Question

For what value of $m_1$ will $8kg$ mass be at rest?

• A. $25/3kg$
• B. $5/3kg$
• C. $10/3kg$
• D. $20/3kg$

Answer 1

The correct option is C $10/3kg$

Since $8kg$ is at rest, tension in upper string is $T=8g=80N$.
Thus, Tension in bottom strings will be $40N$ ($T_1+T_1=T=80$),

Thus $T_1=40$
Now, using free body diagram for $5kg$:

$5a=50-T=50-40=10$

Therefore, $a=2$
Using FBD of mass $m$:

$ma=T-mg$,

$⟹2m=40-10m$

$⟹m=40/12=10/3$