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 $\mathrm{m}/{\sec }^2$):

If the plane makes an angle $\mathrm{\theta }$ with horizontal

$\tan \mathrm{\theta } = \frac{8}{15} \mathrm{If} \mathrm{R} \mathrm{is} \mathrm{the} \mathrm{normal} \mathrm{reaction}$

$\text{R = 17g cos θ = 170×10×(}\frac{15}{17}) = 1500 \mathrm{N}$

Force of friction on A = 1500 $\times 0.2 = 300 \mathrm{N}$

Force on friction on B = 1500 $\times$0.4 = 600 N

Considering the two blocks as a system, the net force parallel to the plane

= $2\times 170\mathrm{g} \sin \mathrm{\theta } - 300 - 600 = 1600-900= 700 \mathrm{N}$

$\therefore \mathrm{Acceleration} = \frac{700}{340} = \frac{35}{17} \mathrm{m}/{\mathrm{s}}^2$

Consider the motion of A alone.

170g sin $\mathrm{\theta } \sin \mathrm{\theta } -300-\mathrm{P} = 170\times \frac{35}{17}$

(where P is pull on the bar)

P = 500-350 = 150 N