In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs 400 N , block B weighs 300 N , and the strut A B weigh 200 N . If μ = 0.25 under B

Question

In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs 400 N , block B weighs 300 N , and the strut A B weigh 200 N . If μ = 0.25 under B , determine the minimum coefficient of friction under A to prevent motion.

TestHub · Accepted Answer

Consider FBD of structure.
Applying equilibrium equations, $A v + B v = 200 N$ $A_{H} = B_{H} \dots$ (ii) From FBD of block $B$ ,
$B_{H} + F_{B} cos 6 0^{\circ} - N_{B} sin 6 0^{\circ} = 0$ $N_{B} cos 6 0^{\circ} - B_{V} - 300 + F_{B} sin 6 0^{\circ} = 0$ $F_{B} = 0.25 N_{B}$ $B_{H} - 0.74 N_{B} = 0$ (iii) $- B_{V} + 0.71 N_{B} = 300$ FBD of block $A$
$F_{A} - A_{H} = 0$ $N_{A} - A_{V} = 400 \dots$ $F_{A} = μ_{A} N_{A}$ $∴ μ_{A} N_{A} - A_{H} = 0 \dots$ on solving the above equation $N_{A} = 650 N, F_{A} = 260 N, F_{A} = μ_{A} N_{A}$ $∴ μ_{A} = \frac{260}{250} = 0.4$

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JEE Advanced 2027 (Class 12+)