# Parallel forces

1. Dec 5, 2007

### schang

[SOLVED] parallel forces

A uniform 2.50-N meterstick is hung from the ceiling by a single rope. A 500.0-g mass is hung at the 25.0-cm mark and a 650.0-g mass at the 70.0-cm mark. (a) what is the tension of the rope? (b) where is the rope attached to the meterstick?

Well, first I converted all the weights into Newtons and then added 1.25 N at the 50 cm mark to account for the mass of the meterstick. But....

How can I find the tension of the rope if I don't know where it is? Even if I place the pivot point at the rope, I can't calculate the torques of the other forces because I don't have the torque arms...

2. Dec 5, 2007

### dotman

Hello,

If the system is in equilibrium (the meter stick isn't tipping over to one side) then you do know where the rope is-- it must be at the center of gravity of the meter stick+weights system. You don't need to have the torque arms to calculate the force gravity is imparting on the weights.

Do you know how to calculate where this is on the stick? Also, what was the reasoning behind adding 1.25 N, as opposed to 2.50 N?

3. Dec 5, 2007

### schang

oh my mistake, i did that on my work, i don't know why i wrote that.
i've got it now, thanks!