How Do You Calculate the Horizontal Force on a Pendulum?

[nx01]

Homework Statement

A 2.0 kg pendulum ball is pulled 30° from the vertical by a horizontal force Fpull. What is the magnitude of Fpull?

Homework Equations

F=ma

The Attempt at a Solution

I'm rather lost with this one - I'm not even sure how to classify it. Any help at all - even a point in the right direction - would be greatly appreciated.

 

[physics&math]

Unless the problem says otherwise, assume Fpull is at a 90° angle from the vertical.

Force diagram. You'll have three forces: tension, Fpull, and the force of gravity. The tension will have two components.

 

[physics&math]

sorry, misread the post. it said horizontal. So it is definitely a right angle from the vertical

 

[Windadct]

Yes -- diagram it.

 

[nx01]

Hi all, thanks for your input. Is it right to assume that the force of gravity = the y component of the tensional force, and that the force applied to the ball = the x component of the tensional force? If so, how can I calculate Tx without knowing the magnitude of T? (If my assumptions are correct, I can calculate Ty using Fg, but I don't know how that value could help me.)

 

[haruspex]

Hi all, thanks for your input. Is it right to assume that the force of gravity = the y component of the tensional force, and that the force applied to the ball = the x component of the tensional force?

Yes.

If so, how can I calculate Tx without knowing the magnitude of T? (If my assumptions are correct, I can calculate Ty using Fg, but I don't know how that value could help me.)

If T is the total tension and the angle is 30^o, what in terms of T are Tx and Ty?

 

[nx01]

If the ball is 30o from the vertical to the right, then T is oriented 130o from the positive direction of the x-axis. Then Tx =Tcos130, and Ty=Tsin130?

 

[haruspex]

If the ball is 30o from the vertical to the right, then T is oriented 130o from the positive direction of the x-axis. Then Tx =Tcos130, and Ty=Tsin130?

No, not 130. How did you calculate that?

 

[nx01]

Oops, I meant 120. I've attached a picture (.jpg) showing how I calculated the angle.

Thanks for your responses so far!

 

[haruspex]

Oops, I meant 120. I've attached a picture (.jpg) showing how I calculated the angle.

Thanks for your responses so far!

Yes, that's better.

 

[nx01]

I've got the answer :)

For anyone who might come across this post in the future:

Ty=Fg=mg=2*9.8=19.6=Tsin120
So T=22.63
|Fpull|=|Tx|=|Tcos120|=|-11.3|=11.3