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# What is the value of acceleration of blocks A and B if F = 40N and mass of block A is 5 kg and mass of block B is 10 kg? The arrangement of two blocks is shown below in the figure.

Last updated date: 15th Sep 2024
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Hint At first, find the value of limiting friction on all contact surfaces which is denoted by ${f_{s\max }}$.
Then, by using the below formula, find out the acceleration for both blocks one by one from given values in question,
$\Rightarrow F - {f_{s\max }} = ma$
where, ${f_{s\max }} =$limiting value of friction
$m =$mass of a block
$a =$acceleration of a block.

Complete step by step Solution
The force which opposes motion of an object moving relative to another is called Friction. It is not a fundamental force.
The maximum value of static friction that occurs when the body is just at the sliding point of the surface of another body. In other words, we can say that it is a product of coefficients of limiting friction and normal force. We can express it mathematically as,
$\Rightarrow {f_{s\max }} = \mu N$
Normal force can also be written as,
$\Rightarrow$ $N = mg$(where, $g = 10m{s^{ - 2}}$)
To find out the acceleration from limiting friction we need to use the below expression,
$\Rightarrow$ $F - {f_{s\max }} = ma \cdots (1)$
So, let ${m_a}$ and ${m_b}$ be the mass of blocks A and B respectively and let ${a_a}$ and ${a_b}$ be the acceleration of blocks A and B respectively.
So, at first, we will find the acceleration for block A
Therefore, from equation $(1)$ we get,
$\Rightarrow F - {f_{s\max }} = {m_a}{a_a}$
$\Rightarrow F - \mu \times N = {m_a}{a_a}$
$\Rightarrow F - \mu \times {m_a}g = {m_a}{a_a}$
$\Rightarrow 40 - 0.5 \times 5 \times 10 = 5{a_a}$
$\Rightarrow 40 - 25 = 5{a_a}$
$\Rightarrow 15 = 5{a_a}$
$\Rightarrow {a_a} = 3m/{s^2}$
Let ${N'}$ be the force acting upon block A from block B -
$\therefore {N'} = N + {m_b}g$
So,
$\Rightarrow {\mu _1}N - {\mu _2}N' = {m_b}{a_b}$
$\Rightarrow 0.5 \times 50 - 0.2(50 + 10 \times 10) = 10{a_b}$
$\Rightarrow 25 - 0.2(150) = 10{a_b}$
$\Rightarrow 25 - 30 = 10{a_b}$
$\Rightarrow {a_b} = \dfrac{{ - 5}}{{10}}m/{s^2}$

So, the acceleration of block B is nearly 0.

Note Limiting friction occurs when the body is just at sliding point over another’s body surface. Limiting friction always opposes the motion of an object. When the surfaces are in contact, the limiting friction always acts tangentially to them.