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Block B has a mass m and is released from rest when it is on top of wedge A, which has mass, Neglect friction 3 m. Determine the tension in cord CD while B is sliding down A. Neglect friction.

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$\frac{m g}{2} \sin 2 θ$

$\frac{m g}{2} \sin 4 θ$

$\frac{m g}{2} \sin 3 θ$

$\frac{m g}{2} \sin 6 θ$

answer is A.

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Detailed Solution

Normal reaction between A and B would be $N = m g \cos θ$ . Its horizontal component is $N \sin θ$ . Therefore, tension in cord CD is equal to this horizontal component.

Hence, $T = N \sin θ = (m g \cos θ) (\sin θ)$

$= \frac{m g}{2} \sin 2 θ$

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