Question

A block of weight $W$ rests on a horizontal floor with coefficient of static friction $\mu$. It is desired to make the block move by applying minimum amount of force. The angle $\theta$ from the horizontal at which the force should be applied and magnitude of the force $F$ are respectively.

• A. $\theta =ta{n}^{-1}\left(\mu \right),F=\frac{\mu W}{\sqrt{1+{\mu }^2}}$
• B. $\theta =ta{n}^{-1}\left(\frac{1}{\mu }\right),F=\frac{\mu W}{\sqrt{1+{\mu }^2}}$
• C. $\theta =0,F=\mu W$
• D. $\theta =ta{n}^{-1}\left(\frac{\mu }{1+\mu }\right),F=\frac{\mu W}{1+\mu }$

Answer 1

Answer: (A)

Let the force $F$ is applied at an angle $\theta$ with the horizontal.
For horizontal equilibrium,
$Fcos\theta =\mu R...\left(i\right)$
For vertical equilibrium,
$R+Fsin\theta =mg$
or, $R=mg-Fsin\theta ...\left(ii\right)$
Substituting this value of $R$ in eq. $\left(i\right),$we get
$Fcos\theta =\mu \left(mg-Fsin\theta \right)$
$=\mu mg-\mu Fsin\theta$
or, $F\left(cos\theta +\mu sin\theta \right)=\mu mg$
or, $F=\frac{\mu mg}{cos\theta +\mu sin\theta }...\left(iii\right)$
For $F$ to be minimum, the denominator
$\left(cos\theta +\mu sin\theta \right)$ should be maximum.
$\therefore \frac{d}{d\theta }\left(cos\theta +\mu sin\theta \right)=0$
or, $-sin\theta +\mu sin\theta =0$
or, $tan\theta +\mu$
or, $\theta =ta{n}^{-1}\left(\mu \right)$
Then, $sin\theta =\frac{\mu }{\sqrt{1+{\mu }^2}}$ and
$cos\theta =\frac{1}{\sqrt{1+{\mu }^2}}$
Hence, $F_{min}=\frac{\mu w}{\frac{1}{\sqrt{1+{\mu }^2}}+\frac{{\mu }^2}{\sqrt{1+{\mu }^2}}}=\frac{\mu w}{\sqrt{1+{\mu }^2}}$