A 45 degree wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge. Find the block’s acceleration. Gravity is directed down.
2. Relevant equations
Equations (p is for parallel and p2 is for perpendicular, eg. x and y directions):
3. The attempt at a solution
It’s embarrassing how this problem seems difficult for me; I’m probably making some fundamental error somewhere.
Newton’s 3rd Law: [itex]N=N'[/itex]
Because the angle is 45 degrees, the tangent of theta is just 1. Solving for the x acceleration, we get:
Now, according to the constraint equation, the acceleration in the y direction is equal to the tangent of theta times the x acceleration of the block minus the x acceleration of the wedge itself. Following the constraint equation, I get:
(tangent of 45 is 1). There is no way to solve for the y acceleration here. If, back when I solved for the normal force, I included the tangent of theta, I would’ve been able to solve for the acceleration but I would have to divide by zero, which means there is no y acceleration. The hint, however, states otherwise: If A = 3g then the y acceleration is g. Where am I going wrong? I’m going to try and solve for the y acceleration instead of the x acceleration first and see where that takes me.