Laws of Motion 3 Question 21
22. A block of mass $m$ rests on a horizontal floor with which it has a coefficient of static friction $\mu$. It is desired to make the body move by applying the minimum possible force $F$. Find the magnitude of $F$ and the direction in which it has to be applied.
(1987, 7M)
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Answer:
Correct Answer: 22. $m g \sin \theta, \theta=\tan ^{-1}(\mu)$ from horizontal
Solution:
- Let $F$ be applied at angle $\theta$ as shown in figure.
Normal reaction in this case will be,
$$ N=m g-F \sin \theta $$
The limiting friction is therefore
$$ f _L=\mu N=\mu(m g-F \sin \theta) $$
For the block to move,
$$ \begin{aligned} F \cos \theta & =f _L=\mu(m g-F \sin \theta) \\ \text { or } \quad F & =\frac{\mu m g}{\cos \theta+\mu \sin \theta} \end{aligned} $$
For $F$ to be minimum, denominator should be maximum.
$$ \begin{aligned} \text { or } & & \frac{d}{d \theta}(\cos \theta+\mu \sin \theta) & =0 \\ \text { or } & & \sin \theta+\mu \cos \theta & =0 \\ \text { or } & & \tan \theta=\mu \text { or } \theta & =\tan ^{-1}(\mu) \end{aligned} $$
Substituting this value of $\theta$ in Eq. (i), we get
$$ F _{\min }=m g \sin \theta $$
