Question 7.4: Supporting a seesaw with two people Find an expression for t......
Supporting a seesaw with two people
Find an expression for the position of the center of mass of a system that consists of a uniform seesaw of mass m_1 and two people of masses m_2 \text { and } m_3 sitting at the ends of the seesaw beam \left(m_2>m_3\right)
Sketch and translate The figure on the next page shows a labeled sketch of the situation. The two people are represented as blocks. We choose the seesaw and two blocks as the system and construct a mathematical equation that lets us calculate the center of mass of that system. We place the x-axis along the seesaw with its origin at some arbitrary position on the left side of the
seesaw. The center of mass of the seesaw beam is at x_1 and the two blocks rest at x_2 \text { and } x_3 \text {. } . At what position x should we place the fulcrum under the seesaw so that the system does not rotate—so that it remains in static equilibrium? At this position, the sum of all torques exerted on the system is zero. This position is the center of mass of the three-object system.
Simplify and diagram We model the seesaw as a rigid body and model each of the people blocks as point-like objects. Assume that the fulcrum does not exert a friction
force on the seesaw. As you can see in the force diagram, we know the locations of all forces except the fulcrum force. We will calculate the unknown position x of the fulcrum so the seesaw with two people on it balances—so that it satisfies the second condition of equilibrium.
Represent mathematically Apply the torque condition of equilibrium with the axis of rotation going through the unknown fulcrum position x. Then determine the torques around this axis produced by the forces exerted on the system. The gravitational force exerted by Earth on the center of mass of the seesaw has magnitude m_1 g, \text { is exerted a distance } x_1-x \text { from } the axis of rotation, and has clockwise turning ability. The gravitational force exerted by Earth on block 2 has magnitude m_2 g \text {, is exerted a distance } x-x_2 \text { from the } axis of rotation, and has counterclockwise turning ability. The gravitational force exerted by Earth on block 3 has magnitude m_3 g \text {, is exerted a distance } x_3-x \text { from } the axis of rotation, and has clockwise turning ability. The torque condition of equilibrium for the system becomes
m_2 g\left(x-x_2\right)-m_1 g\left(x_1-x\right)-m_3 g\left(x_3-x\right)=0
Solve and evaluate Divide all terms of the equation by the gravitational constant g and collect all terms involving x on one side of the equation to get
m_2x + m_1x + m_3x = m_2x_2 + m_1x_1 + m_3x_3or
x(m_2 + m_1 + m_3)= m_2x_2 + m_1x_1 + m_3x_3
Divide both sides of the equation by (m_2 + m_1 + m_3) to obtain an expression for the location of the center of mass of the three-object system:
Let’s evaluate this result. The units of x are meters.
Next check some limiting cases to see if the result makes sense. Imagine that there are no people sitting on the seesaw (m_2 = m_3 = 0). In this case,x = \frac{m_1x_1}{ m_1}=x_1
The center of mass of the seesaw-only system is at the center of mass of the seesaw x_1, as it should be since we assumed its mass was uniformly distributed. Finally, if we increase the mass of one of the people on the seesaw, the location of the center of mass moves closer to that person.
Try it yourself: Where is the center of mass of a 3.0-kg, 2.0-m-long uniform beam with a 0.5-kg object on the right end and a 1.5-kg object on the left?
Answer: 0.8 m from the left end of the beam.