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.
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?
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
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!
Ditto! :-)
THANK YOU 11 YEARS AGO
To solve this problem, we can use the principle of moments, also known as the law of levers. According to this principle, for an object to be balanced, the sum of the clockwise moments must equal the sum of the counterclockwise moments.
In this case, we have the meter stick balanced at the 50.0 cm mark. We also have two coins stacked over the 7.0 cm mark, causing the meter stick to balance at the 41.0 cm mark.
Let's break down the moments acting on the meter stick:
1. Moment from the mass of the meter stick: Let's denote the mass of the meter stick as M and its center of mass as X. The moment created by the meter stick can be calculated as M * (X - 50.0 cm), where (X - 50.0 cm) is the distance between the center of mass and the fulcrum (knife-edge).
2. Moment from the coins: We have two 5.00 g coins stacked over the 7.0 cm mark. The total moment created by the coins can be calculated as (2 * 5.00 g) * (41.0 cm - 7.0 cm), where (41.0 cm - 7.0 cm) is the distance between the coins and the fulcrum.
Since the meter stick is balanced, the sum of these moments must be zero:
(M * (X - 50.0 cm)) + ((2 * 5.00 g) * (41.0 cm - 7.0 cm)) = 0
Now, let's solve this equation for M (the mass of the meter stick). First, convert the masses of the coins from grams to kilograms (1 g = 0.001 kg):
(M * (X - 50.0 cm)) + ((2 * 0.005 kg) * (41.0 cm - 7.0 cm)) = 0
Now, substitute the given values into the equation:
(M * (X - 50.0 cm)) + ((2 * 0.005 kg) * (41.0 cm - 7.0 cm)) = 0
(M * (X - 50.0 cm)) + (0.01 kg * 34.0 cm) = 0
Simplifying the equation further:
M * (X - 50.0 cm) = - 0.01 kg * 34.0 cm
M * (X - 50.0 cm) = - 0.34 kg * cm
Divide both sides by (X - 50.0 cm) to isolate M:
M = (- 0.34 kg * cm) / (X - 50.0 cm)
Now we need to figure out the value of (X - 50.0 cm), which is the distance between the center of mass and the fulcrum. Since the meter stick is balanced at the 50.0 cm mark, we know (X - 50.0 cm) = 0.
Substituting this back into the equation:
M = (- 0.34 kg * cm) / 0
However, division by zero is undefined, so our equation cannot be solved. This means that the problem might have been stated incorrectly or is missing some additional information.
Therefore, without further information, we cannot determine the mass of the meter stick.
1. Two coins stacked at 7 cm mark is
(2)(5g)=10g
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.
CCT=CT
10g(41cm-7cm)=M(50cm-41cm)
10g(34cm)=M(9cm)
340g*cm=9M*cm
(the cm's cancel)
So you have: M=340g/9
M=37.7 or 38g