Q87 / 125 · BITSAT 2015 · Physics
In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block $A$ weighs $400 \mathrm{~N}$, block $B$ weighs $300 \mathrm{~N}$, and the strut $A B$ weigh $200 \mathrm{~N}$. If $\mu=0.25$ under $B,$ determine the minimum coefficient of friction under $A$ to prevent motion.
- Correct answerA0.4
- B0.2
- C0.8
- D0.1
Solution
Consider FBD of structure.
Applying equilibrium equations, $A v+B v=200 \mathrm{~N}$ $A_{H}=B_{H} \ldots$ (ii) From FBD of block $B$,
$B_{H}+F_{B} \cos 60^{\circ}-N_{B} \sin 60^{\circ}=0$ $N_{B} \cos 60^{\circ}-B_{V}-300+F_{B} \sin 60^{\circ}=0$ $F_{B}=0.25 N_{B}$ $B_{H}-0.74 N_{B}=0$ (iii) $-B_{V}+0.71 N_{B}=300$ FBD of block $A$
$F_{A}-A_{H}=0$ $N_{A}-A_{V}=400 \quad \ldots$ $F_{A}=\mu_{A} N_{A}$ $\therefore \mu_{A} N_{A}-A_{H}=0 \quad \ldots$ on solving the above equation $N_{A}=650 \mathrm{~N}, F_{A}=260 \mathrm{~N}, F_{A}=\mu_{A} N_{A}$ $\therefore \mu_{A}=\frac{260}{250}=0.4$
Applying equilibrium equations, $A v+B v=200 \mathrm{~N}$ $A_{H}=B_{H} \ldots$ (ii) From FBD of block $B$,
$B_{H}+F_{B} \cos 60^{\circ}-N_{B} \sin 60^{\circ}=0$ $N_{B} \cos 60^{\circ}-B_{V}-300+F_{B} \sin 60^{\circ}=0$ $F_{B}=0.25 N_{B}$ $B_{H}-0.74 N_{B}=0$ (iii) $-B_{V}+0.71 N_{B}=300$ FBD of block $A$
$F_{A}-A_{H}=0$ $N_{A}-A_{V}=400 \quad \ldots$ $F_{A}=\mu_{A} N_{A}$ $\therefore \mu_{A} N_{A}-A_{H}=0 \quad \ldots$ on solving the above equation $N_{A}=650 \mathrm{~N}, F_{A}=260 \mathrm{~N}, F_{A}=\mu_{A} N_{A}$ $\therefore \mu_{A}=\frac{260}{250}=0.4$
Hence, the correct answer is option: 'A'.