# Video: Determining the Deformation Induced in an Object by Tensile Stress

A nylon rope has a Young’s Modulus of 1.35 × 10⁹ Pa, a length of 35.0 m when not stretched, and a diameter of 0.800 cm. If a mountain climber of mass 65.0 kg hangs on this rope, how much does it extend?

06:10

### Video Transcript

A nylon rope has a Young’s modulus of 1.35 times 10 to the ninth pascals, a length of 35.0 meters when not stretched, and a diameter of 0.800 cm. If a mountain climber of mass 65.0 kg hangs on this rope, how much does it extend.

As we solve this problem, we’ll assume that the rope itself is essentially massless so that it does not exert a gravitational force on itself. We’ll also assume that the acceleration due to gravity 𝑔 is exactly 9.8 meters per second squared. Let’s start by highlighting some of the critical information we’ve been given.

First, we’re told that the nylon rope has a Young’s modulus of 1.35 times 10 to the ninth pascals. We’re told that the rope is 35.0 meters long when it’s not stretched, and that it has a diameter of 0.800 cm. We’re also told that the climber who hangs on the rope has a mass of 65.0 kg. Let’s assign some symbols to represent these values as we work through this problem.

Let’s represent Young’s modulus as a capital 𝐸. We’ll call the length of the rope capital 𝐿. And we’ll call the rope’s diameter lower case 𝑑. And we’ll express it in terms of meters rather than centimeters. 0.800 cm is the same as 8.00 times 10 to the negative third meters. And finally the mass of the mountain climber will represent as 𝑚. And we’re asked to find out under this condition how much does the rope extend or stretch. We’ll refer to that as 𝛥𝐿: the change in the length of the rope due to the mass of the climber stretching it out.

Now the critical relationship in this problem relates Young’s modulus to the other terms we’ve been given. You may recall this relationship. The equation states that 𝐸, Young’s modulus, is equal to the force that is exerted on our object divided by the cross sectional area of our object. In that fraction is then divided by 𝛥𝐿, the change in length of our object, divided by its original length 𝐿.

When we use Young’s modulus equation, we typically assume as we will in this case that the force that is being exerted is expressed equally across the cross section 𝐴. Now let’s use this relationship in our particular scenario to solve for 𝛥𝐿, the change in length of the rope when it’s stretched due to the weight of a climber. A mathematically equivalent way to write this expression is to say that the Young’s modulus is equal to the force divided by area multiplied by the length of our object divided by the change in its length.

Now we want to solve for that 𝛥𝐿, the change in length. So to do that, we’ll isolate it on one side of our equation. Let’s multiply both sides of our equation by 𝛥𝐿. When we do that, that term cancels out of the right side of our equation. If we now divide both sides of our equation by 𝐸, the Young’s modulus will isolate 𝛥𝐿 on the left side of our equation.

Now let’s write out a cleaned up version of this equation. And we find that 𝛥𝐿 is equal to the force exerted divided by the area over which that force is exerted multiplied by the length of our rope divided by 𝐸, the Young’s modulus. Now we can move towards plugging in values to these terms on the right side of our equation. We’ll start with the force 𝐹. That force we’re told is created by a mountain climber hanging on the rope. In other words, this is a gravitational force or weight force. And that force, 𝑊, equals an object’s mass times the acceleration due to gravity, 𝑔. So we can substitute 𝑚, the mass of the mountain climber, times 𝑔 in for 𝐹 in this equation.

Now let’s look at 𝐴, the area. If we draw a sketch of the cross section of our rope, we’re told what the diameter of that rope is. That diameter is given as 8.00 times 10 to the negative third meters. Since we’re working with a cross section which is a circle, we know that overall the area, 𝐴, that we’re working with is equal to 𝜋 times 𝑟, the radius of that circle, squared. So as we look at our equation for 𝛥𝐿, let’s replace the capital 𝐴 with 𝜋𝑟 squared, the cross sectional area of a rope. All we have left in the equation now our capital 𝐿, the length of a rope, which is given to us and capital 𝐸, the Young’s modulus of the rope, which is also given.

We’re now ready to begin plugging in numbers for these values into our equation to solve for 𝛥𝐿. When we plug all these values into the equation and then into our calculator, we find that 𝛥𝐿 is equal to 0.329 meters or 32.9 cm. That’s how much this nylon rope stretches when a climber with mass of 65.0 kg hangs vertically on it.