Torque, equilibrium and center of mass
In this lab you will study torque by observing how various masses balance on a seesaw. You will check the condition for rotational equilibrium, measure the weight of an unknown mass and find the center of mass of a system of three masses.
Learning outcomes:
Understand torque and the condition for rotational equilibrium.
Understand how a balance is used to measure the weight of an object.
Understand how to find the center of mass of a system.
Simulation
Instead of using a physical apparatus, this lab uses the Balancing Act simulation provided by PhET. The simulation can be accessed at:
https://phet.colorado.edu/en/simulation/balancing-act
The simulation can be downloaded and run later without an internet connection.
Torque and rotational equilibrium (theory):
The condition for equilibrium of a point-like object is given by Newton’s first law, F ⃗_total=0. However, considering that objects have a size and shape it is easy to see that Newton’s first law is not enough for complete equilibrium. In addition to linear movement, objects can also rotate. Consider applying two forces to a stick as shown below. In both cases the total force is zero since the two forces have equal strength but are applied in opposite directions. Thus, the center of the stick will not move. In the left picture, the two forces are applied at the same location and the stick will not rotate. In the right picture the forces are applied at different locations and the stick will rotate counter-clockwise.
Torque: Torque is a rotational analogue of force and can be used to determine whether an object is in rotational equilibrium. In general torque depends on the strength of the force and the location where the force is applied relative to the axis the object is rotating around.
There are two simple cases to keep in mind:
Perpendicular: If the force is applied perpendicular to r ⃗ then the torque is given by
τ=r F
This is the most efficient way to apply a torque. For example, this is how you normally open a door.
Parallel: If the force is applied parallel to r ⃗ then the torque vanishes
τ=0
Direction of torque: Torque is a vector and has a direction. In general, each force will push an object either clockwise or counter-clockwise around the axis of rotation. If the force pushes the object clockwise then the corresponding torque is negative. If the force pushes the object counter-clockwise then the corresponding torque is positive.
In the above formula, F is the strength of the force and r is the distance from the axis of rotation to where the force is applied. Since both quantities are magnitudes, they are both always positive.
Conditions for equilibrium:
Newton’s first law:
F ⃗_total=0
Torque:
τ ⃗_total=0
Part 1: Balancing two masses
The simulation consists of a seesaw which you can place masses on. Placing the masses in the correct location will balance the seesaw. Open the simulation and select “Balance Lab”. You will see an empty seesaw with a pillar placed under each side. Select “Rulers” to display a ruler under the seesaw for taking measurements. On the right is a selection of bricks with varying mass you can place on the seesaw. You can click the purple arrow to show additional masses.
Place a 10 kg stack of bricks on the left side of the seesaw at 1.0 m and another 10 kg stack of bricks on the right side at 0.5 m. Once you are done, click the “balance toggle” located at the bottom. This will remove the two pillars and check if your seesaw is balanced. You should observe that it is not balanced, but instead rotates counter-clockwise. Thus, the condition for rotational equilibrium should be violated. Let’s compute the torque for each stack of bricks to check. For both bricks the weight is F_g=(10 kg) (9.8 m/s^2 )=98 N. The force is perpendicular to the seesaw, so the torque is given by τ=r F_g.
For the left brick: τ_L=(1.0 m) (98 N)=98 Nm
For the right brick: τ_R=(0.5 m) (98 N)=49 Nm
The left brick pushes the seesaw counterclockwise and is positive. The right brick pushes the seesaw clockwise and is negative.
τ ⃗_L=98 Nm τ ⃗_R=-49 Nm
The total torque is
τ ⃗_total=τ ⃗_L+τ ⃗_R=98 Nm-49 Nm=49 Nm
Since the total torque does not vanish the system rotates until it hits the ground. Since the torque is positive, the seesaw rotates counterclockwise.
Question 1: To balance two unequal masses, where should the larger one be placed relative to the smaller one? Same distance, closer to the center, or farther from the center?
To check your answer to the question, place a 5 kg stack of bricks at 1.0 m on the left side and a 10 kg stack on the right side. Adjust the 10 kg stack until the seesaw is balanced. Record the balancing position as r_(R,sim) in table 1 below. Remember r is a distance and is always positive. Repeat for the remaining masses listed in table 1. Note you are using the same 10 kg stack of bricks on the right side and moving it until the seesaw is balanced.
Table 1: Balancing two masses
m_L (kg) r_L (m) m_R (kg) r_(R,sim) (m)
5 1.00 10
5 1.50
15 1.00
15 0.50
20 0.50
Computation 1: Compute the torque for each mass in table 1. Make sure to include the correct sign. Record your answers in table 2 below. Show that in each case the total torque is zero.
Table 2: Torque
τ ⃗_L (Nm) τ ⃗_R (Nm)
Part 2: Determining an unknown mass
The triple-beam balance you use in the lab uses torque and equilibrium to determine an unknown mass. To do so, you adjust the position of the known masses until the two sides are balanced. We will use this process to measure the mass of four mystery objects in the simulation. We will place a known mass on the left side of the seesaw and an unknown mass on the right side. We will adjust the position of the known mass until the seesaw is balanced. Once the seesaw is balanced, we know that the left torque is balancing the right torque.
τ_L=τ_R
We can express the torque for each mass in terms of its weight and distance to the center of the seesaw using τ=r F_g=r m g.
r_L (m_L g)=r_R (m_R g)
Finally, we solve this equation for the unknown mass on the right side of the seesaw.
m_R=r_L/r_R m_L
Reset the simulation and place a 20 kg mass on the left side of the seesaw. Use the purple arrow to scroll through the masses until the first set of mystery objects appear. Place mystery object A on the right side of the seesaw at 1.00 m. Adjust the position of the left mass until the seesaw is balanced. Record the balancing position as r_L in table 3 below. Use the above formula to compute the unknown mass of mystery object A and record your result as m_R in table 3 below. Repeat for the remaining mystery objects.
Table 3: Finding mass
Mystery Object m_L (kg) r_L (m) r_R (m) m_R (kg)
A 20 1.00
B
C
D
Question 2: Would you get a different result for the mass of the mystery objects if the experiment was done on the Moon instead of Earth?
Part 3: Center of mass
A natural question to ask is where does the force of gravity act on an object? The answer is that the force of gravity can be thought to act on the center of mass of the object. The center of mass is defined as the average position of all the masses (or parts of the system), where the position of each mass is weighted by the ratio of its mass to the total mass. For a system consisting of three masses, the position of the center of mass is given by
x_CM=m_1/m_tot x_1+m_2/m_tot x_2+m_3/m_tot x_3
where x_1 is the position of mass m_1, x_2 is the position of mass m_2, and x_3 is the position of mass m_3, while m_tot=m_1+m_2+m_3. Note that the x is the objects position and can be positive or negative.
Consider the system of three masses shown below. We will demonstrate that gravity acts on the center of mass. First, we compute the total mass.
m_tot=30 kg+60 kg+20 kg=110 kg
From the total mass, we can figure out the force of gravity acting on the system.
F_g=m g=(110 kg)(9.8 m/s^2 )=1079 N
Next, we compute the center of mass, which will tell us where the above force of gravity acts.
x_CM=30/110 (-1.5 m)+60/110 (0.25 m)+20/110 (1.5 m)=0
We see that the center of mass is located at zero.
Question 3: Arrange the masses in the simulation. Note that additional masses can be found by clicking the purple arrows. Is the system balanced? Explain why the system is or is not balanced using the center of mass.
Computation 2: Compute the total mass and center of mass for the two sets of objects given below in table 4.
Table 4: center of mass
m_1 (kg) x_1 (m) m_2 (kg) x_2 (m) m_3 (kg) x_3 (m) m_tot (kg) x_CM (m)
10 1.25 15 -1.00 5 2.00
20 1.75 10 -1.25 30 0.75
Next, we will demonstrate that the center of mass is located at your computed value. Arrange the first set of masses in the simulation and observe that they are not balanced. If gravity truly acts on the center of mass, we should be able to balance the system by placing an object of equal mass on the opposite side of the seesaw.
Question 4: For the first set of data in table 4, select an object of equal mass to the total mass of the system and place it opposite to the center of mass. Does the mass you place balance the other three masses? Repeat for the second set of three masses.
