Question Video: Studying the Equilibrium of a Horizontal Rod under the Action of Parallel Forces Mathematics

Video Transcript

A uniform rod AB weighs 70 newtons and has a length of 95 centimeters. It is suspended from its ends by two vertical strings where 𝑇 one is the tension of the string at A and 𝑇 two is the tension of the string at B. A weight of 100 newtons is suspended from the rod 30 centimeters away from A, and a weight of 93 newtons is suspended from the rod 20 centimeters away from B. Determine the values of 𝑇 one and 𝑇 two.

Let’s begin by sketching a diagram. Here is our rod AB suspended by the vertical strings. We have tensional forces 𝑇 one and 𝑇 two acting upwards at A and B, respectively. Since the rod is uniform, we can draw the weight force 70 newtons exactly halfway along. So that’s 47.5 centimeters from either A or B. We then have a weight of 100 newtons, which is suspended 30 centimeters away from A, and then a weight of 93 newtons, 20 centimeters away from B. Now, at some point, we’re going to be calculating moments. And since we already know the measurements of the 100 newtons and 70 newtons force from A, let’s calculate the measurement of the 93-newton force from A. The rod is 95 centimeters, so that’s 95 minus 20, which is 75 centimeters. The 93-newton force then is 75 centimeters away from A.

Now we’re trying to find the values of 𝑇 one and 𝑇 two. And so we assume that the rod is in equilibrium, and we can quote two facts. Firstly, if the rod is in equilibrium, the sum of its forces will be zero. More specifically, the sum of the vertical forces must be zero, and the sum of the moments must also be zero. Let’s begin by calculating moments about A. We’re going to take the counterclockwise direction to be positive. And then we recall that moment is equal to 𝐹𝑑, where 𝐹 is the force acting at a point and 𝑑 is the perpendicular distance from the line of action to the force to the point about which the object will turn.

Of course, we have 𝑇 two acting at B. Now that’s 95 centimeters away from A, so the moment here is 𝑇 two times 95. Then acting in the opposite direction, we have that 93-newton force, so its moment is negative. We get negative 93 times 75. Similarly, the force of the weight will be negative, since this is trying to turn the rod in a clockwise direction. The force is 70 and the distance is 47.5. So the moment is negative 70 times 47.5. Finally, we consider the 100-newton force. Again, the moment here is negative, and so we get negative 100 times 30. Of course, the sum of all of these moments is zero, so we form an equation purely in terms of 𝑇 sub two. Note that we haven’t included the moment related to 𝑇 sub one. And this is because 𝑇 sub one is zero centimeters away from A, and so the moment would just be 𝑇 sub one times zero, which is zero.

And so we simplify a little, and we get 95𝑇 sub two minus 6975 minus 3325 minus 3000 equals zero. We then add these constants together, and we get negative 13300. So we then add 13300 to both sides of our equation. To solve for 𝑇 two, we’ll divide by 95. 13300 divided by 95 is 140. So we can say that 𝑇 sub two must be equal to 140 newtons. We still need to calculate, however, the value of 𝑇 sub one. So we’re going to find the sum of the vertical forces. Of course, their sum is going to be equal to zero. Let’s take the upward direction to be positive. And then we have 𝑇 sub one plus 𝑇 sub two. Then acting in the opposite direction, we have the 100 newtons, 70 newtons, and the 93 newtons.

So the sum of our forces acting in a vertical direction is 𝑇 sub one plus 𝑇 sub two minus 100 minus 70 minus 93. And that is of course equal to zero. Now, of course, we just calculated that 𝑇 sub two was 140. And so we simplify this equation, and we get 𝑇 one minus 123 equals zero. To solve for 𝑇 sub one, we’re finally going to add 123 to both sides of our equation. And so 𝑇 sub one is 123 or 123 newtons. 𝑇 sub one is 123 newtons then and 𝑇 sub two is 140 newtons.