# Choosing signs in free body diagrams of subsystems

In an example problem from a book I'm using (David Morin's Introduction to Classical Mechanics with Problems and Solutions), there is a system with a person and a pulley on top of a platform. The pulley is attached to the platform with a rod. The person is standing next to the pulley holding one end of the rope, and the other end of the rope goes up to a ceiling. An equation is given for the subsystem of the platform. The mass of the platform is M, the normal force from the pulley is $$N_\mu$$, and the force from the pulley's rod is f.

The book says the total force on the platform is equal to

$$F=Ma=-Mg+f+N_p$$

I don't understand this. First, am I correct in assuming that the "upward force" from the rod f is simply the normal force from the rod? Second, why would we count the normal forces from these objects instead of their weight on the platform? To me it seems that the normal forces from these objects are being exerted by the platform, not on the platform.

A second equation is given for the subsystem of the person. The mass of the person is m, the normal force from the platform is N, the mass of the platform is M, and the tension in the rope is T, with the equation given as

$$F=ma=N-T-mg$$

The third equation is for the subsystem of the pulley, with $$\mu$$ equal to the mass of the pulley, $$f$$ equal to the force from the rod, and $$T$$ equal to the tension in the string:

$$F=\mu a=2T-f-\mu g$$

Again, I am not sure why the force from the rod is negative here. Is that simply the weight of the rod? Why are we considering the weight of the rod on the pulley, while we considered the normal force from the rod on the platform?

• You really ought to post the text of the problem and its solution as stated in the book. If you are querying what a textbook says, it is essential to have the relevant text reproduced here, including its context. Feb 8, 2020 at 22:55

You are correct in being worried about the given solution in that if you added your three equations you should get the equation of motion for a system comprising the pulley, the person, the platform and part of the left hand rope which is $$(M+m+\mu)a = T- (M+m+\mu)g$$ and this is not so.

The platform equation of motion should be $$F=Ma=-Mg+f-N$$ where $$N$$ is one of the Newton third law pairs of forces: person exerts on platform and platform exerts on person; and $$f$$ is one of the Newton third law pairs of forces: pulley exerts on platform and platform exerts on pulley.

As to the directions of these forces think this way for the forces $$f$$ as an example.
The platform is accelerating upwards (the direction chosen to be positive) so which of the forces acting on the platform are in an upward direction?
Certainly not $$N$$ the force on the platform due to the person and certainly not $$Mg$$ the gravitational attraction of the Earth on the platform as both of those must be negative assuming the person is just wearing ordinary shoes and cannot pull the platform upwards with the shoes.
You are left with the force on the platform due to the pulley, $$f$$, and that must be upwards ie will be $$+f$$ in the equation of motion for the platform so the pulley is pulling the platform upwards.
So the force on the pulley due to the platform must be $$-f$$ in the equation of motion for the pulley ie the platform is pulling the pulley downwards as much as the pulley is pulling the platform upwards.

1. The force $$f$$ is the tension (pull) in the rod which is attached to the platform. Calling it a normal force is misleading because normal contact forces are usually pushes not pulls. But the rod could be attached in such a way that it pushes up on something attached to the platform.

2. Weight is the force of gravity acting on a body. The weights of the pulley and man act on the pulley and man respectively. They do not act on the platform directly.

The essence of a free body diagram is that you consider only the forces which are acting directly on that body. So for the FBD of the platform you consider only the forces which act directly on the platform. For the FBD of the man you consider his weight and the contact forces with the platform and the rope.

Ideally you should draw separate FBDs for each body or sub-system - one for each. You only need to consider external forces acting on the sub-system. You can ignore internal forces acting between different parts of the same sub-system - such as the forces between the man's feet and his shoes or between the rod and the pulley (if they are considered together as one sub-system.)

1. Yes the sign of the contact force from the man on the platform $$N_p$$ is confusing. As written in the first equation it is upwards, like $$f$$. Well it is possible that the man's shoes are clamped to the platform and he does pull up on it. But it is more likely that his shoes are not clamped so he can only exert a downward push. It would be sensible to define $$N_p$$ as positive downwards. However, it usually makes no difference in what direction we define forces. If the man does push down on the platform then $$N_p$$ will later turn out to be negative.

2. $$N$$ (the force which the platform exerts on the man) and $$N_p$$ are paired forces. These should be equal and opposite. If $$N_p$$ acts upwards on the platform then $$N$$ should act downwards on the man. There is no good reason to give them different labels. (Does Morin really do that?)

3. As in 4, the upwards force $$f$$ which the rod exerts on the platform equals the downwards force $$f$$ which the platform exerts on the rod. These are paired forces which are equal and opposite and act on different bodies. You could draw a separate FBD for the rod (in which case the rod pulls up on the platform and down on the pulley) but it works out the same if you include the rod as part of the pulley sub-system (in which case the rod-pulley system pulls up on the platform and the platform pulls down on the rod-pulley system).

Also confusing for any practical-minded person is that, as drawn, the platform is obviously going to turn. This contributes to the difficulty of learning physics. If the man and the COM of the platform are on opposite sides of the rod it could work, but the moments of forces would have to be finely balanced, and the man would have to pull with a constant force, which is difficult to arrange as he switches hands on the rope.