# Tension of a pendulum and force components

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1. Dec 1, 2015

### **Mariam**

1. The problem statement, all variables and given/known data
What is the tension in a pendulum? Is T=mgcos(theta) or is T=mg/(cos(theta)) ?
and is the x component of tension is mgsin(theta) and this is the restoring force?

2. Relevant equations
Fnet=ma
Fg=mg

3. The attempt at a solution
Maybe it depends on the problem given? But I am confused how to know which tension to use

2. Dec 1, 2015

### 24forChromium

You need to provide the position and the speed of the pendulum, for example, a pendulum at rest would just generate a tension equal to its weight. If you know its speed of swinging, you can use the formula ((v^2)/r)=a_c where v is speed, r is radius and a_c is centripetal acceleration to find out the amount of acceleration towards the centre of rotation, then, use the mass of the pendulum to find the net centripetal force, then, subtract the weight of the pendulum (as a vector) from that force and you get the tension force. If you don't understand anything I said (some or all of that) just ask.

3. Dec 1, 2015

### Mister T

Is the pendulum moving?

Well, now I'm confused. You posed two questions in the section where you're asked to place the problem statement, so I'm thinking these are the questions your teacher wants you to answer, and I want to help you but I don't want to just give you the answers.

But then, in the section where you're asked to provide an attempt at a solution you seem to indicate you need to know which to use, so now I'm thinking there's a different question you're trying to answer.

If you're trying to solve a problem it's usually best to resolve the vectors into two components, but not the x- and y-components. Rather, you want one component in the radial direction and the other in the tangential direction. Then apply Newton's 2nd Law in each direction.

Are you trying to show that the restoring force is approximately linear?

Last edited: Dec 1, 2015
4. Dec 2, 2015

### **Mariam**

Yes I got your point, but it's not that helpful in my case because I am currently taking simple harmonic motion and my textbook isn't using and rotational motion in solving pendulums.

5. Dec 2, 2015

### **Mariam**

Actually the questions aren't a homework question but a confusion I had while solving pendulum problems.
For example in the image you had; I could conclude that T=mgcos(theta)... Because there is not vertical motion.

But in a problem I was solving:
T=(mg)/ (cos(theta))

And also I didn't get what you meant by this statement:

6. Dec 2, 2015

### haruspex

The two situations are different.
For a swinging pendulum, the constant length of the pendulum means that the only radial acceleration is centripetal. Resolving in tne radial direction, $T-mg\cos(\theta)=\frac{mv^2}r$. (By conservation of energy we can substitute $mv^2=2mgr(\cos(\theta)-\cos(\theta_0)$.)
For the two hanging charges, there are three forces: tension, gravity and the electric repulsion. If we resolve along the string now we will get a contribution from the repulsion. So instead we resolve vertically and find $T\cos(\theta)=mg$.

7. Dec 2, 2015

### Mister T

No, the direction of $\vec{T}$ is not vertical, it's towards the center. That's what I meant when I mentioned the radial direction.

Also, there is no motion along that direction, but $T \neq mg \cos \theta$ because the net force in the radial direction is not zero. If it were the pendulum bob would move in a straight line perpendicular to the direction of $\vec{T}$. The direction perpendicular to $\vec{T}$ is the tangential direction.

8. Dec 2, 2015

### Mister T

I'm confused, then. If that's you solving that problem then you know that $T=\frac{mg}{cos\ \theta}$. And if that's so, you've answered your own question!

9. Dec 2, 2015

### Mister T

That I can understand! I suggest you post one of those pendulum problems that you are trying to solve. Use the template and show us your attempt at the solution. That is the best way to procede if you're trying to understand why $T$ can equal $mg \cos\theta$ in some cases and $\frac{mg}{\cos \ \theta}$ in others.

There are general rules that apply, but the problem is that we often don't know how to apply them. In those cases we have to work out examples first before we can understand those general rules. The alternative is to memorize some rules we can use to make answers. That's a recipe for failure. The recipe for success is to learn how to make sense.