5. An 80 kg beam of length 4.00 m rests on two support points as shown in the fi
ID: 1285730 • Letter: 5
Question
5. An 80 kg beam of length 4.00 m rests on two support points as shown in the figure. The beam has a length of 4.00 m, and a uniformly distributed mass. Support point 1 is
1.50 m from the left end and support mmmmmmmmmmmmmmmmmmmmmmmm
1 2
point 2 is exactly on the right end.
a) What is the force that each support is applying to the beam?
b) A man of mass 60 kg wants to stand on the beam. At what distance from the left side of the beam must he stand so that each support is applying an equal amount of force to the beam?
c) The man begins walking to the left on the beam. At what distance from the left side of the beam is he standing when the beam begins to tip over?
Explanation / Answer
let F1 and F2 are the forces exerted by supports.
in the equilibrium net torque is zero about any point must be zero.
net torque about point 1 = 0
(2 - 1.5)*m*g - F2*2.5 = 0
F2 = 0.5*m*g/2.5
= 0.5*80*9.8/2.5
= 156.8 N <<<<<<------------Answer
net torque about point 2 = 0
2*m*g - F1*2.5 = 0
F1 = 2*m*g/2.5
= 2*80*9.8/2.5
= 627.2 N <<<<<<------------Answer
b) let x is the distance from left end.
let F1 = F2 = F
2*F = (80+60)*9.8
= 1372 N
so, F = 686 N
net torque about center of beam = 0.
60*9.8*(2-x) - 686*0.5 + 686*2 = 0
(2-x) = 686*1.5/(60*9.8)
2 - x = 1.75
x = 2 - 1.75
= 0.25 m <<<<<----------Answer
c)
net torque about point 1 = 0
60*9.8*(1.5-x) - 0.5*80*9.8 = 0
(1.5-x) = 0.5*80*9.8/(60*9.8)
= 0.667 m
x = 1.5 - 0.6667
= 0.833 m <<<<<----------Answer
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