# Homework Help: Newton's third law question

1. Oct 25, 2012

### Syrus

1. The problem statement, all variables and given/known data

Suppose you have a cantaloupe (C) on a table (T). Let FTC be the force on the table by the cantaloupe, and let FCT be the force on the cantaloupe by the table. Do the magnitudes of these forces increase, decrease, or stay the same if they are placed in an elevator which begins to accelerate upwards?

2. Relevant equations

3. The attempt at a solution

I take the upward direction as positive. Please see the attached photo (my attempt). My solution shows that the magnitude of FCT increases, while the magnitude of FTC decreases. The solution, however, claims they both increase. Where have I gone astray?

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2. Oct 25, 2012

### Staff: Mentor

What free body did you use to do the calculations for the table? Make a list of the forces acting on this free body?

3. Oct 25, 2012

### bossman27

Just using your knowledge that F = ma, can you explain why they must both be the same? What should happen, qualitatively, if the force of C on T increases, and the force of T on C decreases? It always helps to draw a force diagram.

4. Oct 25, 2012

### Syrus

Chestermiller:

Well, there's the (positive) force of the table on the cantaloupe and the (negative) force of gravity which acts on the cantaloupe.

For the table, there's the (positive) normal force on the table, the (negative) force of the cantaloupe on the table, and the (negative) force of gravity on the table.

bossman, I understand the qualitative argument- I am struggling as to why my calculations are disagreeing with my intuition.

5. Oct 25, 2012

### SammyS

Staff Emeritus
Where have you defined FN ?

6. Oct 25, 2012

### Syrus

FN is the normal force acting on the table. That is, the positive force which the ground applies to the table.

7. Oct 25, 2012

### SammyS

Staff Emeritus
It's FN which must be larger, if the table & cantaloupe are to have positive acceleration, rather than zero acceleration.

8. Oct 25, 2012

### Syrus

Is there an explicit way to show this? Again, I understand the hand-waving; can these thoughts be expressed quantitatively?

Last edited: Oct 25, 2012
9. Oct 26, 2012

### SammyS

Staff Emeritus
Yes ... and Yes.

In the case of upward acceleration, FN is the force that the elevator exerts on the table.

Treating the table&cantaloupe as one object with mass mC + MT and recognizing that $\displaystyle \textbf{F}_{(m_C+m_T)g}=\textbf{F}_{m_C g}+ \textbf{F}_{m_T g}\,,$ you can show explicitly that FCT = FTC .

10. Oct 26, 2012

### Syrus

First, I'm interested in showing that both of the (initially equal and opposite) forces increase WITHOUT assuming they are equal and opposite while accelerated. In my picture above, this isn't demonstrated, since the mass of the table times its resultant acceleration is subtracted from the quantity present in the former case (without elevator acceleration), hence decreasing the magnitude of the force on the table from the cantalope (which, allegedly, should increase instead).

Last edited: Oct 26, 2012
11. Oct 26, 2012

### Syrus

Ah, got it. Thank you, SammyS

Last edited: Oct 26, 2012
12. Oct 26, 2012

### bossman27

Since we're assuming that cantaloupe doesn't break through the table, I'd say it's fine to assume that the table applies a normal force to counterbalance the increased downward force applied by the cantaloupe.

I think maybe you need to also think about the normal force from the floor of the elevator on the table. When we're at rest, obviously $F_{Nt} - F_{Gt} = 0$ (talking about F(Gt) just in terms of magnitude, of course)

When the elevator is moving, we need to have $F_{Nt} - F_{Gt} = m_{t}a$

Obviously since the cantaloupe and table aren't accelerating relative to each other, this relationship applies to both the system and to the cantaloupe individually. I think what you may have been neglecting was the force of the floor on the table/system.

Edit: Woops, just saw that you already got it.