Chapter 9

a) $\text{46.8 N·m}$

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.)

23.3 N

Given:

a) Since children are balancing:

So, solving for $r_2$ gives:

b) Since the children are not moving:

So that

$F_{\text{wall}}=1.43\times {\text{10}}^3\text{N}$

a) $\text{2.55}{\mathrm{\times 10}}^3\text{N, 16.3º to the left of vertical (i.e., toward the wall)}$

b) 0.292

$F_{\text{B}}=2.12\times {\text{10}}^4\text{N}$

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

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

a) 21.6 N

b) 21.6 N

350 N directly upwards

25

50 N

a) $\text{MA}=\text{18}\text{.}5$

b) $F_{\text{i}}=\text{29.1 N}$

c) 510 N downward

$1\text{.}\text{3}\times {\text{10}}^3\text{N}$

a) $T=\text{299 N}$

b) 897 N upward

$\begin{array}{lll}{F}_{\text{B}}& =& \text{470 N;}{r}_1=\text{4.00 cm;}{w}_{\text{a}}=\text{2.50 kg;}{r}_2=\text{16.0 cm;}{w}_{\text{b}}=\text{4.00 kg;}{r}_3=\text{38.0 cm}\\ F_{\text{E}}& =& w_{\text{a}}\left(\frac{r_2}{r_1}-1\right)+w_{\text{b}}\left(\frac{r_3}{r_1}-1\right)\\ & =& \left(\text{2.50 kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(\frac{\text{16.0 cm}}{\text{4.0 cm}}–1\right)\\ & & +\left(\text{4.00 kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(\frac{\text{38.0 cm}}{\text{4.00 cm}}–1\right)\\ & =& \text{407 N}\end{array}$

$\begin{array}{c}1.1\times {\text{10}}^3\text{N}\\ \theta =\text{190}\text{º}\text{ccw from positive}x\text{axis}\end{array}$

$F_{\text{V}}=\text{97}\text{N,}\theta =\text{59º}$

(a) 25 N downward

(b) 75 N upward

(a) $F_{\text{A}}=2\text{.}\text{21}\times {\text{10}}^3\text{N}$ upward

(b) $F_{\text{B}}=\text{2.94}\times {\text{10}}^3\text{N}$ downward

(a) $F_{\text{teeth on bullet}}=\text{1.2}\times {\text{10}}^{\text{2}}\text{N}$ upward

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

(a) 147 N downward

(b) 1680 N, 3.4 times her weight

(c) 118 J

(d) 49.0 W

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.