# How to place the axis so that I can calculate the center of mass for the two instances?

I've got this:

A wagon of mass $$M$$, initially at rest, can move horizontally along a frictionless track. When $$t = 0$$, a force $$F$$ is applied to the cart. During the acceleration of M by the force $$F$$, a small mass m slides along the wagon from the front to the rear. The coeﬃcient of kinetic friction between $$m$$ and $$M$$ is $$µ_k$$, and it is assumed that the acceleration of M is suﬃcient to cause sliding.

1. Write two equations of motion, one for m and one for M, and show that they can be combined to give the equation of motion of the mass center of the system of two bodies.

2. Find the displacement of M at the time when m has moved a distance l along the cart.

I don't know how to place the axis so that I can calculate the center of mass. Also, I found only one equation that includes both masses (based on Newton's second law) and I can't figure it out how to get another equation.

• This is a one-dimensional motion, you just have to find the center of mass along the horizontal direction. For the two equations of motion: think about which forces act on the two individual bodies, the wagon and the small mass. This will give two equations of motion. Commented Mar 4, 2019 at 13:56
• But how can I determine the acceleration of the center of the mass, so that I can use it in my equation of motion? Commented Mar 4, 2019 at 14:08

Since the mass $$m$$ is loosely attached to the mass $$M$$, through the time the mass $$M$$ accelerates and the mass $$m$$ slides backwards, the mass $$M$$ is independent of mass $$m$$. According to Newton's third law of motion, the same force $$F$$ act on the mass $$m$$ in the opposite direction. Therefore:

$$\vec F=ma_1=Ma_2$$

Through the time the mass $$m$$ moves from front to rear, the mass $$M$$ should be considered independent of mass $$m$$, so the displacement of the $$M$$ can be calculated with the formula below:

$$d=\frac {\vec F t^2}{M}$$

Now taking into account the friction, things change a little bit. The net force on each mass is different:

Friction force: $$\vec F_k=Nµ_k$$

Net force on mass $$M$$ and $$m$$ ($$\vec F_n$$): $$\vec F- \vec F_k$$

New $$M$$ displacement: $$d=\frac {\vec F_n t^2}{M}$$