# System vs individual objects in the system when calculating forces?

I don't understand the calculation of forces acting on a system vs the forces acting on an individual object in that system.

For example, here are two trolleys (1 and 2), joined with an inextensible rigid towbar in red If trolley 1 starts exerting a force so that the whole system accelerates forwards...

1. The trolley exerts a forward force on the towbar
2. As the towbar pulls the second trolley along, the second trolley exerts a backwards force on the towbar So the whole system is accelerating forwards, and the towbar has this same acceleration, but at the same time, the forces on the towbar cancel because of the opposite forces exerted by the the two trolleys on it: How can an object accelerate within a system if the forces acting on that object by itself cancel?

The forces on the towbar are only exactly equal and opposite if we assume the towbar has no mass.

If $$m = 0$$, then from $$F = ma$$ we can have $$a \ne 0$$ and $$F = 0$$.

For a real towbar the mass is not zero, and the forces at each end are not equal and opposite, except when the towbar is not accelerating.

Another way to think about this, which avoids the fictitious "massless towbar," is to divide the system into just two objects and not three; "truck1 with the towbar" and "truck 2".

You write

1. The trolley exerts a forward force on the towbar
2. As the towbar pulls the second trolley along, the second trolley exerts a backwards force on the towbar


and that's correct, but the second force is slightly smaller than the first one. The towbar's acceleration "consumes" a little bit of the first force for its acceleration. The force that the first towbar exerts on the system of towbar and second trolley accelerates both bodies, and gets split between towbar and second trolley proportional to their resp. masses. Only the amount remaining gets to the second trolley.

How can an object accelerate within a system if the forces acting on that object by itself cancel?

You're absolutely correct, in that case you wouldn't get an acceleration.

One addition: we often neglect the masses of connecting bodies like the towbar, and take both forces to be equal. Then we effectively model the towbar as a zero-mass object accelerated by a zero force, resulting in zero-by-zero division when calculating the towbar's acceleration, which allows for any numerical result desired...

The two horizontal forces acting on the towbar will only cancel out if the towbar has zero mass - in a mechanics problem, it would be described as a "light" towbar.

Suppose the towbar has mas $$m$$ and each trolley has mass $$M$$.

The horizontal force on the left hand trolley is $$F - T_1$$ where $$F$$ is the driving force and $$T_1$$ is the tension on the left hand end of the tow bar. The horizontal force on the tow bar is $$T_1 - T_2$$ where $$T_2$$ is the tension at the right hand end of the tow bar. And finally the horizontal force on the right hand trolley is just $$T_2$$. Both of the trolleys and the towbar must have the same acceleration $$a$$, so we have three equations

$$F - T_1 = Ma \\ T_1 - T_2 = ma \\ T_2 = Ma$$

and three unknowns $$T_1, T_2, a$$. We can solve this system of equations to get:

$$\displaystyle a = \frac {F}{2M+m}$$ (we could have derived this immediately by treating the trolleys and towbar as a single object with mass $$2M+m$$)

$$\displaystyle T_1 = \frac{F(M+m)}{2M+m}$$

$$\displaystyle T_2 = \frac {FM}{2M+m}$$

and you can see that $$T_1 > T_2$$ unless $$m=0$$, when they are equal.