A person with mass m 1 = 53 kg stands at the left end of a uniform beam with mas

Question

A person with mass m1 = 53 kg stands at the left end of a uniform beam with mass m2 = 107 kg and a length L = 3.5 m. Another person with mass m3 = 61 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 10 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.

X-0 A person with mass m 1-53 kg stands at the left end of a uniform beam with mass m2 = 107 kg and a length L = 3.5 m. Another person with mass m3 61 kg stands on the far right end of the beam and holds a medicine ball with mass m4 - 10 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 1) What is the location of the center of mass of the system? Submit Help You currently have 0 submissions for this question. Only 6 submission are allowed You can make 6 more submissions for this question. 2) The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now? m Submit Hepohnlision are alowed. You currently have 0 submissions for this question. Only 6 submission are allowed. You can make 6 more submissions for this question. 3) 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?) m Submit Help You currently have O submissions for this question. Only 6 submission are allowed. You can make 6 more submissions for this question. 4) To return the medicine ball to the other person, both people walk to the center of the beam. At what xposition do they end up? m Submit Help You currently have O submissions for this question. Only 6 submission are allowed You can make 6 more submissions for this question.

Explanation / Answer

A) Center of mass, X_cm = (m1r1 + m2r2 + m3r3 + m4r4) /(m1 + m2 + m3 + m4)

= (53*0 + 107*1.75 + 61*3.5 + 10*3.5)/(53 + 107 + 61 + 10)

X_cm = 1.886 m

B) The center of mass does not change because there is no friction between beam and ground. X_cm = 1.886 m

C) 1.886 = (53*x1 + 107*(x1+1.75) + 61*(x1+3.5) + 10*x1)/(231)

X1 = 0.15 m

4) They all meet at the center of mass that is 1.886 m

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