Question

Block $A$ and $B$ in the figure are connected by a bar of negligible weight. If $A=B=170\text{kg}$ and ${\mu }_A=0.2$ and ${\mu }_B=0.4$, where ${\mu }_A$ and ${\mu }_B$ are the coefficients of limiting friction between the blocks and the plane, calculate the contact force on the bar due to $A$ $(g=10{\text{m/s}}^2)$

• A. $150\text{N}$
• B. $75\text{N}$
• C. $200\text{N}$
• D. $250\text{N}$

Answer 1

The correct option is A $150\text{N}$

If the plane makes an angle $\theta$ with horizontal
$\tan \theta =\frac{8}{15}$
If $R$ is the normal reaction between the plane and any of the block.
$R=170g\cos \theta =170\times 10\times \left(\frac{15}{17}\right)=1500$
Force of friction on $A=1500\times 0.2=300\text{N}$
Force of friction on $B=1500\times 0.4=600\text{N}$
Considering the two blocks as a system, the net force parallel to the plane
$=2\times 170g\sin \theta -300-600$
$=1600-900=700\text{N}$
$\therefore$ Acceleration $=\frac{700}{340}=\frac{35}{17}$
Considering the motion of $A$ alone,
$170g\sin \theta -300-P=\frac{35}{17}\times 170$
(where $P$ is pull on the bar due to $A$)
$\therefore P=500-350=150\text{N}$
Hence, the correct answer is option (a).