16 December, 2013

Construction Supervisor

A construction supervisor of mass M = 82. 2 kg is standing on a board of mass m = 30. 5 kg. The board is supported by two saw horses that are a distance of 4. 10 m apart. If the man stands a distance x1 = 1. 066 m away from saw horse 1 as shown, what is the force that the board exerts on that saw horse? I don't get how to calculate the torque related to the weight of the board . Well, I have your answer right here. Draw the board supported on the two horses at the ends of the board. The left hand end named (1), the right hand end named (2) The supervisor stands 1. 066m to the right hand side of (1) and 3. 034m to the left of (2). The uniformly distributed load (the weight of the board) acts through the Centre of Gravity of the board 4. 10/2 m from each support. The support at (1) exerts an upward reaction of R1 kg and the support at (2) exerts an upward reaction of R2 kg Take moments about (1) R1 acting through the point gives rise to no moment We can say [R2]*4. 1 = [30. 5]*[4. 1/2] + [82. 2]*[1. 066] = 62. 525 + 87. 6252 = 150. 1502 kg. M R2 = 150. 1502/4. 1 = 36. 622 kg Take moments about (2) [R1]*[4. 1] = [30. 5]*[4. 1/2] + [82. 2]*[3. 034] = 62. 525 +244. 3448 = 311. 9198 R1 = 311. 9198/4. 1 = 76. 068 kg R1 + R2 = 76. 068 + 36. 622 = 112. 7 kg The sum of the load exerted by the board and the man on the two supports is 82. 2 + 30. 5 = 112. 7 as we would expect. The answer to the question asked is the value of R1 = 76. 068 kg

Excellent experience in construction site management. Can handle complete civil or construction project with complete efficiency. Has more than 6 years of Mi. . .


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1 Comment

  1. Rachel Terrence December 15, 2013 at 1:11 am #
    Whoa. A smart blonde. The CIA is coming for you.