Question

Blocks A and B in the fig. Are connected by a bar of negligible weight. Mass of each block is 170 kg and ${\mu }_A$ = 0.2 and ${\mu }_B$ = 0.4, where ${\mu }_A$ and ${\mu }_B$ are the coefficients of limiting friction between blocks and plane, calculate the force developed in the bar (g=10 $m{s}^{-2}$)

• A. 150 N
• B. 75 N
• C. 200 N
• D. 250 N

Answer 1

The correct option is A

150 N

If the plane makes an angle $\theta$ with horizontal, then tan $\theta$ = $\frac{8}{15}$.

If R is the normal reaction,

R = 170g cos $\theta$ = 170 $\times$ 10 $\times$ $\left(\frac{15}{17}\right)$= 1500 N

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

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

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

= $2\times 170gsin\theta -300-600=1600-900=700N$

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

Consider the motion of A alone

170g sin $\theta$ - 300 - P = P170 $\times$ $\frac{35}{17}$

(Where P is pull on the bar)

P = 500 - 350 = 150 N