Question

A cube of weight $10N$ rests on a rough inclined plane of slope $3$ in $5$. The coefficient of friction is $0.6$. The minimum force necessary to start the cube moving up the plane is

• A. $5.4N$
• B. $10.8N$
• C. $2.7N$
• D. $18N$

Answer 1

The correct option is B $10.8N$

$w=mg=10Nsin\theta =\frac{3}{5}\mu =0.6F_{min}=mgsin\theta =f=mgsin\theta +\mu mgcos\theta =mg(sin\theta +\mu cos\theta )=10(\frac{3}{5}+\frac{3}{5}\times \frac{4}{5})\frac{10}{5}(3+\frac{12}{5})=\frac{54}{5}{F}_{min}=10.8N$