One end of a uniform beam of mass m b = 47.2 kg and length L = 3.42 m rests on t
ID: 1264725 • Letter: O
Question
One end of a uniform beam of mass mb = 47.2 kg and length L = 3.42 m rests on the ground; the other end is held above the ground by a pivot placed at distance d = 1.09 m from that end (see figure). A man of mass M = 75 kg walks along the beam from the low end toward the high end.
How close (in meters) to the high end of the beam can the man get before that end swings down?
One end of a uniform beam of mass mb = 47.2 kg and length L = 3.42 m rests on the ground; the other end is held above the ground by a pivot placed at distance d = 1.09 m from that end (see figure). A man of mass M = 75 kg walks along the beam from the low end toward the high end. How close (in meters) to the high end of the beam can the man get before that end swings down?Explanation / Answer
distance of center of mass from pivot point = (3.42 / 2 - 1.09)
equating the moments
47.2 * (3.42 / 2 - 1.09) = 75 * x
47.2 * 0.62 = 75 * x
x = 0.39 m
distance from high end = 1.09 - 0.39
distance from high end = 0.7 m
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