Question

Blocks $A$ and $B$ in the figure are connected by a bar of negligible weight. Mass of each block is $170kg$ 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 $(g=10m/se{c}^2)$:

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

Answer 1

Answer: (A)

If the plane makes an angle $\theta$ with horizontal

$tan\theta =8/15$ If $R$ is the normal reaction

$R=170gcos\theta =170\times 10\times \left(\frac{15}{17}\right)$

$=1500N$

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

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

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

$=2\times 170gsin\theta -300-600$

$=1600-900=700N$

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

Consider the motion of A alone

$170gsin\theta -300-P=170\times \frac{35}{17}$

(where P is pull on the bar)

$P=500-350$

$=150N$