Question

A block of mass $m$ is lying on a horizontal surface and the coefficient of static friction between the block and surface is $\mu$. Force $F$ is applied at an angle $\theta$ with the horizontal in two different ways as shown in the figure.


For just moving the block along horizontal direction, identify the correct statement.

• A. If $\mu =0$ , $F$ is lesser in case $\left(a\right)$
• B. If $\mu =0$ , $F$ is lesser in case $\left(b\right)$
• C. If $\mu \ne 0$, $F$ is lesser in case $\left(a\right)$
• D. If $\mu \ne 0$, $F$ is lesser in case $\left(b\right)$

Answer 1

The correct option is C If $\mu \ne 0$, $F$ is lesser in case $\left(a\right)$

Drawing FBD for both cases and considering friction at limiting value$f_{\text{max}}$, for just moving condition of the block:


For Case$\left(a\right)$: Applying equilibrium condition
$N_1+F\sin \theta =mg.....i$
$F\cos \theta =f_1=\mu N_1....ii$
From Eq. $i$ and $ii$
$F\cos \theta =\mu (mg-F\sin \theta )$
$\therefore F=\frac{\mu mg}{\cos \theta +\mu \sin \theta }...iii$

For Case$\left(b\right)$: Applying equilibrium condition
$N_2=F\sin \theta +mg.....iv$
$F\cos \theta =f_2=\mu N_2...v$
From Eq. $iv$ and $v$
$F\cos \theta =\mu (mg+F\sin \theta )$
$\therefore F=\frac{\mu mg}{\cos \theta -\mu \sin \theta }...vi$

From Eq. $iii$ and $vi$, it is evident that value of force $F$ is smaller is case $\left(a\right)$, due to larger denominator.

If we consider $\mu =0$, then block will always move, in the absence of resisting force.
$\therefore$Option $C$ is correct.