ganeshie8
  • ganeshie8
Last question on static equilibrium For the stepladder shown, sides AC and CE are each 2.44 m long and hinged at C. Bar BD is a tie-rod 0.762 m long, halfway up. A man weighing 854 N climbs 1.80 m along the ladder. Assuming that the floor is frictionless and neglecting the mass of the ladder, find (a) the tension in the tie-rod and the magnitudes of the forces on the ladder from the floor at (b) A and (c) E. (Hint: Isolate parts of the ladder in applying the equilibrium conditions.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
|dw:1446721512178:dw|
ganeshie8
  • ganeshie8
I am getting too many variables... not sure about what point I should balance the torques to make it easy..
ganeshie8
  • ganeshie8
|dw:1446725528485:dw|

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ganeshie8
  • ganeshie8
I have shown normal forces from ground, tension, and the weight in above diagram not sure about the point C...
ParthKohli
  • ParthKohli
Actually with the forces you've shown, the second part of the ladder can definitely not be in rotational equilibrium. You'll probably have to add another force exerted on the rod at the hinge.
ParthKohli
  • ParthKohli
Oh, you did add that to your post. Yeah.
ganeshie8
  • ganeshie8
like this ? would it also be like tension ? |dw:1446725812984:dw|
ParthKohli
  • ParthKohli
Again, I'm not sure. But it's not like that because it still means that there is some net torque on the ladder. :|
ganeshie8
  • ganeshie8
yeah, about C, right ?
ParthKohli
  • ParthKohli
It balances out the normal reaction and the tension so it's downwards and to the right.
ParthKohli
  • ParthKohli
It's easier to see that there'd be net torque when you take point B. Two forces are passing through B and the normal reaction would be the only force contributing to the torque.
baru
  • baru
|dw:1446726102185:dw|
ParthKohli
  • ParthKohli
Exactly. \(F_{C} = T \hat i - N \hat j \)
ganeshie8
  • ganeshie8
Ahh, I see...
baru
  • baru
perhaps that way, becuse the hint says consider them seperately, force at C is the force with which the other leg is pushing
ganeshie8
  • ganeshie8
hope this is good |dw:1446726251410:dw|
ParthKohli
  • ParthKohli
|dw:1446726302700:dw|
ParthKohli
  • ParthKohli
\[N_2 \hat j - T \hat i + \vec{F_c}=0\]\[\Rightarrow \vec{F_c} = T \hat i - N_2 \hat j \]Now we probably balance torques?
ParthKohli
  • ParthKohli
The problem with torques is there is way too much geometry/trig involved for a lazy person like me.
ganeshie8
  • ganeshie8
Ohk, so the force at C could be in any direction... makes sense..
mathmate
  • mathmate
|dw:1446725424660:dw|
ganeshie8
  • ganeshie8
I think thats the only useful equation that we can get by balancing forces
ganeshie8
  • ganeshie8
\(854 = F_{ay}+F_{cy} \tag{1}\)
mathmate
  • mathmate
for some reason my drawing for the right leg is not yet showing. You can do three things on the left leg: sum Fx=0, sum Fy=0 sum moments (torques) =0
mathmate
  • mathmate
And on the right leg, by taking moments, you can relate T with Fcy.
mathmate
  • mathmate
|dw:1446727394221:dw|
mathmate
  • mathmate
I'll have to get some chores done and will be back. That leaves you some "quiet" time to work on it! lol
mathmate
  • mathmate
All in all, we need to find Rc, T, since you have already got R in terms of W and Rc.
ganeshie8
  • ganeshie8
I'll try to setup the equations, hope i don't mess up... thanks a lot :)
mathmate
  • mathmate
ttyl!
ganeshie8
  • ganeshie8
Left leg : vertical forces : \(854 = F_{ay}+F_{cy} \tag{1}\) horizontal forces: \(F_{ch}=T\tag{2}\) torque about C : \(854*0.64\sin(18.2) + T*1.22\cos(18.2)-F_{ay}*2.44\sin(18.2)=0\tag{3}\)
ganeshie8
  • ganeshie8
Right leg : vertical forces : \(F_{ey}-F_{cy}=0 \tag{4}\) horizontal forces: \(F_{ch}=T\tag{5}\) torque about C : \(-T*1.22\cos(18.2)+F_{ey}*2.44\sin(18.2)=0\tag{6}\)
ganeshie8
  • ganeshie8
Looks we have it ! http://www.wolframalpha.com/input/?i=solve+solve+854+%3D+a%2Bb%2C+854*0.64%5Csin%2818.2%29+%2B+T*1.22%5Ccos%2818.2%29-a*2.44%5Csin%2818.2%29%3D0%2C+-T*1.22%5Ccos%2818.2%29%2Bb*2.44%5Csin%2818.2%29%3D0&a=TrigRD_D
ganeshie8
  • ganeshie8
\(F_{ay}=539N\) \(F_{ey}=315N\) \(T=207N\)
mathmate
  • mathmate
Yep, exactly what I got, I have T=207.1042, and Rc=Fey=315.0000, R=Fay=539.0000 N. Yay! :)
ganeshie8
  • ganeshie8
matches with textbook answer too xD thanks a lot !

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