How can you balance a 44ft beam with 22kg on the front end.?
Answer:
let:
w1 - 22kg. front end
w2 - 25kg. beam
L - 44ft. length of beam
w - weight needed on the rear end
get center of gravity
wf - center of gravity from fulcrum to front end
wr - center of gravity from fulcrum to rear end
wf=(40ft.*25kg)/44ft.
=22.72kg
wr=w2-wf
=25kg-22.72kg
=2.27kg
equate the summation of force*distance to zero
0=(4ft)w+wr(2ft)-wf(20ft)-w1(4...
(4ft)w=w1(40ft)+wf(20ft)-wr(2f...
=22kg(40ft)+22.72kg(20)-2.27kg... /4ft
=880kg.ft+454.4kg.ft-4.54kg.ft / 4ft
w= 332.465kg
find the fulcrum...
First, let's talk like civilized people:
44 feet = 13.4 m
Because of gravity mass acts like a force (weight) but remember that we are talking about forces, in newtons, N, not masses. (F = m*g)
22*9.8 = 215.6 N
25*9.8 = 245 N
Draw the diagram.
Mark the distances.
13.4 m
6.7 m (half)
4 feet = 1.2 m
12.2 m (6.7 + (6.7-1.2)
Draw F (must be down)
We write the sum of momentum in the fulcrum:
215.6*12.2 + 245*5.5 - F*1.2 = 0
F = 3315 N
(you'll get 332.5 kg with the initial numbers)
For the beam to be supported, the moment of force on the rear end should be zero, meaning it should be in equilibrium or the upward and downward forces should balance out.
You have the following forces:
downward: (load)
1) 22 kgs, 44 ft from the rear
2) 25 kgs beam weight, 22 ft from rear (i.e. at the center)
upward: (support)
1) unknown, 4 ft from the rear
Moment = force* distance = 0 on the rear
(i'm not going to convert ft to metric system since the unit will cancel out, but i'll be consistent in using ft as unit of length)
0 = 22(9.81)*44 + 25(9.81)*22 - 4x
(note: g=9.81 m/s2)
4x = 14,891.58
x = 3,722.895 N (force) to suport the beam 4ft from rear
or divide it by 9.81:
x = 379.5 kgs (mass)
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