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.}

enter image source here

1 Answer
Mar 1, 2018

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.