Question

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

• A. $1.2N$
• B. $10.8N$
• C. $4.8N$
• D. $14.8N$

Answer 1

The correct option is B $10.8N$

Given that,

$\sin \theta =\frac{3}{5}$

$\cos \theta =\frac{4}{5}$

Now, the minimum force is

$F=mgsin\theta +\mu mg\cos \theta$

$F=10\times \frac{3}{5}+0.6\times 10\times \frac{4}{5}$

$F=6+\frac{24}{5}$

$F=\frac{54}{5}$

$F=10.8N$

Hence, the minimum force is $10.8N$