Homework Help: Pondering Newton's Laws

1. Oct 4, 2006

Dorothy Weglend

Hi,

I have a non-homework question, I hope that is ok.

Sometimes, when I solve force problems using Newton's laws, I can easily solve the problem, but I am confused about exactly what law is involved.

For example, I was helping someone solve a pulley problem today. This is problem 34 in Chapter 5 of Serway and Jewett, diagram p5.34. Otherwise, I have attached a (badly drawn) picture, which also includes my freebody diagrams. I want to stress that I have no problem solving this type of problem, but I just feel my understanding is not deep enough.

This is a real beginner problem, with no frictional forces, massless ropes and pulleys and all that.

There is no problem at all with m2. That is just a straightforward application of the second law. Sum of y forces = m2*g - T2 = m2*a2.

The problem arises with m1 and the first pulley. The equations here are not difficult. Obviously, for m1, we have m1*a2 = T1. The pulley is also easy, with 2T1 = T2. And from there the solution is not difficult. When I asked my teacher this question, he showed me the answer in the teachers manual, and this is also how it was solved there.

But I am uncomfortable with this. If the second law truly applies here, then it seems to me the correct equation should be:

m1*g - T1 = m1*a1

Which also reflects the physical reality, I think. If we are on the moon, complete vacuum, no air resistance, we would still have to over come inertia (m1*g, right?) to get a body moving. It certainly seems wrong to me that T1=m1*a1, this should be a net force of zero, and result in equilibrium, and yet the system is accelerating, so there should be some force.

I thought that perhaps this is actually a third law issue, equal and opposite forces. But I have also always thought that the third law applies only to contact forces. "A pin pushes against the finger, the finger pushes against the pin.." That sort of thing. And, of course, the situation with mass 1 is always described as a 2nd law case.

The pulley issue is sort of the same. I can get past this by thinking of it as an internal force. But again, when it is described as a 2nd law application, I balk.

Sorry this is so long, but I have long wondered about it, but didn't want to appear stupid. Too late now, right? :-)

I have another confusion about inertial frames, but I think I have bared my soul enough for one night. I hope someone can clarify this for me, and I deeply appreciate all efforts.

Thanks a lot!
Dorothy

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2. Oct 4, 2006

George Jones

Staff Emeritus
Remember that force and acceleration are vectors, so that Newton's second law is a vector equation. This vector equation can be broken down component equations.

What two forces are missing from your free-body diagram for m1?

Once you put in these forces, you can write Newton's second law as two equations, one for the vertical direction and one for the horizontal direction.

3. Oct 4, 2006

Dorothy Weglend

Hi George, thanks. I left out the m1g vertically down and the normal force because it is a frictionless surface. So is this the answer to my question? m1*a1 = T1 because there is no resistance (either friction or air) to the motion of m1?

I find this runs against my intuition, but I can see that it makes sense. So if I wanted to accelerate a 2kg mass 2 m/s^2, it would take a 4 N 'push' (or pull) on the eart or the moon? (Assuming a frictionless, horizontal surface).

Is this right?

Thanks,
Dorothy