Pendulum question - Vector addition of forces

[daysrunaway]

Homework Statement

A 1.0 kg pendulum bob is suspended from a string 1.0 m long. The pendulum swings until the string makes an angle of 22° with the vertical. At this point, if the pendulum is held with a string fastened to the bob and held horizontally, what would be the tension in the original string (not the horizontal one)?

Homework Equations

Force of gravity parallel to the string = Force of tension = mgcos22
Force of gravity perpendicular to the string = mgsin22

The Attempt at a Solution

I assumed that the horizontal rope would only affect the force of gravity perpendicular to the string, and that the tension in the original string would be the same as it was without a horizontal string holding it. Then my teacher made me question that assumption. Was I right, and she was just trying to make sure I was confident with my answer, or am I wrong?

Thanks!

 

[collinsmark]

Hello daysrunaway,

Welcome to Physics Forums!

Homework Equations

Force of gravity parallel to the string = Force of tension = mgcos22
Force of gravity perpendicular to the string = mgsin22

You should rethink your equations for the case where the horizontal string is attached. For that case, the above equations are not quite right (you're multiplying by something when you should be dividing). It might help by drawing a free body diagram (FBD), for the case where the horizontal string is attached.

The Attempt at a Solution

I assumed that the horizontal rope would only affect the force of gravity perpendicular to the string, and that the tension in the original string would be the same as it was without a horizontal string holding it. Then my teacher made me question that assumption. Was I right, and she was just trying to make sure I was confident with my answer, or am I wrong?
Thanks!

Your instructor was not trying to make sure you were confident in your answer. http://www.websmileys.com/sm/happy/045.gif.

Consider the case where a free-swinging pendulum reaches its peak at 90^o with the vertical. What is the tension on the string at the moment the angle is 90^o? Now consider the hypothetical case where a second string is attached to the bob. How hard would you have to pull on the second string such that both strings are held at 90^o with respect to the vertical (in static equilibrium)? (I'll give you a hint on the second question: the answer is $\infty$ N). My point of all that is the presence of the second string makes a difference.

 

[daysrunaway]

Hi collinsmark,

Thanks for your help, but I'm not sure I completely understand. I drew a FBD and split the force due to the second string into its component vectors. This is what I found:

(F_R is the force due to the second rope, while F_T is due to the first.)

$$\Sigma$$F_Y = F_Ry - F_Gy = F_T (because the forces are balanced)

F_T = F_Rcos(90 - 22) - mgcos22
F_T = F_Rcos68 - 9.1 N

This is where I get stuck, though. How can I find F_R?

 

[collinsmark]

Hi collinsmark,

Thanks for your help, but I'm not sure I completely understand. I drew a FBD and split the force due to the second string into its component vectors. This is what I found:

(F_R is the force due to the second rope, while F_T is due to the first.)

$$\Sigma$$F_Y = F_Ry - F_Gy = F_T (because the forces are balanced)

F_T = F_Rcos(90 - 22) - mgcos22
F_T = F_Rcos68 - 9.1 N

This is where I get stuck, though. How can I find F_R?

Uh... :uhh: ...

Again, I think you're multiplying by something when you should be dividing. Another way of putting it is I think you are attaching your trigonometric functions to the wrong terms.

Start with the up/down direction. (Come back to the horizontal direction later.) One of the vectors is mg, which points straight down. There is only one other vector that has a component in the up/down direction: the tension (of the pendulum string, not the second string). Since nothing is accelerating, mg must be equal to the y-component of the tension.

But in your equations, you have terms such as mgcos(22^o), but that doesn't make much sense in the static equilibrium case with the string. Look at your free body diagram again. Sum the forces in the vertical (up/down) direction and consider,

Tcos(22^o) - mg = 0

or rearranging,

Tcos(22^o) = mg

Do you see where I got that?

[Edit: Earlier you wrote that you "split the force due to the second string into its component vectors." But the second string is held horizontally. That means its tension is completely in the x-direction. The second string has no y-component of force! ]

 

[daysrunaway]

That clears it up perfectly! I think my misunderstanding was the way I'd assigned the axes: parallel to the string for y and perpendicular for x. I re-did the problem with your axes and found it much easier. thanks!