Question

Figure shows two blocks connected by a light string placed on the two inclined parts of a triangular structure. The coefficients of static and kinetic friction are $0.28$ and $0.25$ respectively at each of the surfaces. Find the minimum and maximum values of $m$ for which the system remains at rest. (Both in $\text{kg}$)

• A. $\frac{9}{8},\frac{32}{9}$
• B. $\frac{9}{8},\frac{32}{10}$
• C. $\frac{3}{8},\frac{32}{9}$
• D. $\frac{3}{8},\frac{32}{10}$

Answer 1

The correct option is A $\frac{9}{8},\frac{32}{9}$

When minimum value of $m$ is selected, it has to be less than $(M=2\text{kg})$ block. Then $m$ has a tendency to slip upwards and $M$ to slip downwards. Hence, friction will act on the two blocks as shown in figure.


Taking components perpendicular to the incline for the mass $M$
$N=Mg\cos {45}^{\circ }=\frac{Mg}{\sqrt{2}}$
Taking components parallel to the incline
$T+f=Mg\sin {45}^{\circ }=\frac{Mg}{\sqrt{2}}$
or $T=\frac{Mg}{\sqrt{2}}-f$
As it is a case of limiting equilibrium,
$f={\mu }_{s}N$
or $T=\frac{Mg}{\sqrt{2}}-{\mu }_{s}N=\frac{Mg}{\sqrt{2}}(1-{\mu }_s)....\left(i\right)$

Now consider the other block as the system. The forces acting on this block are shown in figure $\left(b\right)$
Taking components perpendicular to the incline
$N^{\prime }=mg\cos {45}^{\circ }=\frac{mg}{\sqrt{2}}$
Taking components parallel to the incline
$T=mg\sin {45}^{\circ }+f^{\prime }$
$=\frac{mg}{\sqrt{2}}+f^{\prime }$
As it is the case of limiting equilbrium,
$f^{\prime }=u_s{N}^{\prime }=u_s\frac{mg}{\sqrt{2}}$
Thus, $T=\frac{mg(1+{\mu }_s)}{\sqrt{2}}........\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$m(1+{\mu }_s)=M(1-{\mu }_s).......\left(iii\right)$
or, $m=\left(\frac{1-{\mu }_s}{1+{\mu }_s}\right)M=\left(\frac{1-0.28}{1+0.28}\right)\times 2\text{kg}$
$=\frac{9}{8}\text{kg}$

When maximum possible value of $m$ is selected, the directions of friction forces are reversed becasue $m$ has the tendency to slip down and $2\text{kg}$ block to slip up.

The force equations for the two blocks will be:
$T=\frac{Mg}{\sqrt{2}}(1+{\mu }_s)---\left(iv\right)$ &
$T=\frac{mg}{\sqrt{2}}(1-{\mu }_s)---\left(v\right)$
$\Rightarrow M(1+{\mu }_s)=m(1-{\mu }_s)---\left(vi\right)$

Equating the two, the maximum value of $m$ can be obtained from by putting $\mu =0.28$
Thus, the maximum value of $m$ is
$m=\frac{1+0.28}{1-0.28}\times 2\text{kg}=\frac{32}{9}\text{kg}$