Free moving inclined plane

Homework Statement

Consider a small box mass m initially at the bottom of an inclined plane mass M, length L with angle of inclination of $$\theta$$. The surface between the plane and the block and the plane and the horizontal are both frictionless. A force F is applied horizontally to the small box. Need to find the time when the small box reached the top of the inclined plane.

The Attempt at a Solution

I have:
F - Nsin$$\theta$$ = mam,x
Ncos$$\theta$$ = mam,y
Nsin$$\theta$$ = MaM,x

for the relative motion of the small box to the inclined plane:
tan$$\theta$$ =$$\frac{am,y}{am,x - aM,x}$$

then i try to use the distance travelled in y direction, so
1/2 * am,y * t^2 = L sin $$\theta$$

I am not sure they are the correct equations.

the answer given is t = $$\sqrt{\frac{2L(1+(m/M)(sin\theta)^2)}{(F/m)cos\theta - g(1+m/M)sin\theta}}$$

Thanks for any help given :-)

Last edited:

well i have solved it. thanks for anyone who tried or trying to do this question.

cheers,
weesiang_loke