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Prelab for Lab 8 - Forces and Torques in Equilibrium Questions Due at the Beginning of Lab Read the lab and refer to your textbook as needed to answer the following questions before coming to lab. The meter stick, pictured below, is only able to rotate about the pin. The meter stick is in static equilibrium. It's not moving in any direction, and it's not rotating. Therefore, all of the forces in the x-direction add up to zero, and all of the forces in the y-direction add up to zero, and all of the torques add up to zero. The masses m? and m? represent hanging masses while F represents a force applied at the end of the meter stick. This prelab is a guide to solving for situations similar to this. cg pin m1 m2 Figure 1 You may assume the following about this situation: 1. The pin is 20.0 cm from the left edge of the meter stick. 2. The unknown mass, m?, is placed a distance 5cm from the left edge. 3. m? = 12 kg placed 40.0 cm from the left edge. 4. The meter stick has a mass, $m_{cg}$ = 1.0 kg and a center of gravity 50cm from the left edge. 5. F = 15 N at 30° at a position 1m from the left edge. For this system to be in equilibrium, the clockwise torque must equal the counter-clockwise torque. First we compute the clockwise torque. The masses that are rotating the meter stick clockwise are m? and the center of gravity of the meter stick (located 50cm from the left edge). $\tau_{cw} = \tau_2 + \tau_{cg}$ (1) where $\tau_{cg}$ is the torque in the clockwise direction around the pin.

          Prelab for Lab 8 - Forces and Torques in Equilibrium
Questions Due at the Beginning of Lab
Read the lab and refer to your textbook as needed to answer the following questions before coming to lab.
The meter stick, pictured below, is only able to rotate about the pin. The meter stick is in static equilibrium.
It's not moving in any direction, and it's not rotating. Therefore, all of the forces in the x-direction add up
to zero, and all of the forces in the y-direction add up to zero, and all of the torques add up to zero. The masses
m? and m? represent hanging masses while F represents a force applied at the end of the meter stick. This
prelab is a guide to solving for situations similar to this.
cg
pin
m1
m2
Figure 1
You may assume the following about this situation:
1. The pin is 20.0 cm from the left edge of the meter stick.
2. The unknown mass, m?, is placed a distance 5cm from the left edge.
3. m? = 12 kg placed 40.0 cm from the left edge.
4. The meter stick has a mass, $m_{cg}$ = 1.0 kg and a center of gravity 50cm from the left edge.
5. F = 15 N at 30° at a position 1m from the left edge.
For this system to be in equilibrium, the clockwise torque must equal the counter-clockwise torque. First
we compute the clockwise torque. The masses that are rotating the meter stick clockwise are m? and the
center of gravity of the meter stick (located 50cm from the left edge).
$\tau_{cw} = \tau_2 + \tau_{cg}$ (1)
where $\tau_{cg}$ is the torque in the clockwise direction around the pin.

Prelab for Lab 8 - Forces and Torques in Equilibrium
Questions Due at the Beginning of Lab
Read the lab and refer to your textbook as needed to answer the following questions before coming to lab.
The meter stick, pictured below, is only able to rotate about the pin. The meter stick is in static equilibrium.
It's not moving in any direction, and it's not rotating. Therefore, all of the forces in the x-direction add up
to zero, and all of the forces in the y-direction add up to zero, and all of the torques add up to zero. The masses
m? and m? represent hanging masses while F represents a force applied at the end of the meter stick. This
prelab is a guide to solving for situations similar to this.
cg
pin
m1
m2
Figure 1
You may assume the following about this situation:
1. The pin is 20.0 cm from the left edge of the meter stick.
2. The unknown mass, m?, is placed a distance 5cm from the left edge.
3. m? = 12 kg placed 40.0 cm from the left edge.
4. The meter stick has a mass, mcg = 1.0 kg and a center of gravity 50cm from the left edge.
5. F = 15 N at 30° at a position 1m from the left edge.
For this system to be in equilibrium, the clockwise torque must equal the counter-clockwise torque. First
we compute the clockwise torque. The masses that are rotating the meter stick clockwise are m? and the
center of gravity of the meter stick (located 50cm from the left edge).
τcw = τ2 + τcg (1)
where τcg is the torque in the clockwise direction around the pin.

Added by Kenneth S.

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