Search any question & find its solution
Question:
Answered & Verified by Expert
Two bodies $A$ and $B$ of masses $5 \mathrm{~kg}$ and $10 \mathrm{~kg}$ in contact with each other rest on a table against a rigid wall. The coefficient of friction between the bodies and the table is 0.15. A force of $200 \mathrm{~N}$ is applied horizontally at $\mathcal{A}$. What are (a) the reaction of the partition (b) the action-reaction forces between $A$ and $B$ ? What happens when the partition is removed? Does the answer to (b) change, when the bodies are in motion? Ignore difference between $\mu_s$ and $\mu_{k^*}$
Solution:
2688 Upvotes
Verified Answer
Here, Mass of $A, m_A=5 \mathrm{~kg}$ and Mass of $B$, $m_B=10 \mathrm{~kg}$; Coefficient of friction, $\mu=0.15$; Force acting on $A, F=200 \mathrm{~N}$
(a) reaction of the partition $=-200 \mathrm{~N}$
$\because\left(\mathrm{F}_{\text {Block wall }}=-\mathrm{F}_{\text {wall }}\right.$ Block $)$
(b) Since A presses B with a force of $200 \mathrm{~N}$ towards right $\therefore$ B presses $\mathrm{A}$ with a force of $200 \mathrm{~N}$ towards right. When the wall of the partition is removed, the block system will move due to the applied force \& kinetic friction will come into the picture.
$$
\therefore 200-\mu\left(m_A+m_B\right) g=\left(m_A+m_B\right) a
$$
$$
\begin{aligned}
\therefore a &=\frac{200-\mu\left(m_A-m_B\right) g}{m_A+m_B} \\
&=\frac{200-0.15(5+10) 10}{5+10} \\
&=\frac{117.5}{15}=11.8 \mathrm{~ms}^{-2}
\end{aligned}
$$
If the bodies are in motion, this force will overcome the force of friction $(f)$ \& the reaction of block B on A $\left(f_{\mathrm{BA}}\right)$. So under equillibrium
$$
\begin{aligned}
&200-f=m_{\mathrm{A}} a+F_{\mathrm{BA}} \\
&\begin{aligned}
\mathrm{F}_{\mathrm{BA}} &=200-\mu \mathrm{m}_{\mathrm{A}}-\mathrm{m}_{\mathrm{A}} \mathrm{a} \\
&=200-(\mu \mathrm{g}-a) \mathrm{m}_{\mathrm{A}} \\
&=200-(0.15 \times 10-11.8) 5 \\
&=200+51.5=251.5 \mathrm{~N}
\end{aligned}
\end{aligned}
$$
(a) reaction of the partition $=-200 \mathrm{~N}$
$\because\left(\mathrm{F}_{\text {Block wall }}=-\mathrm{F}_{\text {wall }}\right.$ Block $)$
(b) Since A presses B with a force of $200 \mathrm{~N}$ towards right $\therefore$ B presses $\mathrm{A}$ with a force of $200 \mathrm{~N}$ towards right. When the wall of the partition is removed, the block system will move due to the applied force \& kinetic friction will come into the picture.
$$
\therefore 200-\mu\left(m_A+m_B\right) g=\left(m_A+m_B\right) a
$$
$$
\begin{aligned}
\therefore a &=\frac{200-\mu\left(m_A-m_B\right) g}{m_A+m_B} \\
&=\frac{200-0.15(5+10) 10}{5+10} \\
&=\frac{117.5}{15}=11.8 \mathrm{~ms}^{-2}
\end{aligned}
$$
If the bodies are in motion, this force will overcome the force of friction $(f)$ \& the reaction of block B on A $\left(f_{\mathrm{BA}}\right)$. So under equillibrium
$$
\begin{aligned}
&200-f=m_{\mathrm{A}} a+F_{\mathrm{BA}} \\
&\begin{aligned}
\mathrm{F}_{\mathrm{BA}} &=200-\mu \mathrm{m}_{\mathrm{A}}-\mathrm{m}_{\mathrm{A}} \mathrm{a} \\
&=200-(\mu \mathrm{g}-a) \mathrm{m}_{\mathrm{A}} \\
&=200-(0.15 \times 10-11.8) 5 \\
&=200+51.5=251.5 \mathrm{~N}
\end{aligned}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.