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 rn/2 and of B is m.

a.	The bodies will move together if F = O .25 mg
b.	The A will slip with B if F = 0.5 mg
c.	The bodies will move together if F = 0.5 mg
d.	The bodies will be at rest if F = 0.1 mg
e	The maximum value of F for which the two bodies will move together is 0.45 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)

(1) Hint: Apply Newton's laws of motion.

Step 1: Find the combined acceleration.

Consider the adjaænt diagram. The frictional force on B(

f_{1}

) and frictional force on A(

f_{2}

) will be as shown.

Let A and B are moving together

a_{common} = \frac{F - t_{1}}{m_{A} + m_{B}} = \frac{F - f_{1}}{(m / 2) + m} = \frac{2 (F - f_{1})}{3 m}

Step 2: Find the pseudo force on A.

\begin{array}{l} Pseudo force on A = (m_{A}) \times a_{common} \\ = m_{A} \times \frac{2 (F - f_{1})}{3 m} = \frac{m}{2} \times \frac{2 (F - f_{1})}{3 m} = \frac{(F - t_{1})}{3} \end{array}

The force (F) will be maximum when
Pseudo force on A = Frictional force on A

\begin{array}{l} \Rightarrow \frac{F_{\max} - t_{1}}{3} = {μm}_{A} g \\ = 0 .2 \times \frac{m}{2} \times g = 0 .1 mg \\ \Rightarrow F_{\max} = 0 .3 mg + t_{1} \\ = 0 .3 mg + (0 .1) \frac{3}{2} mg = 0 .45 mg \end{array}

Step 3: Analyse each option using Newton's second law of motion.

\Rightarrow

Hence, the maximum force unto which bodies will move together is

F_{\max} = 0.45 mg

\begin{array}{l} (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. \end{array}

\begin{array}{l} (f_{1})_{\max} = {μm}_{B} g = (0 .1) \times \frac{3}{2} m \times g = 0 .15 mg \\ (f_{2})_{\max} = {μm}_{A} g = (0 .2) (\frac{m}{2}) (g) = 0 .1 mg \end{array}

Hence, the minimum force required for the movement of the system (A + B)

\begin{array}{l} F_{\min} = (f_{1})_{\max} + (f_{2})_{\max} \\ = 0.15 m g + 0.1 m g = 025 m g \end{array}

\begin{array}{l} (d) Given, force F = 0 .1 mg < F_{min} . \\ Hence, the bodies will be at rest. \\ (e) Maximum force for combined movement F_{\max} = 0 .45 mg . \end{array}

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