Newton, freefall acceleration problem

[bengt2665]

The weight of the cart is 4 N. The weight of the hanging mass is 4 N.
Friction of the cart on the track is a constant force of 2 N magnitude.
The mass of the string, the mass of the pulley, and the friction of the pulley can be neglected.
What is the magnitude of the acceleration of the cart if g is the free-fall acceleration?
Picture attached.

The answer is to be given in terms of g.


Okay, here's what I have so far:

By drawing a free body diagram I manage to get these two equations for the forces acting on both the cart and the hanging mass:

I'm assuming positive acceleration to the right of the cart and down for the mass.

For the cart: T - Ff = ma
For the mass: T - mg = ma
(since they have the same weight, there is only one m)
We know that the weight of these two is 4N each so for both, 4 = ma (i'm predicting this might not be right...)

I'm having trouble applying their given information (4N weight for both masses) with the equations I'm coming up with, for example I don't know what this 4N corresponds to...
Is it 4N = m.a for the cart and 4N = mg for the hanging mass?

How do I apply the information they are giving me?

Thanks!

 

[rl.bhat]

For mass ma = mg - T.
For cart 4N is the normal reaction. In this problem it is not needed.
For the cart: T - Ff = ma
For the mass: mg- T = ma .Solve for T.
Hence find a by substituting m = 4/g

 

[bengt2665]

This doesn't really help me understand...
I end up having 2 equations:
1) T - Ff = ma
2) -T + mg = ma

Then I end up with

mg - Ff = 2ma (by combining 1 and 2)

and I'm stuck there.

 

[bengt2665]

Just figured it out...

we know that m1g=4 and m2g=4 => m1 = 4/g, m2 = 4/g

So in the end, after substituting in 1 and 2 we get

1) T1 - 2 = (4/g)a
2) -T1 + 4 = (4/g)a

By adding these 2 equations (to get rid of the T) we end up with
2 = 8a/g
which gives a = g/4.

 

[rl.bhat]

That is the answer.