Accelerating a Mass up an incline by pushing said Incline

[splitinferno]

Homework Statement

The coefficient of kinetic friction between the inclined face of a wedge of mass M and a block of mass m sitting on its inclined face is μk (see figure below). The vertical face of the wedge is pushed with a horizontal force such that m slides up the inclined surface with an acceleration a with respect to the inclined surface. Derive an expression for the magnitude of the force. (Assume the angle of inclination is some angle θ. Use any variable or symbol stated above along with the following as necessary: g. Assume SI units.)
Diagram: http://i.imgur.com/OLuNbv7.png


2. Homework Equations (my thought process; equations are further down the post)

Since m is accelerating up the incline, the force of friction on m is down the incline. The normal would be perpendicular to the incline and the gravity would be straight down.

Also, since the m is accelerating up the incline at an angle of θ with an acceleration of a, we can say that the horizontal acceleration of the m (with respect to the ground) would be ax - acosθ and the vertical acceleration of m would be asinθ, where ax is the acceleration of the system. Using this, I just added the horizontal components and set them equal to m(ax - acosθ). I also added up the vertical components and set them equal to m(asinθ).

But the answer I'm getting is wrong... apparently. Any thoughts?

The Attempt at a Solution

My most recent attempt at the problem:Attempt #... I have no clue.

 

[NihalSh]

ax is the acceleration of the system.

you are very close, but ax is acceleration of the Wedge. also draw and balance forces acting on the wedge.

You should get three useful equations, block (horizontal and vertical) and wedge (horizontal acceleration only).

that should help.

 

[splitinferno]

you are very close, but ax is acceleration of the Wedge. also draw and balance forces acting on the wedge.

You should get three useful equations, block (horizontal and vertical) and wedge (horizontal acceleration only).

that should help.

The ax thing was my mistake, but I don't think it should change the equations I've set up already (could be wrong though). Also, I'm sorry to be asking this but did you notice any blatant errors in my setup (aside from the balancing of the forces on the wedge/incline)? And I'll try balancing the forces acting on the wedge as well. From what I understand, there should be the following forces:

Force of friction acting up the incline (thanks to Newton's third law)
A component of the force of gravity of m on the incline.
Force of gravity of the wedge/incline itself.
Normal force acting on the wedge/incline...

That seems to be all I could think of at this hour. Any thoughts?

 

[NihalSh]

The ax thing was my mistake, but I don't think it should change the equations I've set up already (could be wrong though). Also, I'm sorry to be asking this but did you notice any blatant errors in my setup (aside from the balancing of the forces on the wedge/incline)?

ax thing was not a mistake, you have to take that into account. ax is the acceleration of wedge. Both the equations are perfect, no adjustment is needed. But remember that you have three unknown quantity (ax, N, a), so to solve it you need three equations. You are just missing one more equation.

Force of friction acting up the incline (thanks to Newton's third law)
A component of the force of gravity of m on the incline.
Force of gravity of the wedge/incline itself.
Normal force acting on the wedge/incline

A component of the force of gravity of m on the incline. no, this is not correct. you can only assume that when the wedge and the block are stationary but in this situation neither of them are stationary. the Newton's third law pair of normal force acting on mass m. you will have to figure out the angle though, from simple triangle properties.

and of course the force that you are trying to calculate

that is all.

 

[Nickie]

I've been trying for the past hour and a half to no avail.

First, I drew a free body diagram of the block m on the inclined surface. From the problem statement, we know that there is a normal force N and a force of friction Ff acting on the block. The force of gravity mg is also acting on the block, but since the block is not sliding down the incline, we can ignore the component of mg parallel to the surface.

Next, I broke down the forces into their x and y components. In the x direction, we have a horizontal force pushing the inclined surface, which we will call F. We also have a horizontal component of the force of friction, Ffx, acting in the opposite direction. In the y direction, we have the normal force N and the vertical component of the force of friction, Ffy, acting in the opposite direction of the force of gravity mg.

Using Newton's second law, we can set up the following equations:

ΣF x = max - Ffx = max - μkN = max - μkmgsinθ = m(ax - acosθ)
ΣF y = may - N - Ffy = may - mgcosθ - μkmgcosθ = may - mg(1 + μk)cosθ = masinθ

Solving for F in the first equation, we get:

F = m(ax - acosθ) + μkmgsinθ

Substituting this into the second equation, we get:

may - mg(1 + μk)cosθ = masinθ - m(ax - acosθ) - μkmgsinθ

Simplifying, we get:

ma = m(asinθ + axcosθ) + mg(μkcosθ - cosθ)

Finally, we can solve for F by substituting in the values for ax and asinθ:

F = m(a + acosθ) + mg(μkcosθ - cosθ)

This is the final expression for the magnitude of the force needed to accelerate the mass m up the incline by pushing the inclined surface.