#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# In the above figure, the wedge is acted on by a  smooth constant horizontal force F. The wedge is moving on a smooth horizontal surface. A ball of mass m is at rest relative to the wedge.  The wedge is not allow to move assuming no friction b/w the wedge and the ball. Show that the ratio of the force exerted on m by the wede when F is acting and F is withdrawn is F / [(M+m) g sin theta cos theta].

12 years ago

Hi,

when force f is not acting

normal by wedge on ball , N1 = mg cos Ø .....................(1)

when f is applied,

total acceleration of wedge & ball,          a   =    F/ (M+m)

we have to make the fbd of ball with respect to the vedge

in this there will be three forces mg,  N &  ma (  becoz we are seeing w.r.t. wedge )

ball is at rest with respect to wedge,

so,   horizontal forces,  N2 sinØ =  ma

N2 =  ma / sinØ

N2 = mF/(M+m) sinØ............(2)

(2)/ (1) .....

N2 /N1 = F/(M+m)g sin Ø cos Ø