Question

A cube weighing $10N$ is lying on a rough inclined plane of slope $3$ in $5$. The coefficient of friction between the plane and the cube is $0.6$. The force necessary to move the cube up the plane will be

• A. $6.4N$
• B. $10.16N$
• C. $21.6N$
• D. $108N$

Answer 1

The correct option is B $10.16N$

Here, $mg=10\Rightarrow m=1kg$
Slope$=\tan \theta =\frac{3}{5},\sin \theta =\frac{3}{\sqrt{34}}=0.5,\cos \theta =\frac{5}{\sqrt{34}}=0.86$
from FBD,
The necessary force, $F=ma=mg\sin \theta +\mu mg\cos \theta =10\left(0.5\right)+\left(0.6\right)10\left(0.86\right)=10.16N$