# Having trouble figuring a problem out

1. Jun 9, 2005

### frankfjf

Hi, I'm new to this forum, and I must admit that I'm in dire straits when it comes to solving this problem. I have tried reading through the textbook for help but it isn't clear enough for my liking and my professor is unavailable at this time. So without further ado, here is the problem that's been plaguing me:

A little girl is sitting on a playground swing. Her father pulls her to the right with just enough force to hold her stationary. If the little girl's weight is 35 pounds, what force is the father exerting?

I cannot figure a way to draw the picture the problem gives on here, but it gives a further bit of information: The angle is 25 degrees to the vertical.

It then asks me to draw the diagram (That much I figured out) and to do the calculations. I'm stumpted as to how exactly the formula works for a problem like this though, please help!

2. Jun 9, 2005

### Staff: Mentor

The girl is in equilibrium. Which means that the net force on her is zero.

Start by identifying all the forces that act on the girl and what direction they act. (Hint: I count three forces.) Draw yourself a clear diagram illustrating these forces.

Find the horizontal and vertical components of each force. Then apply the conditions for equilibrium: The sum of the vertical components equals zero and the sum of the horizontal components equals zero. You'll get two equations that you can solve together to get your answer.

3. Jun 9, 2005

### frankfjf

So, would those three forces be: The tug of both chains on the swing, her weight (gravity), and the father's pull?

And if so, would the components be found in this way?

Swing Chains: 35lbs.(cos(25))
Weight/Gravity: 35lbs.(cos(25))
Father's Pull: 35lbs.(sin(25))

Last edited: Jun 9, 2005
4. Jun 9, 2005

### Staff: Mentor

When I counted three forces I was counting the pull of both chains as a single force (since they act together).

Not exactly.

Let's start by identifying the three forces:
(a) The weight, which is given as 35 lbs. Which way does it act? What are its vertical and horizontal components?
(b) The force of the chains. This is an unknown, so give it a label. I'll call it T (for tension in the chains). Which way does it act? What are its vertical and horizontal components?
(c) The force of the father's pull. This is also an unknown, so give it a label. I'll call it F (for Father's Force). Which way does it act? What are its vertical and horizontal components? ​

5. Jun 9, 2005

### frankfjf

(a) Well, the weight is acting downward, as gravity is pulling the girl towards the ground. Its vertical component, if I've calculated it correctly is 32 lbs. (35lbs.(cos(25)), and its horizontal component is 15 lbs. (35lbs.(sin(25)).

For b and c, I am uncertain as to how to proceed.

6. Jun 9, 2005

### Staff: Mentor

Since the weight acts downward, it's vertical component is -35 lbs. It's horizontal component is 0.

Since "F" acts to the right (that's given in the problem statement), what must be its horizontal and vertical components?

Only the chain force "T" will require you to use cos(25) and sin(25), since that force acts along the direction of the chains. (I won't tell you which is which.)

7. Jun 9, 2005

### frankfjf

Ahhh, now I see what you mean. I'm tempted to bonk my head against my desk now. Often, in my desperation to get a problem right, I overlook the basics.

Well, the weight acts downward, has a horizontal component of 0 and a vertical component of -35lbs, like you said.

"F" is acting to the right. Therefore, it has no vertical component and has a horizontal component of 15lbs. (35lbs.(sin(25)).

"T" is the one that is diagnol. So it has both components. 35lbs.(sin(25)) + 35lbs.(cos(25)) = 47lbs.

8. Jun 10, 2005

### Staff: Mentor

Right.

It has no vertical component--correct; but its horizontal component is simply F!

35 lbs is the weight--nothing to do with the father's force, which is unknown! And the 25 degrees? That's the angle that the chains make; again, nothing to do with F.

Like "F", "T" is an unknown. It has nothing to do with the 35 lb weight of the girl. But, yes, the angle 25 does come into play here. Here's what I'm looking for: vertical component = T cos(25), horizontal component = -T sin(25)

Realize that T and F are unknowns (represented by letters) that you have to solve for.

9. Jun 10, 2005

### frankfjf

But can't I still use the weight to calculate the tension in the chains since the girl's weight is still pulling on them?

10. Jun 10, 2005

### Staff: Mentor

Yes, but only by setting up the proper equations and solving for the tension in terms of the weight.

11. Jun 10, 2005

### frankfjf

I think I've got it as far as the tension is concerned:

Vertical Component:

T(cos(25)) + T(cos(25)) = -35lbs

1.82T = -35lbs

-35lbs/1.82 = -19lbs.

Vertical of T = -19lbs.

Horizontal Component:

-T(sin(25)) + -T(sin(25)) = 35lbs.

.84-T = 35lbs.

35/.84 = 42

Horizontal of T = -42lbs.

12. Jun 10, 2005

### Staff: Mentor

Remember that T represents the total chain force (both chains). I'll show you how I'd do it.

T cos(25) -35 = 0
thus, T = 35/(cos(25)) = 38.6 lbs

So... that's the tension! You can easily find the horizontal component, once you have the magnitude of the force: T sin(25) = 38.6 sin(25) = 16.3 lbs (and its negative, since it points to the left).
Sanity check: The component of a vector cannot exceed the magnitude! So this can't be correct.

So now that you have the horizontal component of the chain tension, what's the force F?

13. Jun 10, 2005

### frankfjf

Sorry, sorry. Didn't take Physics in high school so am just touching this stuff for the first time. Okay, so by that logic, to get the force the father needs...

F = 38.6(sin(25))

F = 16 lbs.

14. Jun 10, 2005

### Staff: Mentor

Good!

The logic is that the horizontal forces must be in balance, since the girl is in equilibrium. And the only horizontal forces are the the tension pulling to the left and the father pulling to the right. And since the horizontal component of the tension is 38.6 sin(25) = 16 lbs (to the left), F must be 16 lbs to the right.

15. Jun 10, 2005

### frankfjf

Thank you very much for your help.