91Ó°ÊÓ

Weight, ${}^{W=180\mathrm{N}}$

Coefficient of static friction between the toy chest and floor, ${}^{{\mathrm{μ}}_s=0.42}$

Angle,${}^{\mathrm{θ}=42°}$

To find the mass of the block of C and the acceleration of block A, draw the free body diagram for each block and then apply Newton's 2nd law of equation.According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\underset{}{\overset{}{∑}}ma$

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

Free Body Diagram of the toy chest:

Toy chest is moving with constant velocity that means it has 0 acceleration.By using Newton’s 2nd law of motion along the vertical direction (along y) to the system,(positive x axis along the incline and the positive y along the vertical direction),

$$\begin{gathered} N+F\sin \left(42\right)-W=0 \\ N=W-F\sin \left(42\right) \end{gathered}$$

Thus, the frictional force is,

$$\begin{gathered} f_s={\mathrm{μ}}_{s}N \\ ={\mathrm{μ}}_s\left(W-F\sin \left(42\right)\right) \end{gathered}$$

Now, applying Newton’s 2nd along horizontal direction to the Block(A+C) having mass m,

localid="1654577455841" $$\begin{gathered} F\cos \left(42\right)-f_s=0 \\ F\cos \left(42\right)-{\mathrm{μ}}_s\left(W-T\sin \left(42\right)\right)=0 \\ F\left(\cos \left(42\right)\right)+{\mathrm{μ}}_s\sin \left(42\right)-{\mathrm{μ}}_{s}W=0 \\ F=\frac{{\mathrm{μ}}_{sW}}{(\cos \left(42\right)+{\mathrm{μ}}_s\sin \left(42\right)} \\ =\frac{\left(0.42\right)\left(180\right)}{\cos \left(42\right)+\left(0.42\right)\sin \left(42\right)} \\ =74\mathrm{N} \end{gathered}$$

Hence, the magnitude of the force exerted by child is ${}^{74\mathrm{N}}$

In this case, consider the angle made by the rope to toy chest as $\mathrm{θ}$, then, from above calculations,

$$\begin{gathered} F=\frac{{\mathrm{μ}}_{s}W}{\cos \mathrm{θ}+{\mathrm{μ}}_s\sin \mathrm{θ}} \\ =\frac{\left(0.42\right)\left(180\right)}{\cos \mathrm{θ}+\left(0.42\right)\sin \mathrm{θ}} \end{gathered}$$

Hence, the expression for the magnitude of F or T applied by the child have been found.

Minimize the above expression for F by working through the condition,

$$\begin{gathered} \frac{dF}{d\mathrm{θ}}=\frac{d}{d\mathrm{θ}}\left(\frac{{\mathrm{μ}}_{s}W}{\cos \mathrm{θ}+{\mathrm{μ}}_s\sin \mathrm{θ}}\right) \\ =\frac{{\mathrm{μ}}_{s}W\left(\sin \mathrm{θ}-{\mathrm{μ}}_s\cos \mathrm{θ}\right)}{{\left(\cos \mathrm{θ}+{\mathrm{μ}}_s\sin \mathrm{θ}\right)}^2} \\ =0 \\ \sin \mathrm{θ}-{\mathrm{μ}}_s\cos \mathrm{θ}=0 \\ \sin \mathrm{θ}={\mathrm{μ}}_s\cos \mathrm{θ} \\ \frac{\sin \mathrm{θ}}{\cos \mathrm{θ}}={\mathrm{μ}}_s \\ \tan \mathrm{θ}={\mathrm{μ}}_s \\ \mathrm{θ}={\tan }^{-1}\left(0.42\right) \\ =23° \end{gathered}$$

Hence, the${}^{\mathrm{θ}}$at which F or T is minimum is ${}^{23°}$

$$\begin{gathered} Put\mathrm{θ}=23°, \\ F=\frac{{\mathrm{μ}}_{s}W}{\cos \mathrm{θ}+{\mathrm{μ}}_s\sin \mathrm{θ}} \\ =\frac{\left(0.42\right)\left(180\right)}{\cos \left(23\right)+\left(0.42\right)\sin \left(23\right)} \\ =70\mathrm{N} \end{gathered}$$

Hence, the minimum magnitude of F is ${}^{70\mathrm{N}}$