An 80 kg construction worker sits down on a horizontal uniform steel beam of mas

Question

An 80 kg construction worker sits down on a horizontal uniform steel beam of mass 1450 kg and length 6.0 m to eat lunch. The left end of the beam is fixed to a wall by a frictionless pivot. The right end of the beam is held up by a steel cable that makes an angle 35o with the beam and is attached to the wall somewhere above the pivot. The construction worker sits 2.0m from the steel cable’s attachment point. Assume the cable is massless and this situation is static.

(a) What is the tension in the cable?

(b) What is the vector valued force on the beam by the pivot?

Explanation / 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 from horizontal

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