# Video: Finding the Tension in and the Extension of Springs Supporting a Load

Two identical springs, each with the spring constant 20 N/m, support a 15.0-N weight, as shown. What is the tension in spring 𝐴? What is the amount of stretch of spring 𝐴 from the rest position?

07:26

### Video Transcript

Two identical springs, each with spring constant 20 newtons per meter, support a 15.0-newton weight as shown. What is the tension in spring 𝐴? What is the amount of stretch of spring 𝐴 from the rest position?

In this problem, we’ll assume that the springs are ideal, in other words, with a constant relationship between their spring constant and their displacement, and we’ll also assume that 𝑔, the acceleration due to gravity, is exactly 9.8 meters per second squared.

Let’s start by highlighting some of the critical information we’re given in this problem. First we’re told the spring constant of each of the springs, 20 newtons per meter, and we’re also told that the weight that these springs support is 15.0 newtons.

This is a two-part problem, where in part one, we’re asked to solve for the tension in spring 𝐴; we’ll call that capital 𝑇 sub 𝐴. And in part two of the problem, we wanna solve for the amount of stretch that spring 𝐴 experiences from its initial rest position; we’ll call that Δ𝑥 sub 𝐴.

Now let’s begin on part one of this problem, and we’ll start by recalling what Hooke’s law states. Hooke’s law tells us that there is a certain class of elastic materials or springs that obey this relationship.

The force 𝐹 used to stretch or compress the spring is equal to the displacement of the spring from its equilibrium, called 𝑥, multiplied by a constant, called 𝑘. In this problem, we’ll assume that the springs we’re working with are Hookean springs, meaning they obey Hooke’s law.

With that as background, let’s draw a free body diagram of our mass that’s suspended by the springs. We have our mass, and gravity acts down on it; we’ll call that 𝐹 sub 𝑔. And then the spring forces due to spring 𝐴 and spring 𝐵 act up and out on it at an angle of 30 degrees.

So the forces acting on our mass are 𝐹 sub 𝑔, the force of gravity; 𝑇 sub 𝐴, the tension in spring 𝐴; and 𝑇 sub 𝐵, the tension in spring 𝐵. And we know that because our mass is not in motion, all these forces balance one another out. The acceleration of the mass is zero.

To solve for 𝑇 sub 𝐴, the tension in spring 𝐴, which incidentally will be equal to 𝑇𝐵, let’s write a force balance equation in the vertical direction based on Newton’s second law.

As we use the second law to move forward, let’s also define up as a direction of positive motion. We can now write out a vertical force balance equation like this: the vertical component of 𝑇 sub 𝐴, which is 𝑇 sub 𝐴 times the cosine of 30 degrees, plus the vertical component of 𝑇 sub 𝐵, or 𝑇 sub 𝐵 times the cosine of 30 degrees, minus the force of gravity, which is the mass of our block times 𝑔, is equal to zero.

And again that’s true because our block is not accelerating and so 𝑎 in our second law is zero. Now as we look at this equation, we can simplify it a bit based on the fact that our springs 𝐴 and 𝐵 are identical. Therefore, we can write 𝑇 sub 𝐵 as 𝑇 sub 𝐴 because the springs are the same, which means we can simplify the left-hand side of our equation to two times 𝑇 sub 𝐴 cosine of 30 degrees.

We want to rearrange this equation to solve for 𝑇 sub 𝐴; that’s what we’re looking for in part one of this problem. To do that, we can add 𝑚𝑔 to both sides, which cancels out that term on the left-hand side of our equation. And we can then divide both sides of the equation by two times the cosine of 30 degrees.

This results in the factors of two and cosine of 30 canceling on the left-hand side of our equation, leaving us with 𝑇 sub 𝐴 by itself. Rewriting a simplified version of this equation, we see that 𝑇 sub 𝐴 is equal to 𝑚𝑔 divided by two times the cosine of 30 degrees.

Now interestingly, we’re not told the mass 𝑚 that is being suspended by the springs, but we are told that it has a weight of 15.0 newtons. We can call that capital 𝑊. The weight is equal to the product of the object’s mass times the acceleration due to gravity. So in other words, we can replace 𝑚𝑔 in our equation for tension in spring 𝐴 with capital 𝑊, the weight.

We can now substitute in the given value for 𝑊, 15.0 newtons, and enter this fraction on our calculator. This gives us a value for 𝑇 sub 𝐴 of 8.7 newtons. That’s the tension in spring 𝐴, as well as the tension in spring 𝐵 that is experienced when supporting this mass with a weight of 15.0 newtons.

Now that we know 𝑇 sub 𝐴, we can move ahead with solving for Δ𝑥 sub 𝐴, the amount that spring 𝐴 is displaced from equilibrium. To figure this out, we can refer back to Hooke’s law.

We can apply Hooke’s law to our scenario by restating it using our variables. Instead of 𝐹 equals negative 𝑘𝑥, we can write 𝑇 sub 𝐴 equals 𝑘 times Δ𝑥 sub 𝐴. 𝑇 sub 𝐴 is the tension in spring 𝐴 we solved for in part one. 𝑘 is the spring constant which we’re given in the problem statement is 20 newtons per meter. And Δ𝑥 sub 𝐴 is the variable we want to solve for, that is, the displacement from equilibrium that spring 𝐴 experiences under the stress of holding up the weight 𝑊.

So let’s rearrange this equation for Δ𝑥 sub 𝐴 by dividing both sides by 𝑘, the spring constant. When we do that, 𝑘 cancels out from the right-hand side of our equation and we’re left with an equation that reads Δ𝑥 sub 𝐴 is equal to 𝑇 sub 𝐴 divided by 𝑘.

Plugging in for our given and solved for values, we know 𝑇 sub 𝐴 is 8.7 newtons and 𝑘 is 20 newtons per meter. Plugging these numbers into our calculator, we find that Δ𝑥 sub 𝐴 is equal to 0.43 meters. That’s the distance from equilibrium. That spring 𝐴 is stretched under the weight of weight 𝑊.