SATHEE: Laws of Motion 3 Question 20

Laws of Motion 3 Question 20

21. Block $A$ of mass $m$ and block $B$ of mass $2 m$ are placed on a fixed triangular wedge by means of a massless, in extensible string and a frictionless pulley as shown in figure. The wedge is inclined at $45^{\circ}$ to the horizontal on both sides. The coefficient of friction between block $A$ and the wedge is $2 / 3$ and that between block $B$ and the wedge is $1 / 3$ . If the blocks $A$ and $B$ are released from rest, find

(1997C, 5M)

(a) the acceleration of $A$ ,

(b) tension in the string and

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Answer:

Correct Answer: 21. (a) acceleration $= 0$ (b) $\frac{2 \sqrt{2}}{3} m g$ (c) $\frac{m g}{3 \sqrt{2}}$ (down the plane)

Solution:

(a) Acceleration of block $A$

Maximum friction force that can be obtained at $A$ is

$(f_{max})_{A} = μ_{A} (m g \cos 45^{\circ})$

$= \frac{2}{3} (m g / \sqrt{2}) = \frac{\sqrt{2} m g}{3}$

Similarly,

$\begin{aligned} (f_{max})_{B} & = μ_{B} (2 m g \cos 45^{\circ}) \\ = \frac{1}{3} (2 m g / \sqrt{2}) = \frac{\sqrt{2} m g}{3} \end{aligned}$

Therefore, maximum value of friction that can be obtained on the system is

$(f_{max}) = (f_{max})_{A} + (f_{max})_{B} = \frac{2 \sqrt{2} m g}{3} \dots (i)$

Net pulling force on the system is

$F = F_{1} - F_{2} = \frac{2 m g}{\sqrt{2}} - \frac{m g}{\sqrt{2}} = \frac{m g}{\sqrt{2}} \dots (i i)$

From Eqs. (i) and (ii), we can see that

Net pulling force $< f_{max}$ . Therefore, the system will not move or the acceleration of block $A$ will be zero.

(b) and (c) Tension in the string and friction at $A$

Net pulling force on the system (block $A$ and $B$ )

$F = F_{1} - F_{2} = m g / \sqrt{2}$

Therefore, total friction force on the blocks should also be equal to $\frac{m g}{\sqrt{2}}$

$f_{A} + f_{B} = F = m g / \sqrt{2}$

Now, since the blocks will start moving from block $B$ first (if they move), therefore, $f_{B}$ will reach its limiting value first and if still some force is needed, it will be provided by $f_{A}$ .

Here,

${(f_{max})}_{B} < F$

Therefore, $f_{B}$ will be in its limiting value and rest will be provided by $f_{A}$ .

Hence, $f_{B} = {(f_{max})}_{B} = \frac{\sqrt{2} m g}{3}$

and

$\begin{aligned} f_{A} & = F - f_{B} \\ = \frac{m g}{\sqrt{2}} - \frac{\sqrt{2} m g}{3} = \frac{m g}{3 \sqrt{2}} \end{aligned}$

The FBD of the whole system will be as shown in the figure.

Therefore, friction on $A$ is

$f_{A} = m g / 3 \sqrt{2} (down the plane)$

Now, for tension $T$ in the string, we may consider either equilibrium of $A$ or $B$ .

Equilibrium of $A$ gives

$\begin{aligned} T = F_{2} + f_{A} & = \frac{m g}{\sqrt{2}} + \frac{m g}{3 \sqrt{2}} = \frac{4 m g}{3 \sqrt{2}} \\ or T & = \frac{2 \sqrt{2} m g}{3} \end{aligned}$

Similarly, equilibrium of $B$ gives $T + f_{B} = F_{1}$

$or \begin{aligned} T & = F_{1} - f_{B} \\ = \frac{2 m g}{\sqrt{2}} - \frac{\sqrt{2} m g}{3} = \frac{4 m g}{3 \sqrt{2}} \\ or & = \frac{2 \sqrt{2} m g}{3} \end{aligned}$

Therefore, tension in the string is $\frac{2 \sqrt{2} m g}{3}$ .

Laws of Motion 3 Question 20

Answer:

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Laws of Motion 3 Question 20

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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