91Ó°ÊÓ

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

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

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:

$F_{net}=∑ma$

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,

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

Along horizontal direction, $F-mgsin\left(\mathrm{θ}\right)-f_s=0$

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

At

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

At

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

then,

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

role="math" localid="1654581642603" $5.0=(0.50)mg\cos \left(\mathrm{θ}\right)+mg\sin \left(\mathrm{θ}\right)\left(ii\right)$

Using equation (i) in equation (ii),

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

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

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