a 200 gram mass is balanced at one end of a meterstick and a 100 gram mass is at the other. the center of mass is 36.75 cm from the 200 gram mass.
What is the mass of the meterstick?
Thank you for you help
Don't you mean the fulcrum was 36.75 cm from the end with the 200 g mass?
Or do you mean the center of mass of meterstick and both weights (together) is at 36.75 cm? That would also make sense.
Whether it is balanced or not depends upon where the fulcrum is.
The center of mass of the meter stick is probably at the 50 cm mark.
Set the total moment about the fulcrum equal to zero and solve for the mass of the meterstick.
The fulcrum is at 36.75
Set the total moment about the fulcrum equal to zero and solve for the mass M of the meterstick. (You can omit the g factors, which cancel out. I will leave mass in grams, and distances in cm).
200*36.75 + M*13.25 = 100*63.25
M = 79.23 g
To find the mass of the meterstick, we need to consider the balance of torques in this system.
First, let's assign some variables:
M = mass of the meterstick (what we are trying to find)
M1 = mass of the 200 gram mass
M2 = mass of the 100 gram mass
d1 = distance of M1 from the center of the meterstick (in cm)
d2 = distance of M2 from the center of the meterstick (in cm)
d = distance from the center of the meterstick to the center of mass of the entire system (in cm)
We are given:
M1 = 200 grams
M2 = 100 grams
d = 36.75 cm
To balance the torques, we need the sum of clockwise torques to equal the sum of counterclockwise torques.
The torque caused by M1 is calculated as:
Torque1 = M1 * d1
The torque caused by M2 is calculated as:
Torque2 = M2 * d2
Since the system is in balance:
Torque1 = Torque2
To find the mass of the meterstick, M, we need to determine the value of d1 and d2.
We know that the center of mass of the entire system is 36.75 cm from the 200 gram mass. The center of mass is calculated as the weighted average of the individual masses:
d = (M1 * d1 + M2 * d2) / (M1 + M2)
Substituting the given values:
36.75 = (200 * d1 + 100 * d2) / (200 + 100)
Simplifying the equation:
36.75 = (200d1 + 100d2) / 300 ---> Multiplying both sides by 300
36.75 * 300 = 200d1 + 100d2
11,025 = 200d1 + 100d2
We still need another equation to solve for both d1 and d2.
Since the system is in balance, the sum of mass times distance on one side of the pivot must equal the sum of mass times distance on the other side:
M1 * d1 = M2 * d2
Substituting the given values:
200 * d1 = 100 * d2
Now we have a system of two equations:
11,025 = 200d1 + 100d2
200d1 = 100d2
We can solve this system of equations to find the values of d1 and d2. By substituting the value of d1 in terms of d2 from the second equation into the first equation, we can solve for d2:
11,025 = 200(100d2) + 100d2
11,025 = 20,000d2 + 100d2
11,025 = 21,000d2
d2 = 11,025 / 21,000
Simplifying d2:
d2 ≈ 0.525 cm
Now that we have the value of d2, we can substitute it back into the second equation to find d1:
200d1 = 100 * 0.525
d1 ≈ 0.2625 cm
Now that we have the values of d1 and d2, we can find the mass of the meterstick, M:
M = (M1 * d1 + M2 * d2) / d
M = (200 * 0.2625 + 100 * 0.525) / 36.75
M ≈ 3.15 grams
Therefore, the mass of the meterstick is approximately 3.15 grams.