91影视

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

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

Angle,${}^{胃=42掳}$

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:

$F_{net}=\underset{}{\overset{}{鈭�}}ma$

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鈥檚 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,

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

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

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

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 $胃$, then, from above calculations,

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

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,

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

Hence, the${}^{胃}$at which F or T is minimum is ${}^{23掳}$

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

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