Question

A block of mass $1kg$ is at rest on a horizontal table. The coefficient of static friction between the block and the table is $0.50$. If $g=10m{s}^{-2}$, then the magnitude of a force acting upwards at an angle of ${60}^o$ from the horizontal that will just start the block moving is:

• A. $5N$
• B. $5.36N$
• C. $74.6N$
• D. $10N$

Answer 1

Answer: (C)

$74.6N$

Equating all the forces.

In vertical direction, $N=mg-F\sin {60}^o=1\times 10-F\frac{\sqrt{3}}{2}\dots \dots \left(1\right)$

In horizontal direction, $F\cos {60}^o\ge \mu N=0.5\times (10-\frac{F\sqrt{3}}{2})$

$\Rightarrow \frac{F}{2}\ge 5-\frac{F\sqrt{3}}{4}$

$\Rightarrow F\ge 74.64N$

Hence, minimum required force is $74.6N$