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Standard XII

Physics

3rd Equation of Motion

Two men with ...

Question

Two men with masses $m_{1} = 60$ kg and $m_{2} = 50$ kg are climbing on a rope passing over an ideal pulley. If accelerations of $m_{1}$ and $m_{2}$ with respect to the rope are $a_{1} = 4 m / s^{2}$ and $a_{2} = 2 m / s^{2}$ upwards, then tension in the rope is (in Newtons) .

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Solution

Equation from F.B.D
T - $m_{1} g = m_{1} (4 - Q_{R})$
T - $m_{2} g = m_{2} (2 + Q_{R})$
$Q_{R}$ is acceleration of rope.
Solving eqations we get $\Rightarrow$ $a_{R} = \frac{24}{11} m / s^{2}$ & T = 709.09 N

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Two men with masses m1=60 kg and m2=50 kg are climbing on a rope passing over an ideal pulley. If accelerations of m1 and m2 with respect to the rope are a1=4 m/s2 and a2=2 m/s2 upwards, then tension in the rope is (in Newtons) .

Equation from F.B.D T - m1g=m1(4−QR) T - m2g=m2(2+QR) QR is acceleration of rope. Solving eqations we get ⇒ aR=2411m/s2 & T = 709.09 N

Two men with masses $m_{1} = 60$ kg and $m_{2} = 50$ kg are climbing on a rope passing over an ideal pulley. If accelerations of $m_{1}$ and $m_{2}$ with respect to the rope are $a_{1} = 4 m / s^{2}$ and $a_{2} = 2 m / s^{2}$ upwards, then tension in the rope is (in Newtons) .

Equation from F.B.D
T - $m_{1} g = m_{1} (4 - Q_{R})$
T - $m_{2} g = m_{2} (2 + Q_{R})$
$Q_{R}$ is acceleration of rope.
Solving eqations we get $\Rightarrow$ $a_{R} = \frac{24}{11} m / s^{2}$ & T = 709.09 N