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: angle=60 deg, hypotenuse=?, adj=10, opp=? 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.

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.

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 simple pulley problem, like 2 mass on the rope. I've found out sometime in the tension of ther rope you have same tension and some time not. It"s sam when the mass of pulley ignored, and different when using rotaional dynamic. Can someone explain to why, and tell me how to know the tension is the same or not ?

you said it yourself: if the pulley and rope a massless, tension is the same throughout the rope. If not, it's not.

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.

"Σ" what is the meaning of this symbol?

Sum. It is a greek letter called sigma when is showed means a sum of the numbers.

How does this principle work vertically? Is it solved the same way? How is gravity included?

If the string is vertical, the tension is equal in magnitude to the weight force (F = mg)

Main content

Course: Class 11 Physics (India) > Unit 9

Lesson 10: Tension

What is tension?

Google Classroom

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.

If you pull on an object with a rope, the rope will stretch slightly (often imperceptibly). This stretch in the rope will cause the rope to be taut (e.g. under tension) which allows the rope to transfer a force from one side of the rope to the other, roughly similar to how a stretched spring will pull on objects connected to it. The stretch of the rope is usually too small to notice, so we typically ignore the small stretch that occurs in ropes, cables, and wires. However, if the forces involved are too great, the large amount of stretch can cause the rope to break. So it is advisable to check the tension limit for any cable or ropes you plan on using.

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,

Draw the forces exerted on the object in question.
Write down Newton's second law $(a = \frac{Σ F}{m})$ ‍ for a direction in which the tension is directed.
Solve for the tension using the Newton's second law equation $a = \frac{Σ F}{m}$ ‍.

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

2.0 kg

box of cucumber extract is being pulled across a frictionless table by a rope at an angle

θ = 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

3.0 \frac{m}{s^{2}}

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{Σ F_{x}}{m} (use Newton’s second law for the horizontal direction)

3.0 \frac{m}{s^{2}} = \frac{T cos 60^{o}}{2.0 kg} (plug in the horizontal acceleration, mass, and horizontal forces)

We have to break the diagonal tension into vertical and horizontal components as seen below.

The horizontal component of tension

T_{x}

can be found using trigonometry and the definition of

cosine

cos 60^{o} = \frac{adjacent}{hypotenuse} = \frac{T_{x}}{T}

T_{x} = T {cos60}^{o}

The

T

here is the total magnitude of the tension force and

T_{x}

is only the horizontal component of that force (e.g. the amount by which the tension pulls horizontally).

Note that we do not plug in the force of gravity or the normal force into this equation since those forces are directed vertically and we're using Newton's second law for the horizontal direction.

T cos 60^{o} = (3.0 \frac{m}{s^{2}}) (2.0 kg) (get the T isolated on one side)

T = \frac{(3.0 \frac{m}{s^{2}}) (2.0 kg)}{cos 60^{o}} (solve algebraically for T)

T = 12 N (calculate and celebrate)

Example 2: Box hanging from two ropes

0.25 kg

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

θ = 30^{o}

from the horizontal direction as seen below.

What are the tensions $(T_{1}$ ‍ and $T_{2})$ ‍ 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{Σ F_{y}}{m} (use Newton’s second law for the vertical direction)

0 = \frac{T_{2} sin 30^{o} - F_{g}}{0.25 kg} (plug in the vertical acceleration, mass, and vertical forces)

Since we're using Newton's second law for the vertical direction, we only include vertical forces.

The force of gravity (of magnitude

F_{g}

) is directed vertically downward so we include it with a negative sign.

Also, since the tension

T_{2}

is directed diagonally we must break it into horizontal and vertical components as seen below.

We can find the vertical component of the tension

T_{2}

by using the definition of sine.

sin θ = \frac{opposite}{hypotenuse} = \frac{T_{2 y}}{T_{2}}

T_{2 y} = T_{2} sin θ = T_{2} sin 30^{o}

This is the component we include in Newton's second law for the vertical direction. Also, note that this vertical component of the tension

T_{2}

must be equal to the force of gravity so that the vertical forces can cancel, ensuring no vertical acceleration.

T_{2} = \frac{F_{g}}{sin 30^{o}} (solve for T_{2})

T_{2} = \frac{m g}{sin 30^{o}} (use the fact that F_{g} = m g)

T_{2} = \frac{(0.25 kg) (9.8 \frac{m}{s^{2}})}{sin 30^{o}} = 4.9 N (calculate and celebrate)

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{Σ F_{x}}{m} (use Newton’s second law for the horizontal direction)

0 = \frac{T_{2} cos 30^{o} - T_{1}}{0.25 kg} (plug in the horizontal acceleration, mass, and horizontal forces)

We have to use the horizontal component of the tension

T_{2}

since we're using Newton's second law in the horizontal direction.

We can do this by using the definition of cosine.

cos θ = \frac{adjacent}{hypotenuse} = \frac{T_{2 x}}{T_{2}}

T_{2 x} = T_{2} cos θ = T_{2} cos 30^{o}

Note that this horizontal component of the tension

T_{2 x}

must equal the tension

T_{1}

so the horizontal forces can cancel, ensuring that there is no horizontal acceleration.

T_{1} = T_{2} cos 30^{o} (solve for T_{1})

T_{1} = (4.9 N) cos 30^{o} (plug in the value we found for T_{2} = 4.9 N)

T_{1} = 4.2 N (calculate and celebrate)

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Dhruv Makwana
Posted 8 years ago. Direct link to Dhruv Makwana's post “How come in these questio...”
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?
Button navigates to signup pageComment on Dhruv Makwana's post “How come in these questio...”
(11 votes)
Answer
- Vo Thuong
  Posted 7 years ago. Direct link to Vo Thuong's post “But if you don't know the...”
  But if you don't know the hypotenuse the you must use tangent, otherwise you can't calculate your answer. For example:
  angle=60 deg, hypotenuse=?, adj=10, opp=?
  tan 60=opp/10 --> 10*tan 60deg=opp=17.320508075688772935274463415059
  Button navigates to signup page
  (7 votes)
SeaSea
Posted 8 years ago. Direct link to SeaSea's post “If two masses are connect...”
If two masses are connected by the same rope, in what circumstances would the force tension differ between the two masses?
Button navigates to signup pageButton navigates to signup page
(11 votes)
Answer
- Banglue Wei
  Posted 5 years ago. Direct link to Banglue Wei's post “The rope is acting as a t...”
  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.
  Comment on Banglue Wei's post “The rope is acting as a t...”
  (5 votes)
Kyle Delaney
Posted 8 years ago. Direct link to Kyle Delaney's post “In example 1, there's an ...”
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?
Button navigates to signup pageButton navigates to signup page
(10 votes)
Answer
- DINESH
  Posted 8 years ago. Direct link to DINESH's post “Normal force exists only ...”
  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.
  Button navigates to signup page
  (3 votes)
Pooja
Posted 8 years ago. Direct link to Pooja's post “Does it necessarily matte...”
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?
Button navigates to signup pageComment on Pooja's post “Does it necessarily matte...”
(6 votes)
Answer
Victoria
Posted 8 years ago. Direct link to Victoria's post “Do you have to account fo...”
Do you have to account for the mass of the rope, too?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Charles LaCour
  Posted 8 years ago. Direct link to Charles LaCour's post “Yes but until you get to ...”
  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.
  Comment on Charles LaCour's post “Yes but until you get to ...”
  (10 votes)
Sarah Anne Groleau
Posted 8 years ago. Direct link to Sarah Anne Groleau's post “I don't understand the -T...”
I don't understand the -T1 and -Fg? Why is it there and if so, how do we know if it is + or - ?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- nileshbenz
  Posted 8 years ago. Direct link to nileshbenz's post “you are free to choose si...”
  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.
  Button navigates to signup page
  (6 votes)
gorman.uwf
Posted 8 years ago. Direct link to gorman.uwf's post “In example 1, it says sol...”
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.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Adam Kubriczky
  Posted 8 years ago. Direct link to Adam Kubriczky's post “No, it is solved for the ...”
  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.
  Button navigates to signup page
  (3 votes)
ASHUTOSH
Posted 8 years ago. Direct link to ASHUTOSH's post “"Σ" what is the meaning o...”
"Σ" what is the meaning of this symbol?
Button navigates to signup pageComment on ASHUTOSH's post “"Σ" what is the meaning o...”
(1 vote)
Answer
- Diego Alves
  Posted 8 years ago. Direct link to Diego Alves's post “Sum. It is a greek letter...”
  Sum. It is a greek letter called sigma when is showed means a sum of the numbers.
  Comment on Diego Alves's post “Sum. It is a greek letter...”
  (5 votes)
hanifyuandi14
Posted 8 years ago. Direct link to hanifyuandi14's post “in simple pulley problem,...”
in simple pulley problem, like 2 mass on the rope. I've found out sometime in the tension of ther rope you have same tension and some time not. It"s sam when the mass of pulley ignored, and different when using rotaional dynamic. Can someone explain to why, and tell me how to know the tension is the same or not ?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Andrew M
  Posted 8 years ago. Direct link to Andrew M's post “you said it yourself: if...”
  you said it yourself: if the pulley and rope a massless, tension is the same throughout the rope. If not, it's not.
  Comment on Andrew M's post “you said it yourself: if...”
  (3 votes)
Caroline Reinstadtler
Posted 8 years ago. Direct link to Caroline Reinstadtler's post “How does this principle w...”
How does this principle work vertically? Is it solved the same way? How is gravity included?
Button navigates to signup pageButton navigates to signup page
(0 votes)
Answer
- Sam Woon
  Posted 8 years ago. Direct link to Sam Woon's post “If the string is vertical...”
  If the string is vertical, the tension is equal in magnitude to the weight force (F = mg)
  Button navigates to signup page
  (7 votes)