A toy chest and its contents have a combined weight of $\mathbf{180}\mathbf{}\mathbf{N}$A toy chest and its contents have a combined weight of $\mathbf{0}\mathbf{.}\mathbf{42}$.The child in Fig. 6-35 attempts to move the chest across the floor by pulling on an attached rope. (a) If $\mathit{\theta }$is $\mathbf{42}\mathbf{°}$whatis the magnitude of the force $\overrightarrow{\mathit{F}}$that the child must exert on the rope to put the chest on the verge ofmoving? (b) Write an expression for the magnituderequired to put the chest on the verge of moving as a function of the angle $\mathit{\theta }$. Determine (c) the value of $\mathit{\theta }$for which $\mathit{F}$is a minimum and (d) that minimum magnitude.

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

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

Angle,${}^{\theta =42°}$

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:

$F_{net}=\sum _{}^{}ma$

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,

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

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

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

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 $\theta$, then, from above calculations,

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

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,

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

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

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

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