posted by Robert .
A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.00 g coins stacked over the 7.0 cm mark, the stick is found to balance at the 41.0 cm mark. What is the mass of the meter stick? The answer is 38g but I don't understand how to get it.
1. Two coins stacked at 7 cm mark is
2. Choose pivot at 41 cm, because that's where it's balanced.
3. Then, CCT=CT
CCT would be the weight of the coins at the 7 cm mark times the distance from the pivot.
CT would be the weight of the meter stick (M), which is what you're looking for, time the distance from the pivot.
(the cm's cancel)
So you have: M=340g/9
M=37.7 or 38g
Thank you so much. I'm confused about one thing though, what does CCT and CT stand for? is it Torque? Or is that another way for center of mass
CCT is Counter-Clockwise Torque
CT is Clockwise Torque
These two must equal each other in order for equilibrium to exist.
It is the same as saying:
The sum of the Torques = 0
The sum of the torques meaning the
counter clockwise torques minus the clockwise torques = 0.
Does that help?
Ahhh, thank you, a lot clearer :)
Another question in regards to CT. I know CT refers to the clockwise turn of the meter stick but how can you picture it going in that direction? With no weight on the ruler, isn't it at an equilibrium?
1. If you think of the pivot point as the center of a clock, the forces on the left would be going in a counter-clockwise motion whereas the forces on the right would be going in a clockwise motion.
The counter-clockwise motion is positive and the clockwise motion is negative.
2. There is weight on the ruler; it's the weight of the ruler itself (which is what you were looking for). If the meter stick had no weight, it would not balance and would therefore turn in the counter-clockwise direction.
Hope this helps.
After reading your comment and picturing a see saw, now I get it. Thank you soooo much and have a happy holiday :)
Yes, just like a seesaw!