Tension

Chang Xu (Fall 2017)

This topic covers Tension.

Main Idea

What is Tension?

Tension is the force exerted by a rope (or anything that can be used to hang another object) on the object that is hanging from it. Usually, it is ropes and cables that have the tension force. In general, anything that is flexible can pull an object and have tension force. In consequence, the tension force can only be a pulling force. The rope will eventually go slack if someone tries to push with a rope, and it will act like just an object. Later we will see that this concept will help with draing force diagrams with the force of tension always pulling the object.

Tension is considered a contact force which means that the force is exerted when objects are touching. Usually, the force of tension is the force that is transmitted through a rope. If someone is pulling on a block with a rope, the person exerting force on the rope which transmits that force to the block. In problems, the ropes and cables will usually be massless, which perfectly transfers the force.

Mathematical Modeling

When calculating for tension force, tension force should be treated as one of the forces that is in the problem. The most important formula that will be used in this will be:

Fnet = Mass x Acceleration

This is based on Newton's Second Law, which applies because there will be multiple forces acting on the object so the forces must be added together. Because of this, we will also have to add all the forces together to find Fnet:

Fnet = F1 + F2 + ... + Fn for n number of forces.

A lot of times for simple force of tension problems, the acceleration may be 0, which causes our Fnet to be 0 (because Fnet = Mass x Acceleration). Acknowledging this will make calculating forces easier. This case will have to do with Newton's Third Law which sates that every force has an equal and opposite force; therefore, Fnet will be 0.

When acceleration is not 0, we will have to solve for Fnet and then for Ft (force of tension).

How To Calculate Tension Force

Because most situations of tension are different, there is no set way of calculating the force of tension. There are two situations in which we need to consider: zero acceleration and nonzero acceleration. Both will need the same steps, but zero acceleration will be easier to understand at first.

Tension Force with Zero Acceleration

1) Draw a free body diagram

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The force of tension is shown by "T". There seems to be two forces of tension but they are actually the same magnitude because they come from the same string.

Whenever there is a question that has anything to do with force, the best thing to do first is to draw a free body diagram. This will help visualize the situation so you can see exactly where the forces will go. Common mistakes for force questions is that the direction of forces are wrong, so the student's solution turns out to be wrong. When drawing a free body diagram, make sure that the forces are all drawn in the right direction. Common forces to keep in mind when doing force of tension questions are:

• Ft = Force of Tension
• Fg = Force of Gravity
• Ff = Force of Friction
• Fn = Normal Force

2) Write equations that solve for Fnet

Because Fnet is 0 in this case because acceleration is 0, all the forces should add up to be 0. Make sure that the forces that are going in negative directions are negative. What could make this process easier is if you treat every force as the magnitude of the force and then add the magnitudes of the positive forces and subtract the magnitudes of the negative forces. That way, the negative signs will not get confused. Generally the equation would look like:

Fnet = 0 = Ft + F1 + F2 + ... + Fn (and then any other applied forces if there are any)

In this equation, the forces are not treated as magnitudes because the directions are not known. It is best to keep in note that the force of friction is usually going against velocity, and the force of gravity is usually pointing down in the y direction.

3) Solve for Ft

Once you have the equation done, it is possible to plug in forces to get to the solution for force of tension. If the problem does not give direct forces, here are some calculations of the forces:

• Fg = mg
• Ff = μFn
• Fn is perpendicular to the surface of the object

Tension Force with Nonzero Acceleration

The way to solve this is very similar to the way tension force is solved with zero acceleration.

1) Draw a free body diagram

You should always draw a free body diagram to get started with visualizing the problem. Again, common forces to keep in mind when doing force of tension questions are:

• Ft = Force of Tension
• Fg = Force of Gravity
• Ff = Force of Friction
• Fn = Normal Force

2) Write equations that solve for Fnet

In a case where there is acceleration, the problem is basically stating that all the forces act together on an object to make it accelerate at that acceleration. Because Fnet = mass x acceleration, solving for the magnitude of Fnet is very easy:

Fnet = mass x acceleration

Then you would do the same thing for zero acceleration and set Fnet equal to all the forces added together. It should look something like this:

Fnet = mass x acceleration = Ft + F1 + F2 + ... + Fn

You will notice that this actually looks a lot like the equation before with zero acceleration. Actually, it is pretty much the same. Fnet is really just a variable for the total amount of force on the object and in each case it is like that. For zero acceleration, Fnet = 0, so we set the equation equal to zero. For a nonzero acceleration, we actually have to solve for Fnet before we move on.

3) Solve for Ft

Once you have the equation done, it is possible to plug in forces to get to the solution for force of tension.

Example Problems

Example 1: Angled rope pulling on a box

A 2.0kg box of toy box is being pulled across a table by a rope at an angle θ=60º as seen below (ignore friction). The tension in the rope causes the box to slide across the table to the right with an acceleration of 3.0 m/s^2. What is the tension on the rope?

First always draw a free body diagram to visualize the situation better.

• Fg: negative y direction

In most cases, the force of gravity is always present, and in this case, the force of gravity (Fg) is pointing in the negative y direction because gravity pulls objects down.

• Fn: positive y direction

Because there is also a surface keeping the box standing up, there is a normal force (Fn) pushing the box up; therefore the direction of Fn is going in the positive y direction.

• T: 60 degrees from x axis

In the image, the rope is pulling the block 60 degrees from the x axis; therefore the force of tension (T) is pointing that way.

Because the acceleration is nonzero in this situation, Fnet does not equal 0. So, which direction is net force going? The answer to this question is to always think about what direction acceleration is going. Fnet will be going in whatever direction acceleration is going. In this case, acceleration is going in the horizontal direction, so Fnet is pointing to the positive x direction. This can be confusing sometimes because it seems like Ft should be making the box go up as well, but the box does not move up; therefore Fnet is going in the positive x direction. So:

• Fnet = mass x acceleration = 2 x 3 = 6N

It also can be noticed that Ft has two components: the y direction and the x direction. It would also help to differentiate between these two to see how T affects Fnet. So we can break down T into two parts:

• T-y = Tsin(60)
• T-x = Tcos(60)

Now if we separate the forces into y direction and x direction, we can see:

• Y direction: Fn, Fg, T-y
• X direction: T-x

It can also be noted from before that Fnet is going in the positive x direction, and the only force in the x direction is T-x. This means that all the y direction forces all add up to 0, and we only need to focus on T-x, and set it equal to Fnet. We found before that Fnet = 6N, so:

• Fnet = T-x = Tcos(60) = 6
• T = 12N

So, we get that the force of tension is 12N. Here is a quick version of all the calculations together:

1. a= ​Fnet/m (use Newtons's second law in the horizontal direction)
2. Fnet = (2.0kg)(3.0m/s^2) = 6N
3. 6N=Tcos(60) ​​ (plug in the horizontal acceleration, mass, and horizontal forces)
4. T=6N/cos(60)
5. T=12N

Example 2: Box hanging from two ropes

A 0.25 kg container hangs at rest from two strings secured to the ceiling and wall respectively. The diagonal rope under tension T1 is directed at an angle θ=30º from the horizontal direction as seen below.

What are the tensions (T1 and T2) in the two strings?

First draw a force diagram of all the forces acting on the container.

Now 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 there is force of gravity (a vertical force), start with Newton's second law in the vertical direction.

1. a=​ΣF/m (use Newton's second law for the vertical direction)
2. 0=(T2*sin30º-Fg)/0.25kg
3. T2=Fg/(sin30º)
4. T2=mg/(sin30º)
5. T2=[(0.25kg)(9.8m/s²)]/(sin30º)
6. T2=4.9N

Now that we know T2 we can solve for the tension T1 using Newton's second law for the horizontal direction.

1. a=​ΣF/m (use Newton's second law for the horizontal direction)
2. 0=(T2*cos30º-T1)/0.25kg (plug in the horizontal acceleration, mass, and horizontal forces)
3. T1=T2*cos30º
4. T1=(4.9N)*cos30º
5. T1=4.2N

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Connectedness

1. How is this topic connected to something that you are interested in?

When I was little, I went to a department store and I saw a transparent elevator. I never knew how elevator operated but through the transparent elevator, I realized that the pulling force of the ropes was what was keeping the elevator moving. This pulling force is tension and I later realized that this tension force is evident everywhere in my daily life.

1. How is it connected to your major?

I am not exactly sure how tension is connected to industrial engineering, which is my major. However, I can say that tension is a very basic concept in physics related to force and it is important to understand physics mechanism in studying industrial engineering.

1. Is there an interesting industrial application?

Tension force can be seen in everyday life, just like the elevator example I mentioned above. Tension force is applied when I pull a clothing tags.

See also

1. Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

I think it will be interesting to make connection between tension force and friction. Use both tension and friction in pulling a box in ice skating environment and concrete environment. This would be a good example to explore more about the topic.

Further reading

Books, Articles or other print media on this topic

Chabay, Ruth W., and Bruce A. Sherwood. Matter & Interactions. Hoboken, NJ: John Wiley & Sons, 2015. Print.

External links

Internet resources on this topic

References

This section contains the the references you used while writing this page

Chabay, Ruth W., and Bruce A. Sherwood. Matter & Interactions. Hoboken, NJ: John Wiley & Sons, 2015. Print. https://www.khanacademy.org/science/physics/forces-newtons-laws/tension-tutorial/a/what-is-tension http://www.physicsclassroom.com/Class/newtlaws/U2L2b.cfm#tension http://hyperphysics.phy-astr.gsu.edu/hbase/mlif.html http://hyperphysics.phy-astr.gsu.edu/hbase/elev.html http://www.softschools.com/formulas/physics/tension_formula/70/ http://physics.stackexchange.com/questions/36175/understanding-tension http://www.brightstorm.com/science/physics/newtons-laws-of-motion/tension/