If the beam is 90 kg and the man weighs 60 kg, how far can he go before the beam tips over? The problem I am having is incorporating the center of mass into the problem. Maybe I am doing it wrong. I keep getting 3 meters. Please explain (answer is 1.5 meters).
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I am assuming that the center of mass lies at the center of the rod. If so, since the length of the rod is 8 + 6 = 14m, the center of mass of the road lies 7m from the left of the rod and 1m to the left of the pivot point.
There are two forces that will provide torque: the normal force acting on the board and the weight of the man. The normal force acts to tip the board counter-clockwise and the weight of the man acts to tip the board clockwise.
If the man is a distance of d away from the pivot, the net torque is:
Στ = (90)(1.0) - 60d = 90 - 60d.
The maximum value of Στ such that the board doesn't tip is 0, so the board tips when:
90 - 60d = 0 ==> d = 90/60 = 1.5 m.
I hope this helps!
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Yes, you just need to consider the CM. Remember, all bodies act as if all of its mass is concentrated at its CM.
There are two forces that will provide torque: the normal force acting on the board and the weight of the man. The normal force acts to tip the board counter-clockwise and the weight of the man acts to tip the board clockwise.
If the man is a distance of d away from the pivot, the net torque is:
Στ = (90)(1.0) - 60d = 90 - 60d.
The maximum value of Στ such that the board doesn't tip is 0, so the board tips when:
90 - 60d = 0 ==> d = 90/60 = 1.5 m.
I hope this helps!
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Yes, you just need to consider the CM. Remember, all bodies act as if all of its mass is concentrated at its CM.
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Mb = mass beam
Mm = mass man
can you say where the pivot point is for the beam
Mm = mass man
can you say where the pivot point is for the beam