An 80 kg construction worker s
An 80 kg construction worker sits down on a horizontal uniformsteel beam of mass 1450 kg and length 6.0 m to eat lunch. The leftend of the beam is fixed to a wall by a frictionless pivot. Theright end of the beam is held up by a steel cable that makes anangle 35o with the beam and is attached to the wall somewhere abovethe pivot. The construction worker sits 2.0m from the steel cable’sattachment point. Assume the cable is massless and this situationis static.
(a) What is the tension in the cable?
(b) What is the vector valued force on the beam by thepivot?
Answer:
Forces up and down
80*9.8+1450*9.8= T sin 35+ Npy
Forces right and left
T cos35= Npx
Moments about the pivot
T sin35*6 = 1450*9.8*3+ 80*9.8*4
a) T= 79790.6 N
b) 80*9.8+1450*9.8= T sin 35+ Npy
Npy= -30771 N (Downwards)
Nx=T cos35 = 65360.14 N
Net force= sqrt( 30771^2+65360^2) = 72241.28 N
theta= tan^-1(30771/65360) = 25.21 degrees downwar fromhorizontal
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