All Solutions
Page 128: Practice Problems
and acceleration with $mu_k$ $times$ 52
mu_k = dfrac{36}{52}
$$
Known:
$$
begin{align*}
F_g&=52 mathrm{N} \
F_p&=36 mathrm{N} tag{force exerted on sled by girl} \
end{align*}
$$
Unknown:
$$
begin{align*}
{mu } &=? \
\
F_f&={mu}cdot F_N tag{$F_N$ normal force} \
mu&=dfrac{F_f}{F_N} \
\
F_f&=F_p \
F_N&=F_g \
\
mu&=dfrac{F_p}{F_g} \
&=dfrac{36 mathrm{N} } {52 mathrm{N}} \
&=boxed{0.692}
end{align*}
$$
Notice that $mu$ is ratio and that it has no unit of measurement.
mu=0.692
$$
105 x 9.8 = 1029 N
so to convert it we multiply the mass with 9.8
a.k.a left = right
up = down
102 = 1029 times mu_s
$$
mu_s = dfrac{102}{1029}
$$
Known:
$$
begin{align*}
F_g&=134 mathrm{N} \
mu&=0.55
end{align*}
$$
Unknown:
$$
begin{align*}
F_p&=?
end{align*}
$$
$$
begin{align*}
F_p&=F_f \
F_g&=F_N
end{align*}
$$
$$
begin{align*}
F_p&=mucdot F_g \
&=0.55cdot 134 mathrm{N} \
&=boxed{73.98 mathrm{N}}
end{align*}
$$
F_p=73.94 mathrm{N}
$$
$$
begin{align*}
mu_k &= 0.12\
W_{person} &= 650 ,mathrm{N}\
W_{sled} &= 52 , mathrm{N}
end{align*}
$$
intertext{Given the formula}
F_f &= mu_kF_N
intertext{such that}
F_N &= W_{person} + W_{sled}
end{align}
where $F_f$ is the force needed to needed to move the sled, $F_N$ is the normal force for the system, and $mu_k$ is the coefficient of kinetic friction acting against the system.
$$
begin{align*}
F_f &= mu_kF_N\
&= mu_k(W_{person} + W_{sled})\
&= (0.12)(650.,mathrm{N} + 52.0,mathrm{N})\
&= boxed{84.0 ,mathrm{N}}
end{align*}
$$
F_f = 84.0 ,mathrm{N}
$$
$$
begin{align*}
F_f&=5.8 mathrm{N} tag{Force when oil is not applied.} \
mu_{d}&=0.58 \
mu_{wo}&=0.06\
F’_f&=?tag{Force after oil application.}\
\
F_f&=mucdot F_N \
F_N&=dfrac{F_f}{mu_{d}} \
&=dfrac{5.8 mathrm{N}}{0.58} \
&=10 mathrm{N} \
\
F’_f&=mu_{wo}cdot F_N \
&=0.06cdot 10 mathrm{N} \
&=0.6 mathrm{N}
end{align*}
$$
F’_f=0.6 mathrm {N}
$$
