Question Video: Finding the Mass Attached to a Horizontal Rod Suspended by Two Strings given the the Ratio between the Tensions in Both Mathematics

Video Transcript

𝐴𝐵 is a uniform rod having length of 111 centimeters and weighing 78 newtons. The rod is suspended horizontally from its ends 𝐴 and 𝐵 by two vertical strings. Given that a weight of 111 newtons is suspended 𝑥 centimeters away from the end 𝐴 so that the tension at 𝐴 is twice that at 𝐵, determine the tension 𝑇 of 𝐵 at 𝐵 and the value of 𝑥.

We will begin by sketching a diagram of the rod 𝐴𝐵 along with the forces acting upon it. Since the rod is uniform, its weight will act at the midpoint. As the weight of the rod is 78 newtons, we have a 78-newton force acting vertically downwards at the midpoint of the rod. We are told that the rod is 111 centimeters long. The distance from point 𝐴 where the weight force acts is half of this, which is equal to 55.5 centimeters. We are told that the rod is suspended by two vertical strings at points 𝐴 and 𝐵. We will call the tension in these strings 𝑇 sub 𝐴 and 𝑇 sub 𝐵.

Next, we are told that a weight of 111 newtons is suspended 𝑥 centimeters away from end 𝐴. This results in the tension at 𝐴 becoming twice the tension at 𝐵. We can therefore replace 𝑇 sub 𝐴 with two 𝑇 sub 𝐵. Our aim in this question is to find the tension 𝑇 sub 𝐵 and the value of 𝑥. In order to do this, we will recall two key facts about bodies in equilibrium. Firstly, for a body to be in equilibrium, the sum of the forces acting on the body must equal zero. Secondly, the sum of the moments must also equal zero, where the moment of a force is calculated by multiplying 𝐹 by 𝑑, where 𝐹 is the force acting at a point and 𝑑 is the perpendicular distance to the point at which we are taking moments.

Let’s begin by considering the sum of the forces. If we let the positive direction be vertically upwards, the tension forces two 𝑇 sub 𝐵 and 𝑇 sub 𝐵 are positive. The weight forces 111 newtons and 78 newtons are negative. And as the sum of the forces equals zero, we have two 𝑇 sub 𝐵 plus 𝑇 sub 𝐵 minus 111 minus 78 equals zero. Collecting like terms, this simplifies to three 𝑇 sub 𝐵 minus 189 equals zero. We can then add 189 to both sides of our equation. And then dividing through by three, we have 𝑇 sub 𝐵 is equal to 63 newtons. The tension in the string at 𝐵 is equal to 63 newtons.

Whilst it is not required in this question, we could calculate the tension in the string at 𝐴 by multiplying this value by two. And as 63 multiplied by two is 126, the tension in the string at 𝐴 is 126 newtons. We will now consider moments to calculate the value of 𝑥. We can take moments about any point on the rod. However, in this question, we will take moments about point 𝐴 and consider positive moments to be in the counterclockwise direction. The 111-newton force acts in a clockwise direction about point 𝐴. And it will therefore have a negative moment. This is equal to negative 111 multiplied by the perpendicular distance 𝑥. The 78-newton force also acts in a clockwise direction, so the moment here will also be negative. It is equal to negative 78 multiplied by 55.5.

Finally, the tension at 𝐵 acts in a counterclockwise direction, so will have a positive moment equal to 𝑇 sub 𝐵 multiplied by 111. We know that the sum of these moments equals zero. The first term simplifies to negative 111𝑥. 78 multiplied by 55.5 is equal to 4329. So we have negative 4329. As we have already worked out that 𝑇 sub 𝐵 is 63 newtons, we need to multiply 63 by 111, giving us 6993. Adding 111𝑥 to both sides of our equation, we have 2664 is equal to 111𝑥. We can then divide through by 111, giving us 𝑥 is equal to 24. The weight of 111 newtons is suspended 24 centimeters away from the end 𝐴. Our two answers are 𝑇 sub 𝐵 equals 63 newtons and 𝑥 equals 24.