A man of mass m m = 95 kg decides to paint his house. To do this, he builds a pl

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.

1)

F =

2)

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 =

3)

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 =

cans

Explanation / Answer

Your diagram shows a 9 m beam, yet the question states it is 7 m long. I will assume the beam is actually

__2m__ ____5m_______
______A_____________B

you can adjust the calculations if I don't have it right

a) summing torque about B
95*g*3.5+86*g*0-FA*5=0
solve for FA
FA=95*g*3.5/5

b)
Sum torque about A
86*g*x>95*g*1.5
solve for x

x>95*1.5/86

c) I will assume he removes support B
sum torque about A
N*2.8*g*2=95*g*1.5
solve for N
N=95*1.5/(2*2.8)

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