Identifying sign conventions for torque and tension

[droidofthevoid]

My question arises from trying to get the correct sign convention for a pulley that has two boxes attached by a massless rope. one box is to the left of the pulley and the other box is hanging down below the pulley.

If I take the counter clockwise direction to be negative, and set up my coordinate system as having the pulley at the origin, will the tensions as seen by the pulley both be negative because one is in the negative x direction and the other is in the negative y direction?

 

[ElijahRockers]

Don't think of the tension as being in the x or y coordinate system. There's no gravity pulling on the rope, or other forces besides the tension, so you can just say that the tensions have opposite signs.

Perhaps it might help to think that one is CCW, and one is CW.

It's been awhile since I did mechanics, but hopefully this helps.

 

[droidofthevoid]

Thank You

 

[jtbell]

Are you acquainted with the notion of torque as a vector, defined by a vector cross product?

The torque produced by $\vec F_1$ is $\vec \tau_1 = \vec r_1 \times \vec F_1$ which points in the +z direction (out of the screen), according to the right-hand rule.
The torque produced by $\vec F_2$ is $\vec \tau_2 = \vec r_2 \times \vec F_2$ which points in the -z direction (into the screen).

More explicitly in terms of unit vectors, $\vec \tau_1 = (r_1 \hat y) \times (- F_1 \hat x) = - r_1 F_1( \hat y \times \hat x) = - r_1 F_1 (- \hat z) = + r_1 F_1 \hat z$.

I leave $\vec \tau_2$ as an exercise.

 

[droidofthevoid]

Are you acquainted with the notion of torque as a vector, defined by a vector cross product?

The torque produced by $\vec F_1$ is $\vec \tau_1 = \vec r_1 \times \vec F_1$ which points in the +z direction (out of the screen), according to the right-hand rule.
The torque produced by $\vec F_2$ is $\vec \tau_2 = \vec r_2 \times \vec F_2$ which points in the -z direction (into the screen).

More explicitly in terms of unit vectors, $\vec \tau_1 = (r_1 \hat y) \times (- F_1 \hat x) = - r_1 F_1( \hat y \times \hat x) = - r_1 F_1 (- \hat z) = + r_1 F_1 \hat z$.

I leave $\vec \tau_2$ as an exercise.

View attachment 92798

Yes, I am aware of the Torque vector notion. I was just a little thrown off because I had a question where clockwise motion was chosen as positive as opposed to negative. Instinctively I assumed that any answer found in the positive clockwise direction would be of opposite value in the negative counterclockwise direction, however, I made a sign error in my algebra when solving.

To answer your question regarding $\vec \tau_2$

$\vec \tau_2 = (r_2 \hat x) \times (- F_2\hat y) = - r_1 F_2( \hat x \times \hat y) = - r_1 F_1 ( \hat z) = - r_2 F_2\hat z$.

To be honest, I just copied your code for the equation you wrote since I don't really know how to write equations in the forum yet. Is there a simpler way to write equations with vector hats or is this the basic format?Thank you again for your input!

 

[jtbell]

For a tutorial on doing equations in LaTeX, go to the Info menu at the top right of this page, choose "Help/How-To" and then "LaTeX Primer". To see the LaTeX code for a posted equation, control-click it if you're on a Mac or (I think) right-click it if you're on Windows, then choose Show Math As --> TeX Commands. (might be different on Windows).

 

[droidofthevoid]

Awesome! Thank you again!