Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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 Show that the ratio of the force exerted on m by the wede when F is acting and F is withdrawn is F / [(M+m) g sin theta cos theta].
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !