# Torque on a shelf

Say I have a shelf anchored to the wall (how it's anchored doesn't matter here I think), with a mass (m) taking up most of the space on the shelf:

Say the shelf is x meters long, and the mass is y meters long, such that y < x, and (x - y) << x. The mass is pushed right up to the edge of the shelf. Assume the shelf itself is weightless, and that it won't break away from the wall, fracture, or move in any way.

Question: How do I figure out the torque exerted on point A, where the shelf meets the wall?

I know I used to know how to do this, but there's enough missing from my previous knowledge that I can't figure it out and I can't form a good enough search query to find more information.

I know I could find the torque on A if the mass was skinny enough that the distance y was trivial, it would be m * x. Given that y is not trivial, my brain tells me there should be an integral involved, but I can't figure out what it should be.

• You're probably remembering the beam deflection calculations. – JMac Oct 31 '18 at 11:25
• Does the mass have uniform distribution for weight? – ja72 Oct 31 '18 at 13:08
• @ja72 Yes, uniform distribution of weight. – OneWholeBurrito Oct 31 '18 at 16:29
• The official method involves taking the load distribution curve and finding its centroid. Then apply the total load through the centroid. For uniform distribution, the centroid is at $x-y/2$ from the support. – ja72 Oct 31 '18 at 17:29

Your intuition is right, the solution does involve an integral as shown in PiKindOfGuy's solution. The expression he derives for torque simplifies to $$mg(x-\frac y2)$$ If you compare this to the basic equation for torque $$\mathbf\tau = r \times F$$ You can see that you can treat the weight of the extended object as a single force acting at the center of gravity. In this case we're assuming the mass of the object is evenly distributed so the center of gravity is just the geometric center which would be located $$\frac y2$$ meters from the end of the shelf.

• Good simple answer and good explanation. This answer could be improved by explaining how PiKindOfGuy's solution simplifies to mg(x-(y/2)). – OneWholeBurrito Oct 31 '18 at 16:45
• I included the simplification in my answer. I didn't notice it before. (Yikes!) – PiKindOfGuy Oct 31 '18 at 22:25

Let $$z$$ be the distance from the point A and divide the mass into columns of width $$dz$$. Then the torque due to each column is given by $$dN = dmgz = \dfrac{dz}{y}mgz$$. Thus the torque is $$N = \frac{mg}{y}\int_{x - y}^x z \; dz = \frac{mg}{2y}(2xy - y^2) = mg\left(x - \frac{1}{2}y\right).$$

• I mostly follow this, but could you further explain how you get from dN = dmgz to dN = (dz/y)mgz? – OneWholeBurrito Oct 31 '18 at 16:41
• dm = (dz/y)m because dm is a fraction of m and that fraction is dz/y = 1/(the number of dz's that fit into y). – PiKindOfGuy Oct 31 '18 at 22:22

Given that y is not trivial, my brain tells me there should be an integral involved.

Actually, there is no need to an integral - you can just use the COG of the mass. Your resulting torque will be $$\tau=m(x-\frac y 2)$$.

It is easy to show why this approach works.

For instance, if we have two masses, $$m_1$$ and $$m_2$$, at distances $$d_1$$ and $$d_2$$, respectively, from the axis of rotation, the combined torque will be:

$$\tau=m_1 d_1+m_2 d_2$$

Multiplying and dividing by $$m_1+m_2$$, we get:

$$\tau=(m_1+m_2) \frac {m_1 d_1+m_2 d_2} {m_1+m_2}=md$$

where $$m=m_1+m_2$$ is the total mass and $$d=\frac {m_1 d_1+m_2 d_2} {m_1+m_2}$$ is, by definition, the COG of the total mass.

So, we can conclude that the torque of a distributed mass is equivalent to the torque of a point mass with the same mass value as the distributed mass, located at the COG of the distributed mass.

• Thanks for the alternate explanation as to why I can just use the COG here. – OneWholeBurrito Oct 31 '18 at 16:39
• @You are welcome. – V.F. Oct 31 '18 at 17:04