# Uniform plank - Supporting force

1. Apr 11, 2010

### mybrohshi5

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

2. Apr 11, 2010

### rock.freak667

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. Apr 11, 2010

### PhanthomJay

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. Apr 11, 2010

### 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

5. Apr 11, 2010

### rock.freak667

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.

6. Apr 11, 2010

### mybrohshi5

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:$\mathbf{\tau}= \mathbf{r} \times \mathbf{F}$

7. Apr 11, 2010

### PhanthomJay

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

8. Apr 11, 2010

### 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

9. Apr 11, 2010

### rock.freak667

Watch your sign convention, if clockwise moments are +ve, then anticlockwise moments are -ve.

10. Apr 11, 2010

### mybrohshi5

Ok thank you :)