Calculating the Force Required for a Moving Wedge and Block System

[tryingtolearn1]

Homework Statement

Suppose a wedge is moving with a horizontal force F applied to a wedge of mass $M$ and incline angle $\theta$, such that a block of mass $m$ on the wedge remains in fixed position i.e. it doesn’t slide up or down the incline. Find an expression for #F# (assume no friction or air resistance).

Relevant Equations

$F=ma$

The entire mass of the wedge is $(M+m)$ therefore $F=(M+m)a$. The forces acting on the small mass are its downward weight $mg$ and the normal force with the contact of the wedge therefore I got that $N=mg\cos\theta$. Similarly the horizontal component is $N=ma\sin\theta$ therefore equating both sides and solving for $a$ we have that the force is $$F=(M+m)g\cot\theta$$ however the answer is $$F=(M+m)g\tan\theta$$ not sure what I did wrong?

 

[TSny]

The error is with the following

$N=mg\cos\theta$.

Try to find the mistake in reasoning that led you to this equation.

 

[tryingtolearn1]

The error is with the following

Try to find the mistake in reasoning that led you to this equation.

Hm, but the normal of the small mass is $N$ therefore the downward force of the normal is $mg\cos\theta$.

 

[Lnewqban]

Could you show us a diagram?

 

[TSny]

Hm, but the normal of the small mass is $N$ therefore the downward force of the normal is $mg\cos\theta$.

I'm not sure what you mean by the "downward force of the normal". The normal force acting on the small block is perpendicular to the sloping surface of the wedge. So, the normal force on the small block has a horizontal component and a vertical component. The vertical component is upward.

 

[tryingtolearn1]

I'm not sure what you mean by the "downward force of the normal". The normal force acting on the small block is perpendicular to the sloping surface of the wedge. So, the normal force on the small block has a horizontal component and a vertical component. The vertical component is upward.

I see my mistake now, ty