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

Explanation

Normal reation between `A` and `B` would be `N = mg cos theta` ,Its horizontal component is `N sin theta`. Therefore , tension in cord `CD` is equal to this horizontal component.
`T = N sin theta = (mg cos theta) (sin theta)`
`= (mg)/(2) sin 2 theta`