At what value of $m_{1}$ will 8 kg mass be at rest.

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Solution

The FBDs of the different masses and pulley are as shown

Since, $8 k g$ block will be at rest,
$\Rightarrow 8 g = 2 T \Rightarrow T = 40 N$
Let the acceleration of $m_{1}$ and $5 k g$ blocks be $a$
$∴$ from the FBDs we get
$5 g - T = 5 \times a \Rightarrow a = \frac{50 - 40}{5} = 2 m / s^{2}$
Now, we have
$T - m_{1} g = m_{1} \times 2 \Rightarrow m_{1} = \frac{40}{10 + 2}$
$\Rightarrow m_{1} = \frac{10}{3} kg$

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At what value of m1 will 8 kg mass be at rest.

The FBDs of the different masses and pulley are as shown Since, 8 kg block will be at rest, ⇒ 8g=2T⇒ T=40 N Let the acceleration of m1 and 5 kg blocks be a ∴ from the FBDs we get 5g−T=5×a⇒ a=50−405=2 m/s2 Now, we have T−m1g=m1×2⇒ m1=4010+2 ⇒ m1=103 kg

At what value of $m_{1}$ will 8 kg mass be at rest.