All Solutions
Page 135: Section Review
$$
begin{align*}
F_T&=1300 mathrm{N} \
theta_6&=8text{textdegree} \
theta_4&=10text{textdegree} \
m&=? \
\
F_G&=6cdot F_Tcdot cos(theta_6cdot ) + 4cdot F_Tcdot cos(theta_4) \
mcdot g&=6cdot F_Tcdot cos(theta_6cdot ) + 4cdot F_Tcdot cos(theta_4) \
m&=dfrac{ F_Tcdot (6cdot cos( theta_6cdot ) + 4cdot cos(theta_4))}{g} \
&=dfrac{1300mathrm{N}cdot (6cdot cos( 8text{textdegree} ) + 4cdot cos(10text{textdegree}))}{9.8text{textdegree}}\
&=1311 mathrm{kg}
end{align*}
$$
m=1311 mathrm{kg}
$$
$$
begin{align*}
m&=63 mathrm{kg} \
theta&=14text{textdegree} \
F_p&=512 mathrm{N} \
mu&=0.27 \
a&= ?
\
F&=F_p-F_f-F_gcdot sin(theta) \
mcdot a&=F_p-mucdot mcdot g-mcdot gcdot sin(theta) \
a&=dfrac{F_p-mucdot mcdot g-mcdot gcdot sin(theta)}{m} \
&=dfrac{512 mathrm{N} – 0.27cdot 63 mathrm{kg}cdot cos(14text{textdegree}) – 63 mathrm{kg}cdot 9.8 mathrm{m/s^2}cdot sin(14text{textdegree})}{63 mathrm{kg}} \
&=3.189 mathrm{m/s^2}
end{align*}
$$
From the above expression we see that the left side is the maximum for the minimum angle $theta$. So the smaller the angle $theta$, the larger the horizontal component of the tension force $cos(theta)cdot F_T$.
A better way of hanging is shown in Figure b.
