# Homework Help: Tension between two accelerating forces.

1. Oct 24, 2008

### Notyou

1. The problem statement, all variables and given/known data
A 6000kg helicopter accelerates upward at .6m/s^2 while lifting a 1500kg car. What is the tension in the cable (ignore the cable mass) that connects the car to the helicopter?

2. Relevant equations
F=MA

3. The attempt at a solution
Alright... so here's what I've been trying to do. The acceleration that is the acceleration due to gravity plus .6m/s^2 which gives an acceleration of -9.2. From there I would multiply the acceleration by mass to get the force of the helicopter and the car. So for the helicopter the force I get is 55200N and for the car it is 13000N. I add those forces together and get... the wrong answer. The answer is supposed to be 15600N and I can't figure out how.

I have a test coming up and I can't seem to do any tension problems to save my life; any help at all would be greatly appreciated.

2. Oct 24, 2008

### Staff: Mentor

Do it step by step. What forces (in what directions) act on the car? Apply F = ma.

3. Oct 24, 2008

### Redbelly98

Staff Emeritus
The acceleration is +0.6m/s^2, as given in the problem statement. That can be used to find the net force on either the car or helicopter.

I suggest looking at the car and drawing a free body diagram for it. What would the values of all the forces be, to give a net force that causes a +0.6m/s^2 acceleration on the car?

edit: Hello Doc, you beat me by 2 minutes on this one.

4. Oct 24, 2008

### Notyou

Alright. The forces working against the car in the negative direction would be the weight of the car x acceleration due to gravity. The forces working on the car in the positive direction would be the force of the helicopter x gravity. That seems to only be the case if the whole scenario were stationary... so I really don't know how to factor in the acceleration of the helicopter.

Edit: Still stuck. I have a force of 58800N acting on the car from the positive y direction and a force of 14700N acting on the car in the negative y direction. That's without taking the acceleration of the helicopter into consideration. If I take the acceleration of the helicopter into consideration I have no clue what to do.

Last edited: Oct 24, 2008
5. Oct 24, 2008

### Staff: Mentor

Edit: Oops... that's not right. The force is the weight of the car, which is its mass x the acceleration due to gravity. That's what I assumed you meant, but that's not what you said. Careful!
When analyzing the car, we care about forces on the car, not the helicopter. What's pulling up on the car? Hint: It's what you're trying to solve for!
The acceleration of the helicopter is only relevant when you analyze forces on the helicopter. But we're analyzing forces on the car.

Last edited: Oct 24, 2008
6. Oct 24, 2008

### Redbelly98

Staff Emeritus
No, the force in the negative direction is the weight of the car, period. "weight of the car x acceleration due to gravity" is not even a force, so that statement makes no sense.

[EDIT] unless you meant "mass of the car x acceleration due to gravity", which is the weight, and is correct.

As for the positive direction: what is in actual, physical contact with the car?

7. Oct 24, 2008

### Notyou

The rope is pulling on the car in the positive direction... and that force would be equal to the force of the weight due to Newton's third law correct?

8. Oct 24, 2008

### Staff: Mentor

Right. That's the tension in the rope, which is what you're trying to find.
Nope! The 3rd law pair to "the rope pulling on the car" tension force is "the car pulling down on the rope" with an equal force. Since weight is "the earth's gravitational force on the car", the third law pair is "the car's gravitational force on the earth".

Just call the tension "T" and set up Newton's 2nd law. What's the sum of those two forces on the car?

9. Oct 24, 2008

### Notyou

I got it now. The weight of the helicopter does not matter in this question. The acceleration of the car is 9.8 plus .6 acting in the positive direction. So... force would be the mass of the car times 10.4. Thank you so much!

10. Oct 24, 2008

### Redbelly98

Staff Emeritus
I repeat:

11. Oct 26, 2008

### Staff: Mentor

This is true.

As Redbelly98 emphasizes, the acceleration is given as 0.6 m/s² upward. Here's how to properly get the answer:

What forces act on the car? (1) Tension, acting up (+T); (2) weight, acting down (-mg).

The net force on the car: T - mg

The acceleration of the car: +0.6 m/s²

Apply Newton's 2nd law: T - mg = ma

Thus: T = ma + mg