Ncert Exercises & Solutions

In the figure, the coefficient of friction between the floor and body $B$ is $0.1.$ The coefficient of friction between bodies $B$ and $A$ is $0.2.$ A force $F$ is applied as shown on $B.$ The mass of $A$ is $m/2$ and of $B$ is $m.$

(a)	The bodies will move together if $F = 0.25\text{mg}$
(b)	The $A$ will slip with $B$ if $F = 0.5\text{mg}$
(c)	The bodies will move together if $F = 0.5\text{mg}$
(d)	The bodies will be at rest if $F = 0.1\text{mg}$
(e)	The maximum value of $F$ for which the two bodies will move together is $0.45\text{mg}$

Which of the following statement(s) is/are true?
1. (a), (b), (d), (e)
2. (a), (c), (d), (e)
3. (b), (c), (d)
4. (a), (b), (c)

Hint: Apply Newton's laws of motion.

Step 1: Find the combined acceleration.

Consider the adjaænt diagram. The frictional force on

(f_{1})

$(f_1)$ and frictional force on

(f_{2})

$(f_2)$ will be as shown.

Let

A

$A$ and

B

$B$ are moving together the common acceleration is given by;

a_{common} = \frac{F - f_{1}}{m_{A} + m_{B}}

$a_\text{common}=\frac{F-f_1}{m_A+m_B}$

a_{common} = \frac{2 (F - f_{1})}{3 m} . . . (1)

$a_\text{common}=\frac{2(F-f_1)}{3m}~~~...(1)$

Step 2: Find the pseudo force on the block

A .

$A.$
The pseudo force on the block

A

$A$ is given by;

F_{p s e u d o} = (m_{A}) \times a_{common}

$F_{pseudo}=(m_A)\times a_\text{common}$

F_{p s e u d o} = (m_{A}) \times \frac{2 (F - f_{1})}{3 m}

$F_{pseudo}=(m_A)\times \frac{2(F-f_1)}{3m}$

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The force

$(F)$ will be maximum when;
The pseudo force on

$A$ = The frictional force on

$A$

$\Rightarrow (m_A)\times \frac{2(F-f_1)}{3m} = \left(μm\right)_{A} g$

$= 0 .2 \times \frac{m}{2} \times g = 0 .1 mg$

$\Rightarrow F_{max} = 0 .3 mg + t_{1}$

$= 0 .3 mg + \left(\right. 0 .1 \left.\right) \frac{3}{2} mg = 0 .45 mg$

Step 3: Analyse each option using Newton's second law of motion.
Hence, the maximum force unto which bodies will move together is

$F_{max} = 0 . 45 mg$
(a) Hence, for

$F = 0 .25 mg < F _{max}$ bodies will move together.
(b) For

$F = 0 .5 mg > F_{ max },$ body

$A$ will slip with respect to

$B.$
(c) For

$F = 0 .5 mg > F _{max} ,$ bodies slip.

$f_{1max}=\mu m_B g \Rightarrow 0.1\times \frac{3m}{2} \times g=0.15mg$

$f_{2max}=\mu m_A g \Rightarrow 0.2\times \frac{m}{2} \times g=0.1mg$
Hence, the minimum force required for the movement of the system

$(A + B)$

$f_{min}=f_{1max}+f_{2max}$

$\Rightarrow f_{min}=0.15mg+0.1mg=0.25mg$
(d) Given, force

$F = 0 .1 mg < F_{min }$
Hence, the bodies will be at rest.
(e) The maximum force for combined movement

$F_{max} = 0 .45 mg$
Hence, option (1) is the correct answer.

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