Question

A vertical rod A of mass m rests on the inclined surface of a wedge B of mass M?

A vertical rod A of mass m rests on the inclined surface of a wedge B of mass M and wedge angle `theta`.the wedge is held in position on a friction less horizontal surface.the rod A can move freely in guide G.what is the reaction of the rod on the wedge if the wedge is released?{assume all surfaces of contact to be friction less.}

Answer 1

See below.

Explanation:

Rod

`P_1 = (0,-m_1 g)`
`N_1 = n_1(-sin theta, cos theta)`
`H_1 = (h_1,0)`
`alpha_1 = (0,-a_1)`

`P_1+N_1+H_1 = m_1 alpha_1`

Wedge

`N_2=(0,n_2)`
`P_2 = (0,-m_2 g)`
`N_1 = n_1(-sin theta, cos theta)`
`alpha_2 = (a_2,0)`

`N_2+P_2-N_1 = m_2 alpha_2`

due to geometrical restrictions `a_1 = -lambda a_2`

Taking the vertical component to the rod and the horizontal component to the wedge

`{(m_1 g - n_1 cos theta=m_1a_1),(n_1 sin theta = m_2 a_2),(a_1 = -lambda a_2):}`

Solving for `n_1,a_1,a_2` we have

`{(a_1 =-(g m_1 lambda Sin theta)/(m_2 Cos theta + m_1 lambda Sin[theta]) ),(a_2 = (g m_1 Sin theta)/(m_2 Cos theta + m_1 lambda Sin theta)),(n_1 =(g m_1 m_2 Sec theta )/(m_2 + m_1 lambda Tan theta) ):}`

but `lambda = -a_1/a_2 = 1/tan theta` then

`n_1 = (g m_1 m_2 Sec theta )/(m_2 + m_1)`

Here `n_1` is the rod reaction magnitude, and it's orientation is normal to the slanted wedge surface.