Question

Two blocks of masses $m_1$ and $m_2$ are connected through a massless inextensible string. Block of mass $m_1$ is placed at the fixed rigid inclined surface while the block of mass $m_2$ hanging at the other end of the string, which is passing through a fixed massless frictionless pulley shown in figure. The coefficient of static friction between the block and the inclined plane is 0.8. The system of masses $m_1$ and $m_2$ is released from rest. (Take $g=10{\text{m/s}}^2)$

• A. The tension in the string is $20N$ after releasing the system
• B. The contact force by the inclined surface on the block is along normal to the inclined surface
• C. The magnitude of contact force by the inclined surface on the block $m_1$ is $20\sqrt{3}N$
  
• D. Frictional force between $4\text{kg}$ and inclined plane is zero

Answer 1

Answer: (A), (B), (C), (D)

Frictional force between $4\text{kg}$ and inclined plane is zero

Net force on the system of masses is $m_{1}g\sin 30-m_{2}g=4g\sin 30-2g=0$

Since net pulling force is zero on the system, static friction adjusts its value to zero
i.e $f=0$
and net contact force by the incline is perpendicular to the incline.
$N=m_{1}g\cos {30}^{\circ }$
For $m_2$
$T-m_{2}g=0$
Putting $a=0$ and $f=0$
$T=20\text{N}$

Why this question? Tip: Since static friction is self adjusting, we should first find the resultant of all the forces parallel to the interface of the bodies, and then one will get an idea as to how much friction is required, or it is even required or not!