Years ago I built a vertical rack to store long lengths of wood. The rack was built using 2x4" pine scrap. Imagine a typical wall assembly with a single top and bottom "plate" (2x4) held together with vertical members spaced approximately 2' apart.

My shop is a unit within a steel framed warehouse and I have not measured the actual gauge of the steel members I have attached to (I know, it's an important detail). Because I slapped this thing together and wanted to quickly get some wood up and out of the way I just used materials I had around the shop which resulted in my using 5" ledgerlock screws drilled through the top plate of my rack, through the sheetrock and ultimately through the steel framing members.

Each rack is approximately 12' wide and 8' tall. It sits on the ground and is held upright with the ledgerlock screws attaching the top plate to the wall. Having forgotten my physics lessons from years ago, my question is: How much outward force is being applied to the screws if, say, 300 pounds of wood is being held 16" away from the wall on each of the 4 horizontal "shelf" arms.

As I said before, these racks were built years ago and have carried a lot of lumber without fail. I recently rearranged the shop and joined two of these racks together with yet another primitive scrap-built shelf to carry even longer bits of wood across the two rack system and over my jointer. Now that I have substantially more lumber on each rack and because I work right beside them, I decided to think a little more about the integrity of the racks. I'm fairly confident it will kill me if the screws fail and the lumber comes crashing down on my head.

  • $\begingroup$ we need a diagram. My guess is that the weak point is the screw holding in the framing. $\endgroup$
    – Tiger Guy
    Jan 16, 2023 at 16:26

1 Answer 1


I think you just need to do moment equilibrium. There will be a horizontal force couple acting on the vertical members, which can be quantified as a product of the unknown force acting on a screw, screw distance from the ground and the number of screws. Against this, the weight of the wood multiplied by the distance from the wall will also create a moment;

$$F_{screw} \cdot 8' \cdot n_{screw} = 4\cdot 300 lb\cdot 16''$$

So the force on each screw is:

$$F_{screw} = \frac{4\cdot 300 lb\cdot 16''}{8' \cdot n_{screw}}$$

There was a similar problem I wrote an answer to a while ago, which you may find useful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.