Question

A block weighing $10kg$ is at rest on a horizontal table. The coefficient of static friction between the block and the table is $0.5$. If a force acts down at ${60}^0$ from the horizontal, how large can it be without causing the block to move? ($g=10m{s}^{-2}$)

• A. $346N$
• B. $446N$
• C. $746N$
• D. $846N$

Answer 1

Answer: (C)

$746N$

Normal force is given by $N=mg+Fsin\theta =100+\frac{F\sqrt{3}}{2}$

Thus maximum static friction, $f_{S,max}=\mu N=0.5(100+\frac{F\sqrt{3}}{2})=50+\frac{F\sqrt{3}}{4}$

This friction force is equal to the horizontal component of applied force at the time motion starts.

Thus, $50+\frac{F\sqrt{3}}{4}=Fcos60^0=\frac{F}{2}$

$F(\frac{1}{2}-\frac{\sqrt{3}}{4})=50F=\frac{50}{0.067}\approx 746N$