Please so work! thanks: A person with mass m1 = 55 kg stands at the left end of
ID: 1464091 • Letter: P
Question
Please so work! thanks: A person with mass m1 = 55 kg stands at the left end of a uniform beam with mass m2 = 105 kg and a length L = 2.7 m. Another person with mass m3 = 56 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 14 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) The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now? Here is my work but the answer is incorrect:
Beam= m2*(L/2)= 105(2.7/2)=141.75 P2= m3*L= 56*2.7= 151.2
Total forces = 141.75+151.2=292.95
Mass center = 292.95/(m1+m2+m3+m4)
Mass center = 292.95/(230)
Mass center = 1.273
2) 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
The center of mass of the system will be the sum of centers of mass of each component times each component
distance to the origin all divided by the total mass of the system
Before the ball is thrown
Cx = (55*0 + 105*2.7/ 2 + 56*2.7 + 14*2.7) / (55 + 105 + 56 + 14)
Cx = 1.438 m
Now figure the center of mass relative to the left end of the beam realizing that the beam may have moved relative to
the floor.
C = (55*0 + 105[2.7 / 2] + 56*2.7 + 14*0) / (55 + 105 + 56 + 14)
C = 1.27 m
As nothing was added or subtracted from the system, the center of mass will not change relative to the floor. As the
center of mass of the system has shifted relative to itself, the system motion relative to the floor will equal the
change in center of mass within the system but in the opposite direction
As the mass of the ball moved from right to left, the system must move left to right to compensate.
D = 1.438 - 1.27
x = 0.168 m to the right of the original position is the answer to the third question.
We can verify this by calculating the center of mass of the new position elements relative to the original position
Cx = (55*0.168 + 105*[(2.7 / 2) + 0.168] + 56*[2.7 + 0.168] + 14[0.168]) / (55 + 105 + 56 + 14)
Cx = 1.44 m
When they meet at the center of the beam to return the ball, all the mass is located at one position relative to the left
end of the beam or
C = 2.7 / 2
C = 1.35 m from the left end
So the end of the beam must have moved
d = 1.35 - 1.27
d = 0.08 m
As the larger mass (55 + 14) kg moved to the right, the common base beam must have moved left. Therefore the
end of the beam is now at
x = 0.168 - 0.08
x = 0.088 m from the original position
Which checks out if we subtract the final center of mass relative to the left beam end from the original center of mass
relative to both beam end and x origin
1.438 - 1.35 = 0.088 m