Question Video: Finding the Tension in the Strings That Keep a Uniform Rod in Equilibrium Mathematics

Video Transcript

A uniform rod of length 50 centimeters and weight 143 newtons is freely suspended at its ends from the ceiling by means of two perpendicular strings attached to the same point on the ceiling. Given that the length of one of the strings is 30 centimeters, determine the tension in each string.

We’re going to begin by drawing a free-body diagram of this scenario. Let’s define the ends of our rod to be 𝐴 and 𝐵. And so we have our rod suspended from two perpendicular strings, one of which measures 30 centimeters. Now, in fact, we can work out the measurement of the second piece of string. It’s 40 centimeters. And we can calculate that using the Pythagorean theorem or just recognizing that we have a multiple of a Pythagorean triple — three, four, five. Since the rod is uniform, we can say that the downwards force of its weight acts at a point exactly halfway along the rod. Now, since the rod exerts a downwards force on the pieces of string, there will be an opposite reaction force of tension. Let’s call that 𝑇 sub 𝐴 for the tensional force in the first bit of string and 𝑇 sub 𝐵 for the tensional force in the second.

We have our free-body diagram, but there’s still an awful lot going on here. So how do we simplify it further? Well, we know that when three coplanar forces acting at a point are in equilibrium, they can be represented in magnitude and direction by the adjacent sides of a triangle taken in order, since the systems in equilibrium will lay the force vectors end to end to make a triangle. 𝑇 sub 𝐴 and 𝑇 sub 𝐵 will be perpendicular to one another, since the pieces of string along which they run are also perpendicular to one another. But in fact, with a little bit of inspection, we can see that these are similar triangles. We’ll define the angle between the rod and the 30-centimeter piece of string to be 𝑥 degrees. This is equal to the angle between the 143-newton force and the tensional force at 𝐵.

This isn’t hugely intuitive, but we can convince ourselves that this is true by adding a line parallel to the 143-newton force on our first diagram and then using the fact that angles in a triangle sum to 180. Since the angles in our force triangle and the angles in the triangle represented by the lengths of each item are equal, we know these triangles are similar. And so we can use scale factor or ratios to find the magnitudes of 𝑇 sub 𝐴 and 𝑇 sub 𝐵. Let’s consider the ratio of 𝑇 sub 𝐴 to 143. This must be equal to the ratio of the 40-centimeter length to the 50-centimeter length. And we chose these pairs of sides in both our diagrams because we’re interested in the pairs of sides that are adjacent to the angle measuring 90 minus 𝑥 degrees.

Multiplying both sides of this equation by 143 and we find that 𝑇 sub 𝐴 is 40 over 50 or four-fifths times 143, which is 114.4. So the magnitude of 𝑇 sub 𝐴 is 114.4 newtons. In a similar way, the ratio of 𝑇 sub 𝐵 to 143 newtons will be equal to the ratio of the 30-centimeter length to the 50-centimeter length. To solve for 𝑇 sub 𝐵 once again, we multiply by 143, which gives us 85.8 or 85.8 newtons. The tensional forces in each string are 85.8 newtons and 114.4 newtons.