What is the force and tension needed to hold a swinging ball?

[sphykik]

Help! Force/Tension Problem

Homework Statement

A 2.4 kg ball tied to a string fixed to the ceiling is pulled to one side by a force F to an angle of 32.1° from the ceiling.

(a) Just before the ball is released and allowed to swing back and forth, how large is the force F that is holding the ball in position?

(b) Just before the ball is released and allowed to swing back and forth, what is the tension in the string?

Homework Equations

The Attempt at a Solution

Hi all, I am new to physics, and this homework is kicking my butt! haha, so first I tried to convert the 2.4kg to Newtons (9.8N/kg), and then use sin and cos to find the force in the x and y directions. I get 19.9 in the x direction, and 12.5 in the y direction. Our homework is submitted electronically, though, and it doesn't think my answer for a) is right... I really don't know how to do b)

If someone could help me that would very much appreciated, this is due at 7 in the morning!

Thanks,
Damian

 

[learningphysics]

Did you draw a freebody diagram? What are the forces acting on the ball? What are the forces acting in the y-direction? In the x-direction?

 

[sphykik]

I drew a diagram, and I thought I found the forces in the x and y directions (by using sin and cos), but I can't seem to think of anything else. My professor very quickly went over this, and I am really stuck (i have work on it for about a half hour...). Any other suggestions to point me in the right direction?

Thanks

 

[learningphysics]

I drew a diagram, and I thought I found the forces in the x and y directions (by using sin and cos), but I can't seem to think of anything else.

Can you describe exactly the forces you found and what you did? We will help you along from there...

 

[sphykik]

Ok, since you are pulling the ball back, and supporting it in the air, you would have to apply a force "backward" (or in the x direction, as I drew it), and I think there has to be a force in the y direction, since the ball would move up some as you support it. i took the sin of 32.1, and multiplied it by 23.52 (which would be the force on the string, converted from kg, i think... ;) to find the y force, and i got 12.5. i did the same with the cosine, to find the x force, and i got 19.92.

I know there are several areas in which i could be wrong. for one, maybe 2.4kg should not be converted to Newtons like that... also, i could be wrong about the force on the hypotenuse, or the string, being 23.52...

beyond that, i really don't know what to do :(

Thanks!

 

[learningphysics]

Ok, since you are pulling the ball back, and supporting it in the air, you would have to apply a force "backward" (or in the x direction, as I drew it), and I think there has to be a force in the y direction, since the ball would move up some as you support it. i took the sin of 32.1, and multiplied it by 23.52 (which would be the force on the string, converted from kg, i think... ;) to find the y force, and i got 12.5. i did the same with the cosine, to find the x force, and i got 19.92.

I know there are several areas in which i could be wrong. for one, maybe 2.4kg should not be converted to Newtons like that... also, i could be wrong about the force on the hypotenuse, or the string, being 23.52...

beyond that, i really don't know what to do :(

Thanks!

I think the important step where you are going wrong is getting the components of the tension... Suppose the tension is T. Now 32.1 is the angle with the horizontal?

So what is the vertical component of T... what is the horizontal component of T? just give the answer in terms of T, sin, cos etc.

 

[sphykik]

http://img210.imageshack.us/img210/9384/untitledxk6.png

So the tension, T, is what I am trying to find, along with the x and y values? so sin(57.9) would be y/t

and cos(57.9) would be x/t ?

do i have it right so far?

 

[learningphysics]

Yes, looks good. So what is the horizontal component of T? What is the vertical componen of T?

Use upwards as positive. Down as negative. Right as positive. Left as negative.

 

[sphykik]

so, assuming that my 2.4kg to 23.52N conversion is right, and T is 23.52, y would be +19.9, and x would be -12.50 ?

edit: oops, i switched x and y on my diagram...

 

[learningphysics]

so, assuming that my 2.4kg to 23.52N conversion is right, and T is 23.52, y would be +19.9, and x would be -12.50 ?

edit: oops, i switched x and y on my diagram...

wait... you're jumping too far ahead... you don't know T yet... in terms of T, what is the horizontal component of T and the vertical component of T...

 

[sphykik]

so...

t(y)=y/(sin(57.9))
and
t(x)=x/(cos(57.9))

oh, and i don't think i mentioned it before, but they want the answer in Newtons,

Thanks

 

[learningphysics]

so...

t(y)=y/(sin(57.9))
and
t(x)=x/(cos(57.9))

?

If you have a vector with magnitude R, acting at an angle theta from the positive x axis... what is the x-component of this vector?

 

[sphykik]

(R)(cos(theta)) ?

 

[learningphysics]

(R)(cos(theta)) ?

exactly. The x component is R cos theta. The y component is Rsin theta.

Do the same type of thing here... except instead of R we have T...

 

[sphykik]

so

x=T(sin57.9)

and

y=T(cos57.9)

 

[learningphysics]

so

x=T(sin57.9)

and

y=T(cos57.9)

exactly.

Now write the $$\Sigma{F} = ma$$ in the x-direction. Also write the f=ma equation in the y-direction.

 

[sphykik]

im sorry, but i don't think we went over this in class, i can't find reference to it in the book either. could you explain? sorry to be so stupid :(

 

[sphykik]

so if force equals mass * acceleration, than the force would be zero, wouldn't it?

 

[learningphysics]

im sorry, but i don't think we went over this in class, i can't find reference to it in the book either. could you explain? sorry to be so stupid :(

No prob. don't feel bad. this stuff takes time to get used to.

Looking at the diagram... I'm going to assume the F force acts on the ball towards the right... the two forces in the horizontal direction are, F acting to the right, and Tsin57.9 acting to the left.

So $$F - Tsin(57.9) = ma_x$$ (where ax is the acceleration horizontally)

but ax =0, so:

F - Tsin(57.9) = 0

Try doing the same type of thing vertically to get another equation.

 

[sphykik]

wouldn't it be F-T cos(57.9) = 0 for the vertical direction?

 

[learningphysics]

wouldn't it be F-T cos(57.9) = 0 for the vertical direction?

No because F doesn't act vertically... also you need to take gravity into account.

 

[sphykik]

would you have T(cos57.9)+23.52, the 23.52 being (2.4)(9.8) to account for gravity?

 

[learningphysics]

would you have T(cos57.9)+23.52, the 23.52 being (2.4)(9.8) to account for gravity?

Yes. But use upwards positive and downwards negative. ie: Tcos(57.9) - 23.52 = 0. Hope the directions make sense.

From these 2 equations you can solve for F.

 

[sphykik]

THANK YOU SO MUCH! YOU ARE GODLY!

thanks so much for all of the help!

I got both parts right! WWOOOOOOHOOOO!

 

[learningphysics]

THANK YOU SO MUCH! YOU ARE GODLY!

thanks so much for all of the help!

I got both parts right! WWOOOOOOHOOOO!

no prob.