Question

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?

Short answer

The plane inclined to${18}^0$

Step-by-step solution

Given

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

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

Determining the concept

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.

Determining the free body diagram

Free Body Diagram of the sled:

Step 4:Determining the angle

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),

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

Then dividing equation (i) by equation (iii),

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

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