In Fig. 6-23, a sled is held on an inclined plane by a cord pulling directly up the plane. The sled is to be on the verge of moving up the plane. In Fig. 6-28, the magnitude $\mathit{F}$required of the cord’s force on the sled is plotted versus a range of values for the coefficient of static friction${\mathbf{\mu }}_{\mathbf{s}}$ between sled and plane: ${\mathbf{F}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$, ${\mathbf{F}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$, and ${\mathbf{\mu }}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}$. At what angle $\mathbf{\theta }$is the plane inclined?

Force on the sled and coefficients of static friction for respective forces are,

$$\begin{gathered} {\mathrm{F}}_1=2.0\mathrm{N} \\ {\mathrm{\mu }}_{\mathrm{s}1}=0 \\ {\mathrm{F}}_2=5.0\mathrm{N} \\ {\mathrm{\mu }}_{\mathrm{s}2}=0.50 \end{gathered}$$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. First, draw the free body diagram of the sled. From that, by using Newton's 2nd law, find the angle of the inclined plane.

Formula:

$F_{net}=\sum ma$

where, F is the net force, m is mass and a is an acceleration.

Free Body Diagram of the sled:

By using Newton’s 2nd law of motion along the vertical direction,

$$\begin{gathered} N+mg\cos \left(\theta \right)=0 \\ N=mg\cos \left(\theta \right) \end{gathered}$$

Along horizontal direction, $F-mgsin\left(\theta \right)-f_s=0$

$$\begin{gathered} F=f_s+mg\sin \left(\theta \right) \\ F={\mu }_{s}N+mg\sin \left(\theta \right) \end{gathered}$$

At

$$\begin{gathered} F=F_1 \\ =2.0N \\ {\mu }_s={\mu }_{s1}=0 \\ {F}_1=mg\sin \left(\theta \right) \\ 2.0=mg\sin \left(\theta \right)\left(i\right) \end{gathered}$$

At

$$\begin{gathered} F=F_2 \\ =5.0N \\ {\mu }_s={\mu }_s=0.50 \end{gathered}$$

then,

role="math" localid="1654581635242" $F_2={\mu }_{s2}N+mg\sin \left(\theta \right)$

role="math" localid="1654581642603" $5.0=(0.50)mg\cos \left(\theta \right)+mg\sin \left(\theta \right)\left(ii\right)$

Using equation (i) in equation (ii),

$$\begin{gathered} 5.0=(0.50)mg\cos \left(\theta \right)+2.0 \\ 3.0=(0.50)mg\cos \left(\theta \right)\left(iii\right) \end{gathered}$$

$$\begin{gathered} \frac{2.0}{3.0}=\frac{mg\sin \left(\theta \right)}{(0.50)mg\cos \left(\theta \right)} \\ \left(\frac{2.0}{3.0}\right)\times \left(0.50\right)=\tan \left(\theta \right) \\ 0.333=\tan \left(\theta \right) \\ ={18}^0 \end{gathered}$$

Therefore, the plane inclined to ${18}^0$