# Uniform plank - Supporting force

by mybrohshi5
Tags: force, plank, supporting, uniform
 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) = Fy(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
 HW Helper P: 6,207 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?
 Sci Advisor HW Helper PF Gold P: 5,920 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?
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
HW Helper
P: 6,207
 Quote by mybrohshi5 The force at the center would be (120N + 120N) = 240N correct? Is the equation used for this dealing with torque? thanks for the help
No, now you have the forces. So now you have force at the center. Now move X to Z (I don't know if you have that distance) and just take moments about any point. For example, take moments about the new position of X i.e. Z.
 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:$\mathbf{\tau}= \mathbf{r} \times \mathbf{F}$
 Sci Advisor HW Helper PF Gold P: 5,920 the terms torque, moment, or moment of a force, are often used interchangeably.
 P: 365 Oh ok thanks for the insight on that :) does this seem right. if i take the torque about point Z then tcenter + ty = tz (240N)(1/4L) + (Fy)(3/4L) = 0
HW Helper
P: 6,207
 Quote by mybrohshi5 Oh ok thanks for the insight on that :) does this seem right. if i take the torque about point Z then tcenter + ty = tz (240N)(1/4L) + (Fy)(3/4L) = 0
Watch your sign convention, if clockwise moments are +ve, then anticlockwise moments are -ve.
 P: 365 Ok thank you :)

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