A person with mass m_1 = 51 kg stands at the left end of a uniform beam with mas

Question

A person with mass m_1 = 51 kg stands at the left end of a uniform beam with mass m_2 = 105 kg and a length L = 3.3 m. Another person with mass m_3 = 58 kg stands on the far right end of the beam and holds a medicine ball with mass m_4 = 8 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor. What is the location of the center of mass of the system? The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now? What is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?) To return the medicine ball to the other person, both people walk to the center of the beam. At what x - position do they end up?

Explanation / Answer

beam spec. weight Wbs = m/l = 105/3.3 = 31.81 kg/m
calling x the distance in m from left side of beam and the system's center of mass

location of the center of mass of the system= 1.76m (as per your answer of part 1)

2)
59*x+x/2*Wbs*x = 58*(3.3-x)+(3.3-x)/2*Wbs*(3.3-x)
59x+x^2*15.9 = 191.4-58x+15.9(3.3-x)^2
59x+x^2*15.9 = 191.4-58x+173.15+15.9x^2-104.9x
59x+58x+104.9x = 364.55
x = 364.55/222 = 1.642 m (quite near the beam midpoint)

3)
beam motion = x = 1.76-1.642 = 0.12 m

4)
they will meet at the static center of mass of condition 2), i.e. 1.642 m from left

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.