Question

Blocks A and B in the figure 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 bar. Calculate the force developed in the bar.$(g=10m{s}^{-2})$

Answer 1

$\cos \theta =\frac{5}{17}$

$\sin \theta =\frac{8}{17}$

$f_A$ and $f_B$ be the frictional force action of block A and B respectively.

Acceleration of blocks is $a$

Then for block A;

$mg\sin \theta -f_A-T=ma$

$\Rightarrow 170\times 10\times \frac{8}{17}-{\mu }_{A}mg\cos \theta -T=ma$

$\Rightarrow 10\times 10\times 8-0.2\times 170\times 10\times \frac{15}{17}-T=170a$

$\Rightarrow 500-T=170a$ ..............(1)

Now, consider block B;

$T+mg\sin \theta -f_B=ma$

$\Rightarrow T+170\times 10\times \frac{8}{17}-{\mu }_{B}mg\cos \theta =ma$

$\Rightarrow T+800-0.4\times 170\times 10\times \frac{15}{17}=170a$

$\Rightarrow T+800-600=170a$

$\Rightarrow T+200=170a$ ..............(2)

From eq. (1) and (2);

$500-T=T+200$

$\Rightarrow 2T=300$

$\Rightarrow T=150N$