Question

A man of mass 50kg is pulling on a plank of mass 100 kg kept on a smooth floor as shown with force of 100 N. If both man & plank move together, find a force of friction acting on the man.

A.$\dfrac{100}{3}$ N towards left
B.$\dfrac{100}{3}$ N towards right
C.$\dfrac{250}{3}$N towards left
D.\[\dfrac{250}{3}\] N towards right

Answer 1

Hint: We will proceed by considering the man and the plank as a single system. From the figure, we can say that the friction between plank and surface is 0 and the surface is smooth. Thus, the friction force acting between the plank and the man will oppose its motion.

Formula used:
Force acting on the body=$F=ma$

Complete step by step solution:
Follow the below steps to solve such questions:
Step 1: Write all the data given in the question
Mass of man=${{m}_{o}}=50kg$
Mass of the plank=${{m}_{p}}=100kg$
Man applies a force of 100N , therefore, F=100N
Step 2: Consider the man and the plank as a single system, let total mass of the system be m
So, $m={{m}_{o}}+{{m}_{p}}$
Substituting the values, we get
$\begin{align}
  & m=50+100 \\
 & \Rightarrow m=150kg \\
\end{align}$
Now, we know that F=100 N
Then acceleration of the system will be given by the formula:
$a=\dfrac{F}{m}$
$\begin{align}
  & \Rightarrow a=\dfrac{100}{150} \\
 & \therefore a=\dfrac{2}{3}m/{{s}^{2}} \\
\end{align}$
Since, man and the plank are moving together as one system, the acceleration of the man will be equal to the acceleration of the system.
Hence, acceleration of man is $a=\dfrac{2}{3}m/{{s}^{2}}$.
Step 3: Now, we will find the friction force acting on the man.
   $\begin{align}
  & \text{ }F={{m}_{o}}a \\
 & \Rightarrow F=50\times \dfrac{2}{3} \\
 & \therefore F=\dfrac{100}{3}N \\
\end{align}$
F is the force which acts on the man as he moves in the right direction.
Since there is friction present between man and the plank, friction force will act between them and the man will move only when the applied force is overcome by the friction force.
We found out that the applied force is $F=\dfrac{100}{3}N$,
Hence an equal force acts in the opposite direction to oppose the motion of the man.
Now we can say that, applied force is in the right direction hence friction force will be towards the left.
So, the friction force acting on man is $F=\dfrac{100}{3}N$ towards the left.

So, Option(A) is correct.

Note:
We can solve this question by considering man and the plank as two separate systems.
So, the tension on the rope will be T=100 N.
Let the friction force on man be f, so we get
$\begin{align}
  & T-f={{m}_{o}}a \\
 & \Rightarrow 100-f=50a\text{ }..........\text{(1)} \\
\end{align}$
And for the plank
$100+f=150a\text{ }.............\text{(2)}$
Solving equation (1) and (2) we get, $f=\dfrac{100}{3}N$.