What is the maximum value of the force such that the block shown in the arrangement does not move? [2003]

Question

What is the maximum value of the force $F$ such that the block shown in the arrangement does not move? [2003]

Smart Achievers · Accepted Answer

(1)

Since the block is not moving forward for the maximum force $F$ applied, therefore

$F \cos 60 ° = f = μ N$

(Horizontal direction)

For vertical equilibrium of the block,

$N = m g + F \sin 60 °$

$∴$ $F \cos 60 ° = μ N = μ (F \sin 60 ° + m g)$

$\Rightarrow F = \frac{μ m g}{\cos 60 ° - μ \sin 60 °}$ $= \frac{\frac{1}{2 \sqrt{3}} \times \sqrt{3} \times 10}{\frac{1}{2} - \frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{2}} = \frac{5}{\frac{1}{4}} = 20 N$

Solution

Q.
What is the maximum value of the force $F$ such that the block shown in the arrangement does not move [2003]

1 20 N

2 10 N

3 12 N

4 15 N

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