
#1
Apr1110, 06:07 PM

P: 365

1. The problem statement, all variables and given/known data
A uniform plank is supported by two equal 120 N forces at X and Y. The support at X is then moved to Z (halfway to the plank center). The supporting force at Y then becomes? _____________________________ X...........Z...........c......................Y i tried to draw it above. The line is the plank, X and Y are the supports, Z is where X is moved to, and (c) is the center of the plank 3. The attempt at a solution I came across this as i am studying for my test and i am a little lost. The solution says to do this: (240 N)(1/4L) = F_{y}(3/4L) and the answer is 80 N I am confused on where this solution comes from. Any explaining would be very helpful :) Thank you 



#2
Apr1110, 07:16 PM

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P: 6,213

Well start off with X and Y being at the ends. If the system is in equilibrium there, what should be the force acting at the center c?




#3
Apr1110, 07:16 PM

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P: 5,963

you are given the initial reactions; what must be the weight of the plank, and at what point does the resultant force of the plank's weight act?




#4
Apr1110, 07:51 PM

P: 365

Uniform plank  Supporting force
The force at the center would be (120N + 120N) = 240N correct?
Is the equation used for this dealing with torque? thanks for the help 



#5
Apr1110, 08:17 PM

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#6
Apr1110, 08:31 PM

P: 365

I dont have the distance so i believe the plank length is considered L.
I am not sure what you mean by moment. When you say moment do you mean Moment of force = Torque:[itex]\mathbf{\tau}= \mathbf{r} \times \mathbf{F}[/itex] 



#7
Apr1110, 08:41 PM

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the terms torque, moment, or moment of a force, are often used interchangeably.




#8
Apr1110, 08:56 PM

P: 365

Oh ok thanks for the insight on that :)
does this seem right. if i take the torque about point Z then t_{center} + t_{y} = t_{z} (240N)(1/4L) + (F_{y})(3/4L) = 0 



#9
Apr1110, 08:57 PM

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#10
Apr1110, 09:41 PM

P: 365

Ok thank you :)



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