Diagonal Forces

For science teachers, the central idea behind building an effective Paper Bridge is vectors. All bridges must deal with forces in two dimensions -- the load is vertical, while the bridge is stretched between two horizontally, separated support points. This means that the paper has to bear diagonal forces. To design a bridge with the best ratio of strength to weight of paper used, students need to understand how these diagonal forces work.

To describe a force that might act on the sheet of paper used for the bridge, we need two pieces of information:
how great the force is, and what direction it acts in. We can state a magnitude and an angle (“10 Newtons diagonally downward to the right at 45 degrees”), or we can state components in two separate directions (“About 7 Newtons to the right and 7 Newtons downward”). In either case this pair of numbers is mathematically called a vector.

A powerful tool for understanding these forces is to draw vector diagrams, where a vector is an arrow that point in the force’s direction and has a length proportional to the forces’ magnitude.

From the method of “seeing the force” as wrinkles in the paper, even in the uncut piece of paper, we can immediately see a fundamental aspect of the forces in this design challenge: as solid-mechanics theory will attest, tensile forces in a flexible material run in straight lines between connection points. A “connection point” is either a support point, a load attachment, or a “Y” branching point within the paper if you cut it in that shape. Unless you design a T-shaped bridge, the forces in the paper are always diagonal, so in addition to vertical tension, the paper must support horizontal tension as well. But how much can it bear?

Combining the wrinkles-in-the-paper insight with the component diagram vectors, we can answer this question. Since the forces line up with the connection points, the relation between the horizontal and vertical force components in a segment of the paper is the same as the ratio of the horizontal and vertical distances between its attachment points.

For the two sides of the bridge, the total of vertical force components is known: it’s the weight of the water bottle. If the bridge is symmetrical, each side gets half. So then the horizontal tension component will be half the load weight, divided by the slope of the segment of paper between attachment points. On the diagram, rather than having to understand “dividing by the slope,” we can simply see that the size of the horizontal component is whatever it takes, at the given slope, to make the diagonal vector long enough that its vertical component is the required value—its share of the weight of the water bottle.

With both vertical and horizontal components known, the total (diagonal) tension force can be found either by the Pythagorean theorem (the square root of the sum of squares of the two components) or can be estimated by using trigonometry, which involves measuring the vectors drawn to call directly on the diagram. Shallow slopes in these circumstances will start to give very high horizontal tension components. The T-shaped bridge, which theoretically has zero slope, demands an infinite horizontal tension component. This is why we see these bridges fail in the Student Work video.