A man of mass mm = 95 kg decides to paint his house. To do this, he builds a pla
ID: 2191461 • Letter: A
Question
A man of mass mm = 95 kg decides to paint his house. To do this, he builds a platform using a uniform beam with a mass of mb = 100 kg and a length of L = 7 meters. The beam is supported by two sawhorses, as shown in the diagram above. How far from the end of the beam (the end closest to support A) does the person have to stand to unbalance the beam? x = Later that day, after thinking about how cool rotational dynamics really is, the man decides to conduct an experiment. He removes one of the supports and places the other as shown in the diagram. Standing at the end of the board, he has his daughter place paint cans, each of mass mc = 2.9375 kg, on the opposite end. How many cans will the girl have to place on the board to provide the best balance? (You may neglect the actual length of the board that both the man and the cans occupy. Assume both are points at the ends of the board.) Also, take the axis of rotation about the sawhorse. Number of cans =Explanation / Answer
The man's mass is totally supported by the saw horse under him.
The only force on A is 1/2 the board weight, = (100 x g)/2 = 490N.
2) There is 5m. beam length from A to the right end, and 2m. left.
(95 x g)/(5 + 2) = 133N. per metre length.
(133 x 2) x (2/2) = 266 N/m. (weight x 1/2 length).
(133 x 5) x (5/2) = 1,662.5 N/m.
Difference = (1,662.5 - 266) = 1,396.5N/m.
1,396.5/(95 x g) = 1.5 metres left of A.
3) Assuming the same diagram, torque difference between ends is the same, = 1,396.5 N/m.
(2.9375 x g) = 28.7875N.
(1,396.5/2m) = 698.25N. at the end of the board required.
(698.25/28.7875) = 24- 1/4 cans approx.
I have done a lot of assuming for (3), and maybe the man is standing on the long end of the board?
(95 x g) x 5m. = extra torque of 4,655N. That makes total right torque (4,655 + 1,396.5) = 6,051.5N.
The difference becomes (6,051.5 - 266) = 5,785.5N/m.
(5,785.5/2m)/28.7875 = 100.5 cans approx.
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