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enter image description hereI have a light hanging from an arm that is attached to the wall. I need to make sure that the fastener at the wall end is able to hold the weight. How do I calculate this?

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    $\begingroup$ It's a lever but L-shaped, where one side of the fulcrum is 27" and the other side is the distance from the fastener to the bottom of the bar in the thickness direction. $\endgroup$
    – DKNguyen
    Aug 8, 2022 at 19:32
  • $\begingroup$ What is the arm made of, what is the arms weight, and is it uniform in cross-section? Is the light hanging from the electrical cord or is it a stout support (chain)? What is the weight of the stout support? $\endgroup$
    – Jim Clark
    Aug 8, 2022 at 19:33
  • $\begingroup$ There will be compression on drywall due to length below the fastener which is TBD insufficient and pressure on drywall due to lower surface area. No doubt it must be put into a stud. F1R1=F2R2 $\endgroup$ Aug 8, 2022 at 20:20
  • $\begingroup$ This is a design typically put forth by designers who go to a design/art school. So typical to furniture designers etc. The problem here is that it typically neglects how a screw is designed to work. Essentially the screw should operate only on foces along the length of the screw leaving the other forces to friction between the mating surfaces. One should try avoid the situation depicted. Essentially now the important bit becomes what kind of material the wall is or what the wall anchor looks like. But this will work if the connection in the wall is strong enough and the screw is long enough. $\endgroup$
    – joojaa
    Aug 9, 2022 at 6:15

2 Answers 2

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Convert mass and radius to Torque. The force would be distributed between fastener wall contact areas unevenly with very high pressure on the bottom edge puncturing the wall with the same torque but increased force due to the reduction in radius ratio.

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This needs a large brace to work., as well as a rigid area for the fastener to withstand the pressure on the wall.

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  • $\begingroup$ However, even if the wall material can withstand the force at the bottom of the arm, there's also (basically the same magnitude) force trying to pull the screw itself out of the wall. That's why anchor bolts are highly recommended for situations like this. $\endgroup$ Aug 9, 2022 at 14:31
  • $\begingroup$ Thanks for all those helpful responses. :) $\endgroup$
    – Mike R
    Aug 9, 2022 at 20:34
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Say the arm is 2 inches vertically at the base where it is hung by the fastener and the distance between the fastener and the point of pressure is 1.5 inches.

$$F_{fastener}=\frac{8lbs*27in}{1.5in}=144lbs$$

This means your fastener is being pulled out by a nominal force of 144# and the wall on the 1.5 in below must resist a nominal compression force of 144#.

Both the tension and compression should be multiplied by a safety factor of say 1.8.

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  • $\begingroup$ Notice also that increasing the vertical dimension of the arm at the wall decreases the torque, an further increasing the horizontal dimension decreases the force per unit area. $\endgroup$ Aug 9, 2022 at 14:35

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