A rod is clamped between two supports and is moving with acceleration a1 . The rod remains in contact with the wedge. Then acceleration of wedge a2 is
Answers
Answer:
g/(tanθ+M/m(cotθ)).
Explanation:
If we take that the wedge is moved by a small distance x and the point where the rod reaches is y then we get that y=xtanθ and of differentiating we have that the dy/dt = dx/dt(tanθ).
So, we get that the velocity of the rod is equal to the velocity of the wedge*tanθ.
Similarly, acceleration of the rod is equal to the acceleration of the wedge*tanθ. So, if we take the mass of rod as m and the wedge as M.
So, if the reaction of the rod by the wedge is N. Hence, by newtons second law of motion we have mg-Ncosθ=ma(of rod) and Nsinθ=Ma(of wedge). Since, acceleration of wedge is a(rod)/tanθ so, Nsinθ=M[a(rod)/tanθ].
Hence, acceleration of wedge = g/(tanθ+M/mcotθ).
Answer:
g/(tanθ+M/m(cotθ)).
Explanation:
If we take that the wedge is moved by a small distance x and the point where the rod reaches is y then we get that y=xtanθ and of differentiating we have that the dy/dt = dx/dt(tanθ).
So, we get that the velocity of the rod is equal to the velocity of the wedge*tanθ.
Similarly, acceleration of the rod is equal to the acceleration of the wedge*tanθ. So, if we take the mass of rod as m and the wedge as M.
So, if the reaction of the rod by the wedge is N. Hence, by newtons second law of motion we have mg-Ncosθ=ma(of rod) and Nsinθ=Ma(of wedge). Since, acceleration of wedge is a(rod)/tanθ so, Nsinθ=M[a(rod)/tanθ].
Hence, acceleration of wedge = g/(tanθ+M/mcotθ).