If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Class 11 Physics (India)>Unit 9

Lesson 10: Tension

# What is tension?

Ropes pull on things! Learn how to handle that kind of force.

## What does tension mean?

All physical objects that are in contact can exert forces on each other. We give these contact forces different names based on the types of objects in contact. If one of the objects exerting the force happens to be a rope, string, chain, or cable we call the force tension.
Ropes and cables are useful for exerting forces since they can efficiently transfer a force over a significant distance (e.g. the length of the rope). For instance, a sled can be pulled by a team of Siberian Huskies with ropes secured to them which lets the dogs run with a larger range of motion compared to requiring the Huskies to push on the back surface of the sled from behind using the normal force. (Yes, that would be the most pathetic dog sled team ever.)
It's important to note here that tension is a pulling force since ropes simply can't push effectively. Trying to push with a rope causes the rope to go slack and lose the tension that allowed it to pull in the first place. This might sound obvious, but when it comes time to draw the forces acting on an object, people often draw the force of tension going in the wrong direction so remember that tension can only pull on an object.

## How do we calculate the force of tension?

Unfortunately, there's no special formula to find the force of tension. The strategy employed to find the force of tension is the same as the one we use to find the normal force. Namely, we use Newton's second law to relate the motion of the object to the forces involved. To be specific we can,
1. Draw the forces exerted on the object in question.
2. Write down Newton's second law $\left(a=\frac{\mathrm{\Sigma }F}{m}\right)$ for a direction in which the tension is directed.
3. Solve for the tension using the Newton's second law equation $a=\frac{\mathrm{\Sigma }F}{m}$.
We'll use this problem solving strategy in the solved examples below.

## What do solved examples involving tension look like?

### Example 1: Angled rope pulling on a box

A box of cucumber extract is being pulled across a frictionless table by a rope at an angle $\theta ={60}^{o}$ as seen below. The tension in the rope causes the box to slide across the table to the right with an acceleration of .
What is the tension in the rope?
First we draw a force diagram of all the forces acting on the box.
Now we use Newton's second law. The tension is directed both vertically and horizontally, so it's a little unclear which direction to choose. However, since we know the acceleration horizontally, and since we know tension is the only force directed horizontally, we'll use Newton's second law in the horizontal direction.
${a}_{x}=\frac{\mathrm{\Sigma }{F}_{x}}{m}\phantom{\rule{1em}{0ex}}\text{(use Newton’s second law for the horizontal direction)}$

### Example 2: Box hanging from two ropes

A container of animal crackers hangs at rest from two strings secured to the ceiling and wall respectively. The diagonal rope under tension ${T}_{2}$ is directed at an angle $\theta ={30}^{o}$ from the horizontal direction as seen below.
What are the tensions $\left({T}_{1}$ and ${T}_{2}\right)$ in the two strings?
First we draw a force diagram of all the forces acting on the container of animal crackers.
Now we have to use Newton's second law. There are tensions directed both vertically and horizontally, so again it's a little unclear which direction to choose. However, since we know the force of gravity, which is a vertical force, we'll start with Newton's second law in the vertical direction.
${a}_{y}=\frac{\mathrm{\Sigma }{F}_{y}}{m}\phantom{\rule{1em}{0ex}}\text{(use Newton’s second law for the vertical direction)}$

Now that we know ${T}_{2}$ we can solve for the tension ${T}_{1}$ using Newton's second law for the horizontal direction.
${a}_{x}=\frac{\mathrm{\Sigma }{F}_{x}}{m}\phantom{\rule{1em}{0ex}}\text{(use Newton’s second law for the horizontal direction)}$

## Want to join the conversation?

• How come in these questions you always use cosine and sine functions? In an example(example 2 i think) you had to find the opposite. Tangent also deals with the opposite side(opp/adj) but we always use the sine/cosine. Why?
• But if you don't know the hypotenuse the you must use tangent, otherwise you can't calculate your answer. For example:
tan 60=opp/10 --> 10*tan 60deg=opp=17.320508075688772935274463415059
• If two masses are connected by the same rope, in what circumstances would the force tension differ between the two masses?
• The rope is acting as a transferring agent of a force, so there should never be a scenario in which two objects connected by one rope experience tension forces of different magnitudes.
• In example 1, there's an expandable explanation for what would happen if you tried to solve for the vertical force. It says "Well, since there are two unknown vertical forces (tension and the normal force), we just wouldn't be able to solve in that direction since there would be two unknowns."

Why is the normal force unknown? Isn't it the same as the downward force which is mass * gravity or approximately 20 newtons in this case?
• Normal force exists only when the object is in contact with the surface.. when upward force acting on the body, it will be going to decrease.
• Do you have to account for the mass of the rope, too?
• Yes but until you get to the point where you able to use calculus on these problems the wire/rope is considered to be massless.
• Does it necessarily matter what direction the forces go? It doesn't change the equation, so what is the point of showing whether the forces go horizontally or vertically?
• I don't understand the -T1 and -Fg? Why is it there and if so, how do we know if it is + or - ?
• you are free to choose sine conventions, its up to you nature don't understand positive or negative sign. The thing you must understand is tension act in the opposite direction of the gravitational force. If your object is hanging it must balanced by tension, otherwise it will accelerate down due to gravity.
• In example 1, it says solve for the tension of the rope. But we only solved for the tension (Tx) in the horizontal direction. How would I find the total tension of the rope with consideration to fg=mg, normal force Nf, and Tsin(60)? Thanks.
• No, it is solved for the total tension. Tx is 6N (from ax = sum(Fx)/m 3m/s^2 = Tx/2kg -> Tx = 6N) and Tx / cos 60 is T, so T is 12N.