# Tension force and a board fixed to a flat surface

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1. Aug 5, 2015

### amateur111

A heavy board is fixed to another flat surface along its entire upper edge and not moving. How can I find the tension force that is acting on the board? Is it the same as if the board would be fixed to a string in the middle of the upper edge and hung and then the tension would be T=mg?

2. Aug 5, 2015

3. Aug 5, 2015

### PhanthomJay

note that if the board were supported by a string, the string tension would be mg but the resultant tension force in the board would be variable from mg at top to 0 at bottom.

4. Aug 6, 2015

### amateur111

I have uploaded a sketch to make it more clear. A board is attached to the wall along its upper edge as the red line indicates. So the point is how can I determine the tension force that is acting upon the board? As PhantomJay has pointed out the tension force in the board is changing but is there such a thing as the total tension? Because I want to compare two situations: the first one that is shown in the picture and the second would be when everything is inclined to the left and there is a certain angle alpha between the supporting wall and the ground (the board would be like a static body on the slope attached in the same way). Does the tension acting upon the board changes by a factor of sin(alpha)?

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5. Aug 7, 2015

### PhanthomJay

well I don't think you can quite call it a tension force in the board because of the eccentric attachment detail, but in the position shown, ignoring eccentricity, the top edge of the board sees an internal axial force of mg and the bot edge has no internal force, and at the cg, the tensile force is mg/2.. I'm mot sure what you mean by total tension: if the top edge attachment were considered a fat string, the tension in the string would be mg but the board tension would still vary as noted above.

Now if you lower the wall to the extreme horizontal flat (floor) position, there is no longer any tensile load, and in between at angle alpha, yes, sure, assuming no friction, tension force at board 'top' is mg sin alpha, where alpha is measured with respect to horizontal ground.

6. Aug 8, 2015

### CWatters

Is there friction between the board and wall? If so I have a feeling this is statically indeterminate. eg I don't think you can work out what fraction of the weight of the board is supported by friction vs the attachment point.

If there is friction... When vertical the off set weight of the board creates a torque about the attachment and hence a normal force.

7. Aug 8, 2015

### CWatters

If we assume there is no friction between the board and wall and the board is thin compared to it's other dimensions then..

1) With the wall & board vertical the force on the attachment will be = mg.
2) With the whole assembly tilted to the left at say 30 degrees to the horizontal the force on the attachment acting down the slope will be = mgSin(30)
3) As the angle approaches horizontal the term Sin(angle) approaches zero so the force on the attachment approaches zero.

8. Aug 9, 2015

### amateur111

The situation I described is like, for example, a thin but heavy picture is glued to the wall so I was thinking if it would be possible to reduce the tension by tilting the supporting wall so that the weight of the object wouldn't damage the area that is glued to the supporting wall. So I believe there is no friction that could influence the results and as you are saying it all comes down to the term sin(alpha). Thanks a lot for helping me!

9. Aug 9, 2015

### stedwards

Assuming the attachment along the upper surface acts as a hing, and everything else is ideal:

The vertical force due to mass is $mg$.

The torque about the hing line is $(1/2)t mg$. This places a force, normal to the supporting wall, of zero at the hing line and maximum at the lower edge. The average normal force is $(1/2)t mg / [(1/2)h] = (t/h)mg$, where $h$ is the height of the board. (The vertical extent of the board.) The upward acting force from static friction is $\mu(t/h)mg$.

The resultant downward force is $mg(1- \mu(t/h)$ where $\mu$ is the coefficient of static friction.

Last edited: Aug 9, 2015
10. Aug 10, 2015

### PhanthomJay

Word of caution regarding the glued attachment: the glue is primarily in shear , not tension, and many glues are notoriously very weak in shear. Don't want that heavy pic come crashing down .

11. Aug 10, 2015

### CWatters

But that's the maximum friction force. The actual friction force could be less. It's not possible to know how the weight of the plate is shared between the attachment and friction.

Consider what happens if you start with the wall vertical and the plate has been swung out the right by 1 degree so that it's almost vertical but not quite in contact with the wall. None of the weight is carried by friction. Now reduce the angle from 1 to 0 degrees. Why would friction suddenly start carrying some of the weight?

12. Aug 10, 2015

### CWatters

Depending on how you hang the plate friction could act downwards rather the upwards giving you a worse (worst) case greater than mg.

13. Aug 10, 2015

### stedwards

Yea, this bothers me somewhat, too. The It's seems more a potential to carry weight.