Question

In given arrangement, $10kg$ and $20kg$ blocks are kept at rest on two fixed inclined plane. All strings and pulleys are ideal. Value(s) of $m$ for which system remain in equilibrium is/are: $(g=10m/s^2)$

• A. $m=6kg$
• B. $m=3kg$
• C. $m=9kg$
• D. $m=22kg$

Answer 1

Answer: (C)

$m=9kg$

From the FBD of 10kg block, we have:
$10gsin\theta =T+10\mu gcos\theta$
$\Rightarrow T=20N$

Now if the mass m has a greater magnitude we see that the friction now acts in the downward direction and we get the maximum Tension force using the following relation
$10gsin\theta +10\mu gcos\theta =T$
$\Rightarrow T=100N$

Thus, for equilibrium condition we see that the range of tension force thus becomes:
$20\le T\le 100\left(i\right)$

Similarly, from the FBD of $20kg$ block, we have
$20gsin\theta =T+20\mu gcos\theta$
$\Rightarrow T=40N$

Similarly for the weight m much greater than the $20kg$ block we see that the tension force now acts in the downward direction giving us the equation as:
$20gsin\theta +20\mu gcos\theta =T$

Thus, for equilibrium we get the range of tension force as:
$40\le T\le 200\left(ii\right)$

From (i) and (ii), we have
$40\le T\le 100$
$40\le \frac{mg}{2}\le 100$
$\frac{80}{10}\le m\le \frac{200}{10}$
$8\le m\le 20$