Question

Two blocks of masses $m_1=5\text{kg}$ and $m_2=6\text{kg}$ are connected by a light string passing over a light frictionless pulley as shown in figure. The mass $m_1$ is at rest on the inclined plane and mass $m_2$ hangs vertically. If the angle of incline $\theta ={30}^{\circ }$, what is the magnitude and direction of the force of friction on the $5\text{kg}$ block. Take $g=10{\text{ms}}^{-2}$.

• A. $35\text{N}$ up the plane
• B. $35\text{N}$ down the plane
• C. $85\text{N}$ up the plane
• D. $85\text{N}$ down the plane

Answer 1

The correct option is B $35\text{N}$ down the plane

Weight of mass $m_2=6\times 10=60\text{N}$. The weight of $m_2$ provides the tension. Thus
$T=60\text{N}$
Opposing this force along the plane is the component $F_1=m_{1}g\sin \theta$ of the force $m_{1}g$. Now $F_1=m_{1}g\sin \theta =5\times 10\times sin{30}^{\circ }=25\text{N}$. Since $F_1$ is less than $T$ and is, therefore, insufficient to balance $T$ (see Fig.), the force of friction $\left(f\right)$ down the plane is necessary to keep block $m_1$ at rest. Thus, $f$ must act down the plane. Since the mass $m_1$ is at rest, the net force on $m_1$ along the plane must be zero. Thus

$T-m_{1}g\sin {30}^{\circ }-f=0$
or $f=T-m_{1}g\sin {30}^{\circ }$
$=60-5\times 10\times \sin {30}^{\circ }$
$=60-25=35\text{N}$
Hence, the correct choice is (b).