$A$ wooden block of mass $2\; kg$ rests on a soft horizontal floor. When an iron cylinder of mass $25\; kg$ is placed on top of the block,the floor yields steadily and the block and the cylinder together go down with an acceleration of $0.1\; m/s^2$. What is the action of the block on the floor $(a)$ before and $(b)$ after the floor yields? Take $g = 10\; m/s^2$. Identify the action-reaction pairs in the problem.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The block is at rest on the floor. Its free-body diagram shows two forces on the block: the force of gravitational attraction by the earth equal to $2 \times 10 = 20\; N$,and the normal force $R$ of the floor on the block. By the First Law,the net force on the block must be zero,i.e.,$R = 20\; N$. Using the third law,the action of the block (i.e.,the force exerted on the floor by the block) is equal to $20\; N$ and directed vertically downwards.
$(b)$ The system (block $+$ cylinder) accelerates downwards with $0.1\; m/s^2$. The free-body diagram of the system shows two forces on the system: the force of gravity due to the earth $(27\; kg \times 10\; m/s^2 = 270\; N)$; and the normal force $R'$ by the floor. Note that the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system:
$270 - R' = 27 \times 0.1\; N$
$R' = 270 - 2.7 = 267.3\; N$
By the third law,the action of the system on the floor is equal to $267.3\; N$ vertically downward.
Action-reaction pairs:
For $(a)$: $(i)$ The force of gravity $(20\; N)$ on the block by the earth (action); the force of gravity on the earth by the block (reaction) equal to $20\; N$ directed upwards. $(ii)$ The force on the floor by the block (action); the force on the block by the floor (reaction).
For $(b)$: $(i)$ The force of gravity $(270\; N)$ on the system by the earth (action); the force of gravity on the earth by the system (reaction),equal to $270\; N$ directed upwards. $(ii)$ The force on the floor by the system (action); the force on the system by the floor (reaction). In addition,for $(b)$,the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.

Explore More

Similar Questions

$A$ bullet of mass $20 \, g$ travelling horizontally with a speed of $500 \, m/s$ passes through a wooden block of mass $10.0 \, kg$ initially at rest on a surface. The bullet emerges with a speed of $100 \, m/s$ and the block slides $20 \, cm$ on the surface before coming to rest. Find the coefficient of friction between the block and the surface. $(g = 10 \, m/s^2)$

Difficult
View Solution

What is the relationship between $W_1$ and $W_2$?

Consider the following statements $A$ and $B$ and identify the correct answer given below:
$(A)$ $A$ body initially at rest is acted upon by a constant force. The rate of change of its kinetic energy varies linearly with time.
$(B)$ When a body is at rest, it must be in equilibrium.

In the figure,a block of weight $60 \, N$ is placed on a rough surface. The coefficient of friction between the block and the surface is $0.5$. Find the maximum weight $W$ such that the block does not slip on the surface.

Difficult
View Solution

Imagine the situation in which the given arrangement is placed inside a trolley that can move only in the horizontal direction,as shown in the figure. If the trolley is accelerated horizontally along the positive $x$-axis with $a_0$,then identify the correct statement$(s)$ related to the tension $T$ in the string.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo