DISHANT
Last Activity: 5 Years ago
hope fully we are going to solve the first probem (D-5) as marked .
consider that the wedge moves along the horizontal in with the acceleration a1 and that of rod in the downward direction is a2 .
now we obviously see that the rod is constrained that it can’t get embedded inside the wedge .
therefore ,
the acceleration of rod along the surface of contact and the velocity of wedge along the suface of contact should
be equal .(we are here going to resolve the given accelerations of wedge and the rod )
- now after resolving the componenets we find that a2 cos0= a1sin0.That is a2=a1tan0. …........1
- equation 1 here is the constrained relation for a2.
- now as a2 is acceleration of rod w.r.t wedge ;but we are interested in finding the acceleration w.r.t ground .fo that use the pythagoras theoram .
- accelertion (net) of rod = sqrt[a1^2+a2^2]. in this put the value of a2 from eq.1.and the acceleration w.r.t ground will appear.