Blocks A ard B in the figure are connected by a bar of negligible weight. Mass of each block is 170 kg and ${\mathrm{\mu }}_{\mathrm{A}} = 0.2 \mathrm{and} {\mathrm{\mu }}_{\mathrm{B}} = 0.4$,  where ${\mathrm{\mu }}_{\mathrm{A}} \mathrm{and} {\mathrm{\mu }}_{\mathrm{B}}$ are the coefficients of limiting friction between blocks and plane, calculate the force developed in the bar (g = 10 $\mathrm{m}/{\sec }^2$):

detailed solution

Correct option is A

If the plane makes an angle θ with horizontaltan θ = 815 If R is the normal reactionR = 17g cos θ = 170×10×(1517) = 1500 NForce of friction on A = 1500 ×0.2 = 300 NForce on friction on B = 1500 ×0.4 = 600 NConsidering the two blocks as a system, the net force parallel to the plane= 2×170g sin θ - 300 - 600 = 1600-900= 700 N∴ Acceleration = 700340 = 3517 m/s2Consider the motion of A alone.170g sin θ sin θ -300-P = 170×3517(where P is pull on the bar)P = 500-350 = 150 N