12 points SerCP11 8.P026. A window washer is standing on a scaffold supported by

Question

12 points SerCP11 8.P026. A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs 210 N and is 2.8 m long. What is the tension in each rope when the 696-N worker stands 1.14 m from one end? smaller tension arger tension Need Help?Read It My Notes Ask Your Submit Answer Save Progress Practice Another Version 5. -15 points SercP11 8.P028 My Notes Ask Your A hungry bear weighing 745 N walks out on a beam in an attempt to retrieve a basket of goodies hanging at the end of the beam (see the figure blow). The beam is uniform, weighs 200 N, and is 5.50 m long, and it is supported by a wire at an angle of = 60.00. The basket weighs 80.0 N. Goodies (a) Draw a force diagram for the beam. (Submit a file with a maximum size of 1 MB.) Choose Fille No file chosern This answer has not been graded yet (b) when the bear is at x = 1.12 m, find the tension in the wire supporting the beam. When the bear is at x1.12 m, find the components of the force exerted by the wall on the left end of the beam. (Assume the positive +x direction is to the right and the positive ty direction is upward. Include the sign of the value in your answer.) (c) If the wire can withstand a maximum tension of 775 N, what is the maximum distance the bear can walk before the wire breaks? Tm Need Help? Readi

Explanation / Answer

1)

assuming the scaffold is uniform weight, take moments around 1 end to get the tension in the rope of the other
the M = 0, take the rope tension direction as positive
696 x 1.14 + 210 x (2.8/2) = T1 x 2.8
T1 = 388.4 N

Now moments around the other end
696 x (2.8 – 1.14) + 210 x (2.8/2) = T2 x 2.8
T2 = 517.6 N

You can check your answers as F = 0, so
388.4 + 517.6 - 210 - 696 = 0

2)

Draw a free body diagram of the beam. From this, you can get the following three equations:

Fx = Tcos(60)
Fy + Tsin(60) = 745 + 200 + 80
Tsin(60)*5.5 = 750*1.12 + 200*2.75 + 80*5.5 (sum of the moments about the left end of beam)

solving these three equations for T, Fx, and Fy (in that order) you get
T = 384.2 N Fx = 192.1 N, Fy = 692.3 N

C) to find the max distance, solve the moment equation again substituting 775 N for T and x for 1.12

Use the same equation, but solve for x instead of T:
M = 0 = 775*5.5*sin60 - 745*x - 80*5.5 - 200*5.5/2
0 = 2701.4 – 745*x
x = 3.63 m

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