# I I need to find out the forces acting on an anchor point

1. Apr 15, 2017

### Davem27

Hello all. i am after some help and hopefully you are the right people for this! i have a problem will a kitchen wall cabinet. i can only get fixings which rate at 40kg, but i work out my cabinet fully loaded (including self weight) could be 95kg. i have attached a drawing which hopefully explains my theory?! Any opinions or calcs would be most welcome!!

2. Apr 15, 2017

### Davem27

Additional info which i forgot! The cabinet is 724mm high, 1200mm wide and 350mm deep. It is fixed at 4 points along the back wall, 95mm from the top. Each fixing point is rated as 10kg.
Thanks!

3. Apr 15, 2017

### CWatters

I believe this problem is indeterminate. You can't tell how much of the weight is carried by the wall fixings or the board.

What is the wall made of? If it's a plasterboard/stud wall consider fixing a batten to the wall studs first and then fixing the cabinet to that.

4. Apr 15, 2017

### Davem27

Thanks for your reply. The wall is 12.5mm plasterboard. I would like to put a batten in the wall but this is not possible. The building is steel framed and fire protected so no timber can be placed in the cavity.
I am sure there must be a way to work this. my thinking is the 10mm board acts as a fulcrum so the force is like a lever. my question from another angle is what force is required to move the 40kg fixing?

5. Apr 15, 2017

### sophiecentaur

You need to know the weight distribution of the cabinet. Is there a heavy door, for instance? If you can deduce where the centre of mass lies (X,Y co ordinates then your ideas about levers and a fulcrum are along the right lines. Treat the contact point with the vertical board as the fulcrum and take moments about it. You have the Weight force X the distance from the wall of the CM (that will be the clockwise moment) and the Force on the anchor point (your unknown) X its distance from the fulcrum. You can say they make two sides of an equation with one unknown, which is easily solved.
Weight force is Mass X g (=950N near enough)
If this is a real practical problem, you can find where the CM lies by finding where it balances, when it's vertical.
Note: one force is horizontal and the other force is vertical but this is fine for calculations with Moments.

6. Apr 15, 2017

### CWatters

I wasn't suggesting putting timber in the cavity. Just fix the batten through the plasterboard into the metal frame? Or isn't the back of the cabinet recessed?

Could also replace the plaster board with a stronger cement board?

7. Apr 16, 2017

### Davem27

Thanks CWatters. This was a thought but we couldn't hit enough steel studs to comply with the FIRA regs we need to.
Sophiecentaur. Thanks for your response. Thats a great answer which i will work with. Much appreciated!

8. Apr 16, 2017

### Davem27

Having researched further,i am perplexed!! can anyone help?! i am sure it is a simple equation, but it has me stumped! i have updated my drawing with further info that may help!!

9. Apr 16, 2017

### sophiecentaur

Now you have the measurements (in particular, the position of the CM) you can use the method in my post #5. You have the relevant distances and the weight force. Write out the equation on paper. On the next line, put in the values you know. Multiply the two numbers on the left and divide by the number on the right (the vertical distance from the fulcrum to the anchor point. The answer will be the Force F you want.
It should come to less than the Weight (950N).
950X200=F X (724-95)
You are asking for a Force in kg. The SI system doesn't do forces in kg. The strengths of materials and components are always quoted in Newtons. But it's easy to remember that 1kg weighs 10N. The shear force (downwards) on that anchor could be anything between zero and 950N, depending on the sharing of the weight between it and the lower pivot. If it is screwed in with some slop, it will be zero when the cabinet has settled onto the lower pivot; the only force will be pulling it out. Value, roughly one third of the weight because of the leverage. (ratio of the two lengths involved).
If 300+N seems a lot for the plasterboard, use a number of fixings to divide the force by two or three. All good fixings have their strength specified and so will plasterboard. Don't forget to take into account the intended contents of the cabinet. (Full of gold bars?) The loads could easily be several times what y ou calculate - even if it's just books or paint cans. [Edit - sorry, I realise you've done that already]

10. Apr 16, 2017

### Davem27

Thank you very much. You put it very logically, and i can certainly follow that! The cabinet's fully laden weight including the self weight is 95kg (actually less, but i have over specified it). the load is spread over 4 fixings which rate at 10kg per fixing.

11. Apr 16, 2017

### sophiecentaur

You just need to ascertain whether this 40kg rating refers to shear load or perpendicular load. That is a relevant concern. Hope it's good enough for your purpose.