All Solutions
Page 104: Practice Problems
begin{align*}
& F_{hand on ball}, F_{Earth on ball} tag{Forces on ball} \
& F_{Ball on hand}, F_{Ball on Earth} tag{Forces exerted by ball}
end{align*}
$$
begin{align*}
& F_g tag{Force of graviti on body by the Earth} \
& F_{s-c} tag{Force exerted by suitcase on cart} \
& F_{c-s} tag{Force exerted by cart on suitcase}\
& F_n tag{force exerted by floor on cart}
end{align*}
$$
Force pair is $F_{s-c} , F_{c-s}$ for which stands $F_{s-c}=-F_{c-s}$.
Known:
$$
begin{align*}
m&=42 mathrm{kg} \
F_{t max}&=450 mathrm{N} \
g&=9.8 mathrm{m/s^2}
end{align*}
$$
Unknown:
$$
begin{align*}
a_{max}&=?
end{align*}
$$
$$
begin{align*}
F_{t max}&=F_g+F \
F_{t max}&=mg+ma_{max} \
a_{max}&=dfrac{F_{t max}-mg}{m}
end{align*}
$$
begin{align*}
a_{max}&=dfrac{F_{t max}-mg}{m} \
&=dfrac{450 mathrm{N}-42 mathrm{kg}cdot 9.8 mathrm{m/s^2}}{42 mathrm{kg}} \
&=boxed{0.9143 mathrm{m/s^2}}
end{align*}
$$
The maximum acceleration by which a worker can lift a bucket without breaking the rope is $a_ {max} = 0.9143 mathrm {m/s^2}$.
a_ {max} = 0.9143 mathrm {m/s^2}
$$
Known:
$$
begin{align*}
F_{Diego}&=31 mathrm{N} \
F_{Mika}&=23 mathrm{N}
end{align*}
$$
Unknown:
$$
begin{align*}
F_{tire-wheel}&=?
end{align*}
$$
$$
begin{align*}
F_{net}&=F_{Mika}+F_{Diego} \
&=23 mathrm{N}+31 mathrm{N} \
&=54 mathrm{N} \
F_{tire-wheel}&=F_{net} \
&=boxed{54 mathrm{N}}
end{align*}
$$
F=54 mathrm{N}
$$
