Video: Tension

In this video we learn about tension as a pulling force and use free-body diagrams and Newton’s Second Law of Motion to calculate it.

12:05

Video Transcript

In this video, we’re going to learn about tension. No, not the kind of tension that you feel in the classroom when the instructor is handing back a graded exam, this is a different type of tension having to do with physical force. To get started with this topic, imagine you are an art student, completing a graduation project for the end of the semester. Your project, an abstract demonstration of the interconnectedness of life on Earth, consists of a paper mache sphere connected by various colored strings to the walls and ceilings of the room. Very proud of your work, you look forward to showing it to the teachers and students in your class. Some of the streams in the project block an emergency exit door in the room. The school safety officer tells you that that doorway needs to be clear. You realize you can cut one of the chords to clear the exit way. But the question is, how will that affect the equilibrium of your model? Understanding tension helps us to understand this question.

When we talk about tension, we’re talking about a force. Tension is a force measured in newtons that acts along the length of a linear segment. That linear segment could be a rope or a chain or a string. Or it could even be a rigid body, like a section of a bridge. Say we had two objects: a rope on the left and a concrete column on the right. Both of these objects are able to transmit an experienced tension force. With something pulling on the ends of each object, to create tension, we could pick any point along their lengths and at that cross section find tension pulling in either direction. This helps us understand tension, because tension is always a pulling force. It never pushes. This fact is helpful to us when we draw free body diagrams.

Imagine that we had a mass π‘š suspended by two cables. When we go to draw the free body diagram of the forces acting on this mass, since tension is always a pulling force, we know that the tension in the cables pulls the mass away from its center. When we go to solve for tension in an example, there are two steps we want to remember to the process. To solve for the force of tension in an example, we take two steps, the first of which is to draw a free body diagram of the object we’re interested in. In our example of the mass suspended by two cables, we could represent the tension in the cables by 𝑇 sub one and 𝑇 sub two. And then there is the weight of the mass pulling it down.

Step two to solving for tension is to apply Newton’s second law of motion, which says that the net force acting on an object equals its mass times its acceleration. Even when an object is not accelerating, as this mass in equilibrium is not, this second law of relationship lets us balance out the forces acting on our object of interest. It’s through that balance or imbalance in the case of an accelerating object that lets us solve for the forces acting on our object and therefore the tension force. Now, let’s get some practice with this two-step process through a couple of examples.

A ball of mass 11.5 grams hangs from the roof of a freight car by a string. While the freight car accelerates, the string makes an angle of 22.7 degrees with the vertical. What is the magnitude of the acceleration of the freight car? What is the magnitude of the tension in the string?

We can call the acceleration magnitude of the freight car π‘Ž and the tension magnitude in the string capital 𝑇. Let’s start by drawing a diagram of our situation. In this situation, we have a freight car accelerating we can say to the right with some acceleration magnitude we’ve called π‘Ž. Inside the freight car is a mass hanging from a string with a mass value of 11.5 grams. When the freight car is accelerating, the string makes an angle πœƒ of 22.7 degrees with the vertical. The first thing we’ll do to start off solving for π‘Ž and then 𝑇 is to draw a free body diagram of the forces acting on our mass. There’s the string acting with a tension we’ve called capital 𝑇 on the mass. And second, there’s the weight force, the force of gravity acting on it, which is equal to its mass times the acceleration due to gravity. We’ll treat that acceleration as exactly 9.8 meters per second squared.

Next, to prepare to write out force balance equations, we’ll put a 𝑦- and π‘₯-axis in our scenario, where positive π‘₯-motion is to the right and positive 𝑦-motion is up. And next, we recall Newton’s second law of motion, which tells us that the net force on an object is equal to its mass multiplied by its acceleration. Applying this relationship to our scenario, if we focus on the vertical forces, the forces along the 𝑦-axis, we can write that 𝑇 times the cos of πœƒ, the vertical component of the tension force, minus the weight force π‘šπ‘” is equal to the mass of the object multiplied by its acceleration. Since the mass is in equilibrium, that acceleration is equal to zero. And we can say that 𝑇 times the cos of πœƒ is equal to π‘š times 𝑔.

Now, let’s apply Newton’s second law to the forces in the π‘₯-direction. In this case, there’s only one force. It’s the horizontal component of the tension force. And by the second law, that equals the mass of the object multiplied by π‘Ž. This tells us that π‘Ž is equal to 𝑇 sin πœƒ over π‘š. We know both π‘š and πœƒ. But we don’t yet know the tension 𝑇. But we do know from our 𝑦 force balance equation that, rearranging the equation, tension is equal to π‘šπ‘” over cos πœƒ. If we substitute this expression for 𝑇 in to our equation for π‘Ž β€” and when we do this substitution, we recall the trigonometric identity sin πœƒ divided by cos πœƒ equals tan πœƒ β€” then we find an expression for π‘Ž, which when the masses cancel out equals 𝑔 times the tan of πœƒ.

When we plug in for these values, 𝑔 and πœƒ, and solve for π‘Ž, we find it’s 4.10 meters per second squared. That’s the acceleration of the freight car that will create this angle πœƒ. Next, we want to solve for the tension, capital 𝑇, in the string. To do that, we can return to our 𝑦-direction force balance equation, which told us that tension 𝑇 is equal to π‘šπ‘” over cos πœƒ. Plugging in for π‘š, 𝑔, and πœƒ, being careful to use units of kilograms for the mass, when we solve for 𝑇, we find that, to three significant figures, it’s 0.122 newtons. This is the magnitude of the tension in the string. Now, let’s look at an example involving a system of masses in motion.

Two blocks are connected across a pulley by a rope as shown. The mass π‘š one of the block on the table is 4.0 kilograms. And the mass π‘š two of the hanging block is 1.0 kilogram. The mass of the rope is negligible. The table and the pulley are both frictionless. Find the acceleration of the system. Find the tension in the rope. Find the speed of the hanging block when it hits the floor. Assume that the hanging block is initially at rest and located 1.0 meters vertically above the floor.

From this last bit of information, we can add some detail to our diagram. We now know that π‘š two is a distance we can call 𝑑 of 1.0 meters above the floor. We’re told previously the masses π‘š one and π‘š two. And we want to solve overall for three pieces of information. We want to solve for the acceleration π‘Ž of the system, the tension 𝑇 in the cable connecting the two blocks, and the final speed, we can call 𝑣 sub 𝑓 of π‘š two, as it hits the ground. As we get started solving for system acceleration π‘Ž, we recall Newton’s second law of motion that the net force acting on an object or a system is equal to the mass of the object multiplied by its acceleration.

When we consider the forces involved with respect to our diagram, we see that the only force responsible for the motion of this system of masses is the gravitational force acting on π‘š two. That’s because our system overall is frictionless and the gravitational force on π‘š one is counteracted by the normal force pushing up on π‘š one. So by Newton’s second law, we can write that the weight force on π‘š two, which is π‘š two times 𝑔, equals the mass of our system, π‘š one plus π‘š two, multiplied by the acceleration of the system, π‘Ž. Rearranging this expression to solve for π‘Ž, we see it’s equal to π‘š two times 𝑔 divided by the sum of the masses π‘š one and π‘š two. 𝑔 we’ll treat as exactly 9.8 meters per second squared. And since we’re given π‘š one and π‘š two, we’re ready to plug in and solve for π‘Ž.

When we enter this expression on our calculator, we find that π‘Ž is 2.0 meters per second squared. That’s the overall acceleration of this system of masses. Next, we want to solve for the tension 𝑇 in the line connecting the two masses. The interesting thing about solving for 𝑇 is that since 𝑇, the tension, is a force on both masses, we could look at either one of the masses to solve for it. Just to pick a mass, let’s choose π‘š two. The two forces acting on π‘š two are the tension force 𝑇 and the weight force π‘š two times 𝑔. If we consider motion down to be motion in the positive direction, then Newton’s second law tells us we can write π‘š two times 𝑔 minus 𝑇 is equal to π‘š two times π‘Ž, where π‘Ž is the acceleration we solved for earlier.

Rearranging to solve for 𝑇, we find it’s equal to π‘š two times 𝑔 minus π‘Ž, or 1.0 kilograms multiplied by 9.8 minus 2.0 meters per second squared. When we multiply these values together, we find that 𝑇 is 7.8 newtons. That’s the tension in the line connecting the masses. And finally, knowing that π‘š two starts from rest and descends a distance 1.0 meters before it reaches the floor, we want to solve for 𝑣 sub 𝑓, its final speed, just as it does reach the floor. Since the mass π‘š two has a constant acceleration, we can use the kinematic equations to solve for its motion.

Specifically, we can use the equation that 𝑣 sub 𝑓 squared equals 𝑣 sub zero squared plus two times π‘Ž times 𝑑. Since π‘š two starts at rest, that means 𝑣 sub zero is zero. So our equation is simplified to 𝑣 sub 𝑓 squared equals two times π‘Ž times 𝑑. When we take the square root of both sides and plug in 2.0 meters per second squared for π‘Ž and 1.0 meters for 𝑑, when we calculate the square root, to two significant figures, it equals 2.0 meters per second. That’s the speed of block π‘š two just as it hits the ground.

Let’s summarize now what we’ve learned about tension. We’ve seen that tension is a force, measured in newtons, that acts along the length of a linear segment, where the segment could be a rope or a cable or a string or a solid object. Tension, we’ve also seen, is always a pulling force. It never pushes. To remember that, just imagine how hard it would be to try to push a rope. And finally, when we solve for tension in an example, we follow a two-step process. First, we draw a free body diagram of the object of interest. And then we apply Newton’s second law of motion. This second law lets us generate the equation that we’ll use to solve for the tension involved.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.