26 November, 2013

Construction Supervisor

A construction supervisor of mass M = 80. 0kg is standing on a board of mass m = 30. 0kg. Two sawhorses at a distance L = 4. 00 m apart support the board. If the man stands a distance x = 1. 25m away from sawhorse 1, what is the force that the board exerts on the other sawhorse. After looking around, I learned — Sum of torques about any chosen pivot axis = 0. Let us choose sawhorse 1 as the pivot axis. That way, F1 drops out of our equation: 1. 25m*80kg*9. 8 + 2m*30kg*9. 8 = 4m*F2 where F2 = force on sawhorse 2. F2 = 392 N answer

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. Mrs Masterson November 25, 2013 at 1:57 am #
    = 0sawhorse 1 as a reference point ( point A). Assuming the force that the board exerts on the other sawhorse =F A = 0 -> F * 4 - 300 * 2 - 800 * 1. 25 = 0 4F - 600 - 1000 = 0 ---> F = 1600/4 = 400 N