# Homework Help: Tension and cent. acc.

1. Mar 17, 2005

### xangel31x

I am working on a problem where i need to find the 2 tensions connectd to a mass that is spinning horizontally in a circle at a constant speed. i know that the tensions are different...the first has the tension and the weight and the second is just the second tesnion. i am having a hard time getting the first tension....please someone help me.

2. Mar 17, 2005

### Shockwave

You haven't done any work?

3. Mar 17, 2005

### xangel31x

no i have i found the cent. acc. the radius, the angle and the lower tension

4. Mar 17, 2005

### arildno

Welcome to PF!
First state your problem CLEARLY, not in a muddled manner (it is basically impossible to figure out the particular problem you are struggling with from what you've written).
Secondly, post a few thoughts on how to approach the problem

5. Mar 17, 2005

### xangel31x

ok sorry...i will try again

6. Mar 17, 2005

### xangel31x

i have a figure that shows 2 wires connected to a single mass. the mass is revolving horizontally in a circle around the vertical pole. at a constant speed. i need to the two tensions. at first i thought they were the same, but they i realized that the top tension has the weight acting on it also. the second tension equals (m*(v^2)/r)/cos(theta)...now for the first tension i have that it equals(m*(v^2)/r)/cos(theta), but i don't know what to do with the y value i have that as t*sin(theta) = m*g.
i hope this is better

7. Mar 17, 2005

### arildno

Allright, just to get this clear:
The vertical pole plus the two wires form a triangle with the mass whirling about the pole as the third vertex?
I asssume the angles the wires make with the pole are equal to each other?

8. Mar 17, 2005

### arildno

I'll assume I'm right.
Now, you need eqilibrium of forces acting upon the mass in the vertical direction, so you must have:
$$T_{top}\cos\theta-T_{bottom}\cos\theta-mg=0$$
That is, the ifference between the "top" tension and tension in the lower wire is:
$$T_{top}-T_{bottom}=\frac{mg}{\cos\theta}$$
where $$\theta$$ is the angle between the wire and the pole.
In the horizontal direction, the tensions must provide the centripetal acceleration.

Last edited: Mar 17, 2005
9. Mar 17, 2005

### xangel31x

yes the angles are equal... the large triangle splits into two smaller triangles with equal everything exp the hyp. with are the tensions. i saw that you used cos(theta) for both the top and the bottom. that is the whole system that you were doing that for right? i mean doesn't the sin(theta) come into play when you are doing the top tension with the weight?

10. Mar 17, 2005

### arildno

I was referring to the geometry, rather than the force diagram.

You are very unclear here, are you sure you understand that the basic law you are to use is F=ma for the mass?

It simply doesn't make sense to allocate the "weight" to the "top tension" as you seem to do.
The sum of forces (both the tensions plus the weight) must yield the centripetal acceleration of the mass.
That is the horizontal component of Newton's 2.law, in the vertical component, the sum of forces must be equal to 0, since the mass doesn't move vertically.

11. Mar 17, 2005

### xangel31x

i thought that the weight went with the top triangle and was used when finding the force of the radius line, which is at the side opposite of the hyp. of the triangle?

12. Mar 18, 2005

### arildno

Now, I'll post a detailed solution here; try to see if you can follow it:
1). Each wire makes an angle $$\theta$$ with the vertical pole, and the tensile forces acting upon the mass from them have directions parallell to the wire.
That is the tensile force from the top-most wire can be written as:
$$\vec{T}_{top}=T_{top}(\cos\theta\vec{k}-\sin\theta\vec{i}_{r})$$
where $$\vec{i}_{r}$$ lies in the horizontal plane (think of it, at a given instant, as $$\vec{i}$$), and $$\vec{k}$$ is along the vertical.
$$T_{top}$$ is magnitude of the tensile force, i.e, the tension.

Similarly, we have for the tensile force in the lower wire:
$$\vec{T}_{bottom}=T_{bottom}(-\cos\theta\vec{k}-\sin\theta\vec{i}_{r})$$

2) The weight of the mass is: $$\vec{W}=-mg\vec{k}$$
3) The acceleration of the mass is: $$\vec{a}=-\frac{v^{2}}{R}\vec{i}_{r}$$
That is, the mass only experiences centripetal acceleration.
4) Hence, Newton's 2.law states:
$$\vec{T}_{top}+\vec{T}_{bottom}+\vec{W}=m\vec{a}$$
Can you take it from here on your own?

Last edited: Mar 18, 2005
13. Mar 18, 2005

### xangel31x

yes, i believe i can take it from there. i was on that path for the most part. thanks for your help