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First key observation here is that the gymnast-beam-supports system is in equilibrium.
Therefore the net torque about "any" point must be 0
Assume for now that the forces from supports are as shown : |dw:1449380941308:dw|
Add up the the torques about the point 1
do we get \[0*F_1 + 3.7*F_2 -(3.7+0.6)*46g=0\] ?
is mine right?
about what point are you taking the torques ?
my pivot point?
can I set my pivot point to point 1?
you need to decide the pivot point first
I'm putting it to point 1
I'll identify first the perpendicular distance?
Can I set my pivot point in the other end of the beam?
the opposite of the girl
it doesn't matter, since the system is in equilibrium, the torques must add up to 0 about any point
okay wait will attempt to try my solution
I got 1650.9027 in F1
Torque = 0 =(mg)(d1)+(Mg)(d2)-(F1)(d3)
torque about what point ?
"torque" makes sense only when you specify the "point" about which you're taking the torque
what I mean is net torque is equal to zero
I took F1 as my pivot point
what is your pivot point ?
could you show in the diagram above ?
okay will do
my answer in F = ? is 1650.9027
Oh wait, you're also considering the mass of slab...
I missed teh mass of slab earlier, your FBD looks correct
Okayyyyy is my F correct?
Yep! I am getting the same : http://www.wolframalpha.com/input/?i=solve+-1.85*230*9.8+%2B+3.7*F+-+%283.7%2B0.6%29*46*9.8%3D0&a=*C.F-_*Variable-&a=UnitClash_*F.*DegreesFahrenheit.dflt--
okay what about the other support? The one in the leftmost
work it the same way
okay will try. I'll post it in here when I'm done
sum the torques about the point 2
im just guessing. Does it mean I will get the same answer or no?
add up the torques about point2 set them equal to 0 and solve \(F_1\)
I am getting \(F_1 = 1053.9\) http://www.wolframalpha.com/input/?i=solve+-0.6*46*9.8+%2B+3.7%2F2*230*9.8+-+3.7*F+%3D+0&a=*C.F-_*Variable-&a=UnitClash_*F.*DegreesFahrenheit.dflt--
yeap I got it too Thanks man! Appreciate it.