anonymous
  • anonymous
A gymnast with mass 46.0 kg stands on the end of a uniform balance beam as shown in the figure. The beam is 4.90 m long and has a mass of 230 kg (excluding the mass of the two supports). Each support is 0.600 m from its end of the beam. In unit-vector notation, what are the forces on the beam due to (a) support 1 and (b) support 2?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Thank youu
ganeshie8
  • ganeshie8
First key observation here is that the gymnast-beam-supports system is in equilibrium.

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ganeshie8
  • ganeshie8
Therefore the net torque about "any" point must be 0
anonymous
  • anonymous
yes yes
ganeshie8
  • ganeshie8
Assume for now that the forces from supports are as shown : |dw:1449380941308:dw|
ganeshie8
  • ganeshie8
Add up the the torques about the point 1
ganeshie8
  • ganeshie8
do we get \[0*F_1 + 3.7*F_2 -(3.7+0.6)*46g=0\] ?
anonymous
  • anonymous
\sum T\:=\:0\:=-\:\left(F_1\right)d_1\:-\:\left(F_2\right)d_2\:+\:\left(mg\right)d_3
anonymous
  • anonymous
oops
anonymous
  • anonymous
\[\sum T\:=\:0\:=-\:\left(F_1\right)d_1\:-\:\left(F_2\right)d_2\:+\:\left(mg\right)d_3\]
anonymous
  • anonymous
is mine right?
ganeshie8
  • ganeshie8
about what point are you taking the torques ?
anonymous
  • anonymous
my pivot point?
anonymous
  • anonymous
can I set my pivot point to point 1?
ganeshie8
  • ganeshie8
you need to decide the pivot point first
anonymous
  • anonymous
I'm putting it to point 1
anonymous
  • anonymous
I'll identify first the perpendicular distance?
anonymous
  • anonymous
Can I set my pivot point in the other end of the beam?
anonymous
  • anonymous
the opposite of the girl
anonymous
  • anonymous
|dw:1449382286210:dw|
ganeshie8
  • ganeshie8
it doesn't matter, since the system is in equilibrium, the torques must add up to 0 about any point
anonymous
  • anonymous
okay wait will attempt to try my solution
anonymous
  • anonymous
I got 1650.9027 in F1
ganeshie8
  • ganeshie8
how ?
anonymous
  • anonymous
Torque = 0 =(mg)(d1)+(Mg)(d2)-(F1)(d3)
ganeshie8
  • ganeshie8
torque about what point ?
ganeshie8
  • ganeshie8
"torque" makes sense only when you specify the "point" about which you're taking the torque
anonymous
  • anonymous
what I mean is net torque is equal to zero
anonymous
  • anonymous
I took F1 as my pivot point
ganeshie8
  • ganeshie8
what is your pivot point ?
ganeshie8
  • ganeshie8
could you show in the diagram above ?
anonymous
  • anonymous
okay will do
ganeshie8
  • ganeshie8
|dw:1449383404202:dw|
anonymous
  • anonymous
|dw:1449383342814:dw|
anonymous
  • anonymous
my answer in F = ? is 1650.9027
ganeshie8
  • ganeshie8
Oh wait, you're also considering the mass of slab...
ganeshie8
  • ganeshie8
I missed teh mass of slab earlier, your FBD looks correct
anonymous
  • anonymous
Okayyyyy is my F correct?
ganeshie8
  • ganeshie8
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--
anonymous
  • anonymous
okay what about the other support? The one in the leftmost
ganeshie8
  • ganeshie8
work it the same way
anonymous
  • anonymous
okay will try. I'll post it in here when I'm done
ganeshie8
  • ganeshie8
sum the torques about the point 2
ganeshie8
  • ganeshie8
|dw:1449384089263:dw|
anonymous
  • anonymous
im just guessing. Does it mean I will get the same answer or no?
ganeshie8
  • ganeshie8
add up the torques about point2 set them equal to 0 and solve \(F_1\)
ganeshie8
  • ganeshie8
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--
anonymous
  • anonymous
yeap I got it too Thanks man! Appreciate it.

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