# Homework Help: Weights on a string problem - is this right?

1. May 9, 2004

### redshift

Weights on a string problem -- is this right??

Hello all,
I'd appreciate it if anyone can check whether I have the right approach to this problem:

"Two weights, A and B, are connected by a string. Weight A weighs 0.7 kg, and B weights 0.8 kg. If A is raised vertically upward with a force of 30N, (1) what is the acceleration of weights A and B? (2) what is the tension of the string?"

First of all, the downward force of weight A is (0.7kg)(9.8m/s2) = 6.86F, and that of B is (0.8kg)(9.8m/s2) = 7.84N. Therefore, I guess the net force would just be 30N - 6.86F - 7.84N = 15.3N.
Since the 2 weights are connected by a string, I assume they have the same acceleration. If so, using F = ma, the accerlation of both would be a = 15.3N/1.5kg = 10.2m/s2.

Since tension, T, =ma, this should be 15.3F too. Correct?

2. May 9, 2004

### arildno

Frankly, I don't understand the phrasing given in the problem text:
"Two weights, A and B, are connected by a string"
HOW??
If for example, B is suspended from the ceiling, with A hanging down in a string connected to B, raising A using a force of 30 N will loosen the string, i.e. the tension in the string will be zero.

My point with this example, is that there is insufficient information in the problem text to solve it.

3. May 9, 2004

### Staff: Mentor

Makes sense.
To find the tension in the string, consider one of the weights separately. Taking weight B for example: What are the forces on B? Find the net force and apply Newton's 2nd law (you know the acceleration).

4. May 9, 2004

### Staff: Mentor

I agree with arildno that you didn't state the problem clearly enough. But, taking the most obvious interpretation (which could be wrong!) I assumed that A and B are in a vertical orientation (A on top) and are being pulling up by a force on A. The string is taut.

Did I guess correctly?

5. May 9, 2004

### redshift

Let me clarify the positions of the weights.
The picture shows A and B vertically arranged -- A being above B -- with one end of the string above A and the other in B. (That is, the string passes through A and terminates in B.)

6. May 9, 2004

### arildno

All right, I was rather nit-picking here..
(The point is, I hate textbook writers who mistakenly assume that understanding of a topic can be measured by a student's ability to construct a meaningful problem out of a meaningless text)

7. May 9, 2004

### Staff: Mentor

Right. Pretty much as I guessed. Remember there are two string tensions to account for: the tension in the string above A and the tension in the string between A and B. The first tension is given: that's the applied force. To solve for the other tension, follow my earlier post.

8. May 9, 2004

### redshift

I think I get it. Since the upward accleration of weight B is 10.2m/s2, and the downward acceleration due to gravity is 9.8m/s2, the difference (0.4m/s2) multiplied by its mass (0.8kg) would be the tension (0.32N) acting on the connecting string. Somehow, this doesn't seem right. That is, if 30N is being applied to the upper string, I would think more force as tension would be transferred to the connecting string.

Last edited: May 9, 2004
9. May 9, 2004

### arildno

You've got the minus sign wrong here:

T-Mg=Ma->T=M*(a+g) (M is B's mass, a its acceleration, g the acceleration of gravity, T the tension)

10. May 9, 2004

### redshift

Thank you. That makes much more sense.

11. May 9, 2004

### UrbanXrisis

So would the tension in Mass A be T=M*(a+g) (M being the mass of both masses?)

12. May 9, 2004

### arildno

Not quite sure what is meant by "tension in mass A", UrbanXrisis, but:
The tension force from the string connecting A and B on A is: -Mb(a+g) (Mb mass of B).
The gravity force on A is Ma*g (Ma mass of A)
Adding together, with M=Ma+Mb, we have:
F=M(a+g), where F is the tension force acting on A (i.e, from the string we are pulling A with, given as 30 N)

I assume this is what you meant by "tension in mass A"?

(Obviously, this must be the correct result, since by considering A, B, interconnecting string as a single system, the only external force is F)

Last edited: May 9, 2004
13. May 9, 2004

### UrbanXrisis

yes, that's correct. But what is M when you wrote

"F=M(a+g), where F is the tension force acting on A (i.e, from the string we are pulling A with, given as 30 N)"

is M the mass of both A and B?

14. May 9, 2004

### arildno

15. May 9, 2004

### UrbanXrisis

Going back to what I posted first...I wrote

"So would the tension in Mass A be T=M*(a+g) (M being the mass of both masses?)"

then that is correct?

16. May 9, 2004

### arildno

Assuming you with the somewhat ambigouous (I hate spelling that word!) expression:
"Tension in mass A" means my F, then your'e correct.

17. May 9, 2004

### Staff: Mentor

arildno explained where you went wrong with your signs, but let me just add one comment: Don't think in terms of separate accelerations acting on the mass. Instead think in terms of separate forces. The net force will create the acceleration.
Think of it this way. The string in the middle only has to accelerate mass B, but the top string must exert enough force to accelerate both masses. Even with zero acceleration, wouldn't you expect the top string to have to pull harder just to balance the weight of both masses?

But the best thing to do, as always, is draw a diagram carefully identifying the forces on each mass. Then solve it step by step.