Laws of Motion 3 Question 10
10. What is the maximum value of the force $F$ such that the block shown in the arrangement, does not move?
(2003, 2M)
(a) $20 N$
(b) $10 N$
(c) $12 N$
(d) $15 N$
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Answer:
Correct Answer: 10. (a)
Solution:
- Free body diagram (FBD) of the block (shown by a dot) is shown in figure.
For vertical equilibrium of the block
$$ N=m g+F \sin 60^{\circ}=\sqrt{3} g+\sqrt{3} \frac{F}{2} $$
For no motion, force of friction
$$ \begin{gathered} f \geq F \cos 60^{\circ} \text { or } \mu N \geq F \cos 60^{\circ} \\ \text { or } \quad \frac{1}{2 \sqrt{3}} \sqrt{3} g+\frac{\sqrt{3} F}{2} \geq \frac{F}{2} \text { or } g \geq \frac{F}{2} \\ \text { or } \\ F \leq 2 g \text { or } 20 N \end{gathered} $$
Therefore, maximum value of $F$ is $20 N$.
