kg;}\phantom{\rule{0.25em}{0ex}}{r}_{3}=\text{38.0 cm}\\ {F}_{\text{E}}& =& {w}_{\text{a}}\left(\frac{{r}_{2}}{{r}_{1}}-1\right)+{w}_{\text{b}}\left(\frac{{r}_{3}}{{r}_{1}}-1\right)\\ & =& \left(\text{2.50 kg}\right)\left(9.80\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}\right)\left(\frac{\text{16.0 cm}}{\text{4.0 cm}}\u20131\right)\\ & & +\left(\text{4.00 kg}\right)\left(9.80\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}\right)\left(\frac{\text{38.0 cm}}{\text{4.00 cm}}\u20131\right)\\ & =& \text{407 N}\end{array}$

$\begin{array}{c}1.1\times {\text{10}}^{3\phantom{\rule{0.25em}{0ex}}}\text{N}\\ \theta =\text{190}\text{\xba}\phantom{\rule{0.25em}{0ex}}\text{ccw from positive}\phantom{\rule{0.25em}{0ex}}x\phantom{\rule{0.25em}{0ex}}\text{axis}\end{array}$

${F}_{\text{V}}=\text{97}\phantom{\rule{0.25em}{0ex}}\text{N,}\phantom{\rule{0.25em}{0ex}}\theta =\text{59\xba}$

(a) ${F}_{\text{A}}=2\text{.}\text{21}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N}$ upward

(b) ${F}_{\text{B}}=\text{2.94}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N}$ downward

(a) ${F}_{\text{teeth on bullet}}=\text{1.2}\times {\text{10}}^{\text{2}}\phantom{\rule{0.25em}{0ex}}\text{N}$ upward

(b) ${F}_{\text{J}}=\text{84 N}$ downward

a) ${\stackrel{-}{x}}_{2}=\text{2.33 m}$

b) The seesaw is 3.0 m long, and hence, there is only 1.50 m of board on the other side of the pivot. The second child is off the board.

c) The position of the first child must be shortened, i.e. brought closer to the pivot.

Both objects are in equilibrium. However, they will respond differently if a force is applied to their sides. If the cone placed on its base is displaced to the side, its center of gravity will remain over its base and it will return to its original position. When the traffic cone placed on its tip is displaced to the side, its center of gravity will drift from its base, causing a torque that will accelerate it to the ground.

**F**_{L}= 7350 N,**F**_{R}= 2450 N- As the car moves to the right side of the bridge,
**F**_{L}will decrease and**F**_{R}will increase. (At exactly halfway across the bridge,**F**_{L}and**F**_{R}will both be 4900 N.)

The student should mention that the guiding principle behind simple machines is the second condition of equilibrium. Though the torque leaving a machine must be equivalent to torque entering a machine, the same requirement does not exist for forces. As a result, by decreasing the lever arm to the existing force, the size of the existing force will be increased. The mechanical advantage will be equivalent to the ratio of the forces exiting and entering the machine.

- The force placed on your bicep muscle will be greater than the force placed on the dumbbell. The bicep muscle is closer to your elbow than the downward force placed on your hand from the dumbbell. Because the elbow is the pivot point of the system, this results in a decreased lever arm for the bicep. As a result, the force on the bicep must be greater than that placed on the dumbbell. (How much greater? The ratio between the bicep and dumbbell forces is equal to the inverted ratio of their distances from the elbow. If the dumbbell is ten times further from the elbow than the bicep, the force on the bicep will be 200 pounds!)
- The force placed on your bicep muscle will decrease. As the forearm lifts the dumbbell, it will get closer to the elbow. As a result, the torque placed on the arm from the weight will decrease and the countering torque created by the bicep muscle will do so as well.