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Advanced Physics

Advanced Physics Question

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Question

Expert Solution

Step 14

Torque about the left end

$\begin{array}{rcl} T_{r} l - m_{s} g \frac{l}{2} - m_{p} g r_{p} - m_{e} g (l - r_{e}) & = & 0 \\ T_{r} & = & \frac{g (m_{s} (\frac{l}{2}) + m_{p} r_{p} + m_{e} (l - r_{e}))}{l} (1) \end{array}$

Torque about the right end

$\begin{array}{rcl} T_{l} l - m_{s} g \frac{l}{2} - m_{p} g (l - r_{p}) - m_{e} g r_{e} & = & 0 \\ T_{l} & = & \frac{g (m_{s} (\frac{l}{2}) + m_{p} (l - r_{p}) + m_{e} r_{e})}{l} (2) \end{array}$

Step 2

Tension in the left cable is twice that of the tension in the right cable

$\begin{array}{rcl} 2 T_{r} & = & T_{l} \\ \frac{2 g (m_{s} (\frac{l}{2}) + m_{p} r_{p} + m_{e} (l - r_{e}))}{l} & = & \frac{g (m_{s} (\frac{l}{2}) + m_{p} (l - r_{p}) + m_{e} r_{e})}{l} \\ m_{s} (\frac{l}{2}) + m_{p} (3 r_{p} - l) + m_{e} (2 l - 3 r_{e}) & = & 0 \\ m_{e} & = & \frac{- m_{s} (\frac{l}{2}) - m_{p} (3 r_{p} - l)}{(2 l - 3 r_{e})} \end{array}$

Step 3

Substitute

$\begin{array}{rcl} m_{e} & = & \frac{- (43 kg) (\frac{5.6 m}{2}) - (76.0 kg) (3 (1.0 m) - 5.6 m)}{2 (5.6 m) - 3 (1.6 m)} \\ = & 12.06 kg \\ \approx & 12.1 kg \end{array}$