In the equilibrium lab, you place the fulcrum (pivot point) at the 25 cm
mark on the meter stick. Then you hang a mass #1 of 75 grams at the 62.5 cm mark, a
mass #2 of 25 grams at the 12.5 cm mark, and a mass #3 at the 0 cm mark. Find the value
of mass #3 in order for the system to be in equilibrium. Show your work with a free body
diagram and algebraic manipulations. Mass of the meter stick is 100 grams.
To find the value of mass #3 in order for the system to be in equilibrium, we need to set up an equation based on the principle of equilibrium.
Let's start by drawing a free body diagram of the system. We have three masses: mass #1 at 62.5 cm, mass #2 at 12.5 cm, and mass #3 at 0 cm.
| |
----------| m1 |----------
d1 d2
------------| |----------------
| m3 |
| |
| |
| |
Additionally, we have the meter stick with its mass distributed along its length. The fulcrum is at the 25 cm mark.
Now, let's consider the torques acting on the meter stick. Torque is the product of the force and the perpendicular distance from the fulcrum. For the system to be in equilibrium, the total torque on the meter stick must be zero.
Considering the clockwise direction as positive, the torques can be calculated as follows:
Torque due to mass #1 = (force due to m1) * (distance from fulcrum to m1)
= (0.075 kg * 9.8 m/s^2) * (0.625 m - 0.25 m) (converting grams to kilograms)
= 0.073125 Nm
Torque due to mass #2 = (force due to m2) * (distance from fulcrum to m2)
= (0.025 kg * 9.8 m/s^2) * (0.125 m - 0.25 m)
= -0.0245 Nm (note the negative sign as the torque is in the opposite direction)
Torque due to mass #3 = (force due to m3) * (distance from fulcrum to m3)
= (m3 * 9.8 m/s^2) * (0 m - 0.25 m)
= -2.45m Nm (note the negative sign as the torque is in the opposite direction)
Torque due to the meter stick itself = (force due to meter stick's mass) * (distance from fulcrum to meter stick's center)
= (0.1 kg * 9.8 m/s^2) * (0.25 m)
= 0.245 Nm
Now, let's sum up all the torques:
0.073125 Nm - 0.0245 Nm - 2.45m Nm + 0.245 Nm = 0
Simplifying the equation:
0.073125 Nm - 0.0245 Nm - 2.45m Nm + 0.245 Nm = 0
Combining like terms:
-2.4m Nm + 0.294625 Nm = 0
-2.4m Nm = -0.294625 Nm
Dividing both sides by -2.4 Nm:
m = (-0.294625 Nm) / (-2.4 Nm)
m ≈ 0.12276 kg
Therefore, the value of mass #3 in order for the system to be in equilibrium is approximately 0.12276 kg.
To find the value of mass #3, we need to determine the weight (force due to gravity) of each mass and set up an equation representing the equilibrium condition. Let's go step by step:
1. First, determine the weight of each mass:
- The weight of mass #1 (75 grams) is calculated as 75g * 9.8 m/s^2 (acceleration due to gravity) = 735 grams * m/s^2.
- The weight of mass #2 (25 grams) is calculated as 25g * 9.8 m/s^2 = 245 grams * m/s^2.
- The weight of mass #3 is unknown, so let's represent it as "w3" grams * m/s^2.
2. Next, set up a free body diagram:
- Draw a diagram of the meter stick with the fulcrum (pivot point) at the 25 cm mark.
- Label the three weights: mass #1 (75g) hanging at the 62.5 cm mark, mass #2 (25g) hanging at the 12.5 cm mark, and mass #3 (w3) at the 0 cm mark.
- Draw arrows representing the weight acting downward on each mass.
- Since the meter stick itself has a mass of 100 grams, draw an arrow representing the weight of the meter stick (100g) acting downward at the 25 cm mark.
3. Write the equilibrium condition equation:
- For the system to be in equilibrium, the sum of the torques (rotational forces) on each side of the fulcrum should balance each other.
- Torque (τ) is calculated as the product of the force (F) and the distance (d) from the fulcrum: τ = F * d.
- On the left side of the fulcrum, we have mass #1 (75g) causing a torque in the clockwise direction, so its torque is (75g * 9.8m/s^2) * (62.5 - 25) cm.
- On the right side of the fulcrum, we have mass #2 (25g), mass #3 (w3), and the meter stick (100g) causing torques in the counterclockwise direction, so their torques are (25g * 9.8m/s^2) * (25 - 12.5) cm, (w3 * 9.8m/s^2) * (-25) cm, and (100g * 9.8m/s^2) * (-25) cm, respectively.
- To balance the torques, we need: (75g * 9.8m/s^2) * (62.5 - 25) cm = (25g * 9.8m/s^2) * (25 - 12.5) cm + (w3 * 9.8m/s^2) * (-25) cm + (100g * 9.8m/s^2) * (-25) cm.
4. Simplify and solve the equation:
- Given that 1 cm = 1/100 m, we can convert the centimeters to meters by dividing by 100.
- Simplifying the equation, we have: (75 * 9.8) * (0.375) = (25 * 9.8) * (0.125) + (w3 * 9.8) * (-0.25) + (100 * 9.8) * (-0.25).
- Further simplifying, the equation becomes: 275.625 = 30.625 - 2.45w3 - 24.5.
- Rearranging, we have: -2.45w3 = 275.625 - 30.625 + 24.5.
- Combining terms, we have: -2.45w3 = 269.5.
- Finally, solving for w3, we divide both sides by -2.45: w3 = 269.5 / -2.45.
By calculating the value of w3, you can determine the mass required at the 0 cm mark for the system to be in equilibrium.