# Find speed with tension, mass given. thanks

1. Feb 20, 2010

### MissPenguins

1. The problem statement, all variables and given/known data
009 (part 1 of 2) 10.0 points
A 33 kg child sits in a swing supported by two chains, each 3.7 m long. If the tension in each chain at the lowest point is 259 N, ﬁnd the child’s speed at the lowest point. (Neglect the mass of the seat.) Answer in unit m/s.
b. Find the force of the seat on the child at the
lowest point. Answer in unit N

2. Relevant equations
F = mv2/r

3. The attempt at a solution
Rearrange equation, and I got sqrt(Fr/m) = sqrt(259N * 3.7m)/33 kg) = 5.3888....
I got it wrong. What did I do wrong? Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 20, 2010

### dr_k

Good Evening Miss Penguins,

You're correct that this is a circular motion problem, but you neglected some of the forces. First you need to draw a picture, and determine the free body diagram for the isolated child.

The isolated child experiences a gravitational attraction, or weight, and a contact force from the seat, the normal force N. Here's a picture to get you started:

http://img109.imageshack.us/img109/6937/swingw.jpg [Broken]

Hint: The normal force N has a magnitude of 2 times the tension, since the seat is massless.

Last edited by a moderator: May 4, 2017
3. Feb 21, 2010

### Delphi51

I don't know if this is the same as already suggested, but my thought is that the forces add up to the centripetal force:
Fc = 2T - mg

4. Feb 21, 2010

### MissPenguins

so I used the wrong equation for it? Thanks.

Last edited by a moderator: May 4, 2017
5. Feb 21, 2010

### MissPenguins

I don't know, but for the first answer, they are asking for speed though, and the equation u suggested doesn't have speed.

6. Feb 21, 2010

### Delphi51

You will of course replace Fc with mv²/R as you did in the first go.
The difference is that mg is involved.

7. Feb 21, 2010

### MissPenguins

I rearranged the equation and ended up with v = sqrt [(2T-mg)r]/m = sqrt [(2(259 N) - (33kg)(9.8m/s2))(3.7)] / (33 kg) = 21.818787......
Does that look reasonable?

8. Feb 21, 2010

### dr_k

That m should be inside the square root, I believe.

9. Feb 21, 2010

### MissPenguins

Yeah it is inside. I submitted the answer, and it is wrong. :(

10. Feb 21, 2010

### dr_k

It doesn't look that way, according to the equation you wrote.

Can you prove that the magnitude of the Normal force is equal to twice the tension?

11. Feb 21, 2010

### MissPenguins

What do you mean by it doesn't look that way? lol.......well, I put it in the calculator that way. So v = sqrt of [((2T-mg)r) / m] = sqrt of [((2(259 N)-(33kg)(9.8m/s2)(3.7m) / 33 kg] = 21.8187878787 WRONG. :(

12. Feb 21, 2010

### MissPenguins

Alright, I was able to get a right thanks to a friend. I use ((259*2) - (33)(9.8)) = 194.6, and then v = sqrt of (Fr/m) => sqrt of ((194.6)(3.7)/33 = 4.671058539.

Then in part 2, it says find the force of the seat on the child at the lowest point. I have no clue how to do it. :( Please help, thanks.

13. Feb 21, 2010

### Delphi51

Two ways to look at part 2.
First, there is no mass other than the child so the tension forces are entirely applied to the child via the seat.
Second, apply "sum of Forces = ma" to the child. The only forces are gravity and the seat force.

14. Feb 22, 2010

### MissPenguins

Nvm, I got it, thanks a lot anyway. :)

Last edited: Feb 22, 2010