A young work crew is trying to quickly finish some touch-up painting to complete

Question

A young work crew is trying to quickly finish some touch-up painting to complete a job. Rather than setting up a secure work area, they decide to try a shortcut. Worker A will hold the far end of a 9.47 m work plank which weighs 112 N and is secured with a hinge on the opposite end. Worker B will stand on the plank 4.55 m from worker A. Worker B also brings up his paint can, mc = 3.46 kg, which he places 0.490 m from where he is standing on the plank (between the two workers). Worker A unfortunately needs to supply 4.70 × 102 N of force to hold the end of the plank. How much does worker B weigh? The acceleration due to gravity is g = 9.81 m/s2.

Explanation / Answer

Since plank is in equilibrium ,Net force acting and net torque actiong is zero

torque due to (Worker A + workerB + can + weight of the plank) = 0

(FA*rA)+(FB*rB) +(Fc*rC)+(Fw*rD) = 0

rA,rB and rC are the distance from the fixed end to the worker A ,worked B and can respectievly

rA = 9.47 m

rB = 9.47-4.55 = 4.92 m

rC = 4.92+0.49 = 5.41 m

rD = 9.47/2 = 4.735 m

Tnet = (FA*rA)+(FB*rB) +(Fc*rC)+(Fw*rD) = 0

(-4.7*10^2*9.47)+(W*4.92)+(3.46*9.8*5.41)+(112*4.735) = 0

weight of worker B = 759.58 N

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