Physics - NLM Question with Solution | TestHub

The forces acting on the block are gravity ($mg$), normal force ($N$), and static friction ($f_s$).

Resolve gravity into components parallel ($mg\sin \theta$) and perpendicular ($mg\cos \theta$) to the plane.

The normal force is $N=mg\cos \theta$.

The maximum static friction is $f_{s,max}={\mu }_{s}N={\mu }_{s}mg\cos \theta$.

The given frictional force $f_s=10\mathrm{N}$ must be less than or equal to $f_{s,max}$.

Also, for the block to remain at rest, the frictional force must balance the component of gravity down the incline, $f_s=mg\sin \theta$.

Given:

$\theta =30^{\circ }$

${\mu }_s=0.8$

$f_s=10\mathrm{N}$

$g=10m/{\mathrm{s}}^2$

From $f_s=mg\sin \theta$:

$10=m\times 10\times \sin (30^{\circ })$

$10=m\times 10\times 0.5$

$10=5m$

$m=\frac{10}{5}=2\mathrm{kg}$

We should also check if this frictional force is less than or equal to the maximum possible static friction.

$f_{s,max}={\mu }_{s}mg\cos \theta =0.8\times 2\times 10\times \cos (30^{\circ })$

$f_{s,max}=0.8\times 20\times \frac{\sqrt{3}}{2}=16\times \frac{\sqrt{3}}{2}=8\sqrt{3}\approx 8\times 1.732=13.856\mathrm{N}$

Since $10\mathrm{N}<13.856\mathrm{N}$, the block is indeed at rest and the frictional force is $10\mathrm{N}$.

The mass of the block is $2.0\mathrm{kg}$.

Solution:

Given $f_s=mg\sin \theta$.

$10\mathrm{N}=m\times 10m/{\mathrm{s}}^2\times \sin (30^{\circ })$

$10=m\times 10\times 0.5$

$10=5m$

$m=2.0\mathrm{kg}$