10182

# Structural Analysis of a Pyramid

A pyramid with a square base of identical spheres fails structurally for certain coefficients of static friction for the stacking surface. As the number and mass of the balls increase, the mass of the pyramid applies an increasing force to each ball on the bottom layer. This force eventually exceeds the capacity of the frictional force, which is limited by the coefficient of static friction between the floor and the balls. The first points of failure are the balls along the perimeter of the base, excluding the corners; these balls are colored red but change to gray when the component of that is parallel to the floor exceeds the capacity of the frictional force.

### DETAILS

Each red ball supports two balls from the layer above. Each of the balls on the upper layer have the same effective mass , and therefore apply the same force to the balls supporting them. The effective mass of each ball is calculated by separating the base layer from the remaining smaller pyramid. Multiplying the number of balls in the smaller pyramid by the mass of each ball gives the total mass of the smaller pyramid. This mass is distributed evenly over a certain number of points of contact with the base layer. Each ball in the base of the smaller pyramid contacts the base layer in four places. Dividing the number of contact points between a given ball and the smaller pyramid by the total number of contact points creates a ratio. When this ratio is multiplied by the total mass of the smaller pyramid, the effective mass of a specific ball can be calculated. The corners of the pyramid have a lower value for , because the corner balls support only one ball, instead of two.
is the force exerted on a single ball in the perimeter of the base layer (excluding corners) by the two balls it supports. Because there are two balls and the force comes in at a angle from the vertical (clear from the symmetry of the pyramid), multiplying gives the value for at from the vertical. The component of that is parallel to the floor opposes the force of friction . This component of can be expressed as . When F μ<, the pyramid falls.
Interestingly, the mass of each ball does not have an effect on the structural integrity of a pyramid with identical balls. When the mass of the pyramid increases, so does the force of gravity . Because the sum of the forces in the direction perpendicular to the floor is always zero, when increases, must also increase to compensate. , so when increases, so does .

### PERMANENT CITATION

 Share: Embed Interactive Demonstration New! Just copy and paste this snippet of JavaScript code into your website or blog to put the live Demonstration on your site. More details » Download Demonstration as CDF » Download Author Code »(preview ») Files require Wolfram CDF Player or Mathematica.

#### Related Topics

 RELATED RESOURCES
 The #1 tool for creating Demonstrations and anything technical. Explore anything with the first computational knowledge engine. The web's most extensive mathematics resource. An app for every course—right in the palm of your hand. Read our views on math,science, and technology. The format that makes Demonstrations (and any information) easy to share and interact with. Programs & resources for educators, schools & students. Join the initiative for modernizing math education. Walk through homework problems one step at a time, with hints to help along the way. Unlimited random practice problems and answers with built-in Step-by-step solutions. Practice online or make a printable study sheet. Knowledge-based programming for everyone.