In Fig. 6-58, force is applied to a crate of mass mon a floor where the coefficient of static friction between crate and floor is ${\mathit{\mu }}_{\mathbf{s}}$. Angle u is initially $\mathbf{0}\mathbf{°}$but is gradually increased so that the force vector rotates clockwise in the figure. During the rotation, the magnitude Fof the force is continuously adjusted so that the crate is always on the verge of sliding. For ${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{70}$, (a) plot the ratio F/mgversus $\mathit{\theta }$and (b) determine the angle ${\mathit{\theta }}_{\mathbf{i}\mathbf{n}\mathbf{f}}$at which the ratio approaches an infinite value. (c) Does lubricating the floor increase or decrease ${\mathit{\theta }}_{\mathbf{i}\mathbf{n}\mathbf{f}}$, or is the value unchanged? (d) What is ${\mathit{\theta }}_{\mathbf{inf}}$for ${\mathit{\mu }}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{60}$?

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

The problem deals with the Newton’s 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,

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

(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’s second law implies,

X direction: $F\cos \theta -f_s=0$ (i)

Y direction: $F_N-F\sin \theta -mg=0$ (ii)

To be “on the verge of sliding” means frictional force

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

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

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

Rearranging to get the ratio of F l mg ,

$\Rightarrow \frac{F}{mg}=\frac{{\mu }_s}{\left(\cos \theta -{\mu }_3\sin \theta \right)}$

This is plotted below ($\theta$in degrees and ${\mu }_s=0.70$given)

(b)

The ratio approaches infinite value when the denominator vanishes:

$$\begin{gathered} \cos \theta -{\mu }_s\sin \theta =0 \\ \Rightarrow \cos \theta ={\mu }_s\sin \theta \\ \Rightarrow \left(1/\tan \theta \right)={\mu }_s \\ \Rightarrow \tan \theta =\left(1/{\mu }_s\right) \\ \Rightarrow {\theta }_{inf}={\tan }^{-1}\left(1/{\mu }_s\right) \\ ={\tan }^{-1}\left(1/0.70\right) \\ ={55}^0 \end{gathered}$$

(c)

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

(d)

$\begin{array}{l}{\theta }_{inf}={\tan }^{-1}\left(1/{\mu }_s\right)\\ \Rightarrow {\theta }_{inf}={\tan }^{-1}\left(\frac{1}{0.60}\right)\\ \Rightarrow {\theta }_{inf}={59}^o\end{array}$