91影视

1. Coefficient of static friction: ${渭}_s=0.70$
2. (d) Coefficient of static friction:${渭}_s=0.60$

The problem deals with the Newton鈥檚 second law of motion which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Also it involves static friction.

Formula:

Frictional force is given by,

${\mathbf{f}}_{\mathbf{k}}\mathbf{=}{\mathbf{渭}}_{\mathbf{k}}{\mathbf{F}}_{\mathbf{N}}$

(a)

The x component of F tries to move the crate while its y component indirectly contributes to the inhibiting effects of friction (by increasing the normal force).

Newton鈥檚 second law implies,

X direction: $F\cos 胃-f_s=0$ (i)

Y direction: $F_N-F\sin 胃-mg=0$ (ii)

To be 鈥渙n the verge of sliding鈥� means frictional force

$\begin{array}{l}{f}_s=f_{s,max}\\ ={渭}_s{F}_N\end{array}$

Solving above equations (i) and (ii) for F (actually, for the ratio of F to mg)

$$\begin{gathered} F\cos 胃={渭}_s{F}_N \\ =F\cos 胃={渭}_s(F\sin 胃+mg) \\ =F\cos 胃={渭}_{s}F\sin 胃+{渭}_{s}mg \\ =F\cos 胃-{渭}_{s}F\sin 胃={渭}_{s}mg \\ =F\left(\cos 胃={渭}_s\sin 胃\right)={渭}_{s}mg \end{gathered}$$

Rearranging to get the ratio of F l mg ,

$鈬�\frac{F}{mg}=\frac{{渭}_s}{\left(\cos 胃-{渭}_3\sin 胃\right)}$

This is plotted below ($胃$in degrees and ${渭}_s=0.70$given)

(b)

The ratio approaches infinite value when the denominator vanishes:

$$\begin{gathered} \cos 胃-{渭}_s\sin 胃=0 \\ 鈬�\cos 胃={渭}_s\sin 胃 \\ 鈬�\left(1/\tan 胃\right)={渭}_s \\ 鈬�\tan 胃=\left(1/{渭}_s\right) \\ 鈬�{胃}_{inf}={\tan }^{-1}\left(1/{渭}_s\right) \\ ={\tan }^{-1}\left(1/0.70\right) \\ ={55}^0 \end{gathered}$$

(c)

Lubricating the floor (reducing the friction) means reducing the coefficient ${渭}_s$, increases the angle ${胃}_{inf}$by the above condition in part (b).

(d)

$\begin{array}{l}{胃}_{inf}={\tan }^{-1}\left(1/{渭}_s\right)\\ 鈬�{胃}_{inf}={\tan }^{-1}\left(\frac{1}{0.60}\right)\\ 鈬�{胃}_{inf}={59}^o\end{array}$