Question Video: Analysis of a Body Resting in Equilibrium by Means of Two Strings Attached to It Mathematics

A body weighing 240 N is attached at point 𝐵 by a string whose other end is fixed to a point 𝐴 on a vertical wall. The length of the string 𝐴𝐵 is 30 cm. The body is also pulled by a horizontal string attached from point 𝐵 until point 𝐵 is 18 cm away from the wall. Determine the tensions 𝑇₁ in the horizontal string and 𝑇₂ in string 𝐴𝐵.

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Video Transcript

A body weighing 240 newtons is attached at point 𝐵 by a string whose other end is fixed to a point 𝐴 on a vertical wall. The length of the string 𝐴𝐵 is 30 centimeters. The body is also pulled by a horizontal string attached from point 𝐵 until point 𝐵 is 18 centimeters away from the wall. Determine the tensions 𝑇 sub one in the horizontal string and 𝑇 sub two in string 𝐴𝐵.

In order to answer this question, we can begin by drawing a simple diagram that represents the scenario.

First, we have this body attached at point 𝐵 by string, where the end of the string is attached to a vertical wall at point 𝐴. The downwards force of the weight of the body is 240 newtons, and this acts vertically downwards like so. The string is 30 centimeters. And then we have this second piece of string attached horizontally to the wall. Let’s say it meets at point 𝐶 at a distance of 18 centimeters. There will be tensional forces keeping this system in place. The first is defined as 𝑇 sub one and that acts in the horizontal direction and 𝑇 sub two in the string 𝐴𝐵.

Notice that what we’ve created is a triangle. This means that we can think about this system as a triangle of forces, and we can use the triangle of forces rule. In the triangle of forces rule, the magnitudes of the forces are proportional to the side lengths of the triangle. In other words, 𝐹 sub one over 𝐴𝐵 equals 𝐹 sub two over 𝐵𝐶, which equals 𝐹 sub three over 𝐴𝐶. This means that if we can establish the length of the third side in the triangle and then use the weight force, we should be able to calculate the magnitudes of 𝑇 sub one and 𝑇 sub two.

In fact, since the triangle is right-angled, we can use the Pythagorean theorem or Pythagorean triples to find the length 𝐴𝐶. 𝐴𝐶 squared plus 𝐵𝐶 squared equals 𝐴𝐵 squared. That means 𝐴𝐶 squared plus 18 squared equals 30 squared. Subtracting 18 squared from both sides and taking the positive square root and 𝐴𝐶 equals root 30 squared minus 18 squared, which equals 24. Hence, 𝐴𝐶 is 24 centimeters.

So, now that we have the lengths of the triangle, how do we find the forces? The system is in equilibrium, and we know that the force with a magnitude of 240 newtons acts in the same direction as the length 𝐴𝐶. Using the triangle of forces rule, we can say that the ratio of this force and the length of the triangle 𝐴𝐶 is 240 over 24. That’s 10. This must be equal to the ratio of 𝑇 sub one and the horizontal length of 18 centimeters. So, 𝑇 sub one over 18 equals 10.

Multiplying through by 18 and we find that 𝑇 sub one equals 180 newtons. We repeat this for the final force. So, 𝑇 sub two over 30 equals 10. Multiplying through by 30 and we find that 𝑇 sub two equals 10 times 30, which is 300. Hence, 𝑇 sub one equals 180 newtons and 𝑇 sub two equals 300 newtons.