Question

The value of mass $m$ for which the $100kg$ block remains in static equillibrium is/are: (take $g=10{m/s}^2$)

• A. $35kg$
• B. $37kg$
• C. $83kg$
• D. $85kg$

Answer 1

Answer: (B), (C)

The correct options are
B $37kg$
C $83kg$

$m_{min}$ would just stop the $100kg$ block from sliding down the incline. So, direction of friction is upwards along the incline.$(100gsin{37}^{\circ }=600,100gcos{37}^{\circ }=800)$

from FBD , for balancing forces
$N_1=800N$
$T=600-f_l$
$T=600-0.3\left(800\right)$
$T=360N$

$T=m_{min}g=360$
$\Rightarrow m_{min}=36kg$

$m_{max}$ would just make the block move up the incline. So, direction of friction is downwards along the incline

from FBD , for balancing forces
$N_1=800N$
$T=600+f_l$
$T=600+800\left(0.3\right)$
$T=840N$
$T=m_{max}g=840$
$\Rightarrow m_{max}=84kg$

For $36kg\le m\le 84kg$ the $100kg$ block is at rest.