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,
  1. Draw the forces exerted on the object in question.
  2. Write down Newton's second law (a=ΣFm)(a=\dfrac{\Sigma F}{m}) for a direction in which the tension is directed.
  3. Solve for the tension using the Newton's second law equation a=ΣFma=\dfrac{\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 2, point, 0, space, k, g box of cucumber extract is being pulled across a frictionless table by a rope at an angle theta, equals, 60, start superscript, o, end superscript as seen below. The tension in the rope causes the box to slide across the table to the right with an acceleration of 3, point, 0, start fraction, m, divided by, space, s, start superscript, 2, end superscript, end fraction.
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.
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.
It's no big deal though. There are really only two directions to choose (e.g. x or y). If you choose the wrong one and find you can't solve, just choose the other direction.
ax=ΣFxm(use Newtonss second law for the horizontal direction)a_x=\dfrac{\Sigma F_x}{m} \quad \text{(use Newtons's second law for the horizontal direction)}
3, point, 0, start fraction, m, divided by, space, s, start superscript, 2, end superscript, end fraction, equals, start fraction, start color purpleD, T, end color purpleD, c, o, s, 60, start superscript, o, end superscript, divided by, 2, point, 0, space, k, g, end fraction, space, left parenthesis, p, l, u, g, space, i, n, space, t, h, e, space, h, o, r, i, z, o, n, t, a, l, space, a, c, c, e, l, e, r, a, t, i, o, n, comma, space, m, a, s, s, comma, space, a, n, d, space, h, o, r, i, z, o, n, t, a, l, space, f, o, r, c, e, s, right parenthesis
We have to break the diagonal tension into vertical and horizontal components as seen below.
The horizontal component of tension start color purpleD, T, start subscript, x, end subscript, end color purpleD can be found using trigonometry and the definition of c, o, s, i, n, e,
c, o, s, 60, start superscript, o, end superscript, equals, start fraction, a, d, j, a, c, e, n, t, divided by, space, h, y, p, o, t, e, n, u, s, e, end fraction, equals, start fraction, T, start subscript, x, end subscript, divided by, T, end fraction
start color purpleD, T, start subscript, x, end subscript, end color purpleD, equals, start color purpleD, T, end color purpleD, c, o, s, 60, start superscript, o, end superscript
The start color purpleD, T, end color purpleD here is the total magnitude of the tension force and start color purpleD, T, start subscript, x, end subscript, end color purpleD 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.
start color purpleD, T, end color purpleD, c, o, s, 60, start superscript, o, end superscript, equals, left parenthesis, 3, point, 0, start fraction, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, left parenthesis, 2, point, 0, space, k, g, right parenthesis, space, left parenthesis, g, e, t, space, t, h, e, space, start color purpleD, T, end color purpleD, space, i, s, o, l, a, t, e, d, space, o, n, space, o, n, e, space, s, i, d, e, right parenthesis
start color purpleD, T, end color purpleD, equals, start fraction, left parenthesis, 3, point, 0, start fraction, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, left parenthesis, 2, point, 0, space, k, g, right parenthesis, divided by, c, o, s, 60, start superscript, o, end superscript, end fraction, space, left parenthesis, s, o, l, v, e, space, a, l, g, e, b, r, a, i, c, a, l, l, y, space, f, o, r, space, start color purpleD, T, end color purpleD, right parenthesis
start color purpleD, T, end color purpleD, equals, 12, space, N, space, left parenthesis, c, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, right parenthesis

Example 2: Box hanging from two ropes

A 0, point, 25, space, k, g container of animal crackers hangs at rest from two strings secured to the ceiling and wall respectively. The diagonal rope under tension T, start subscript, 2, end subscript is directed at an angle theta, equals, 30, start superscript, o, end superscript from the horizontal direction as seen below.
What are the tensions left parenthesis, T, start subscript, 1, end subscript and T, start subscript, 2, end subscript, right parenthesis 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.
ay=ΣFym(use Newtonss second law for the vertical direction)a_y=\dfrac{\Sigma F_y}{m} \quad \text{(use Newtons's second law for the vertical direction)}
0, equals, start fraction, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, s, i, n, 30, start superscript, o, end superscript, minus, start color blueD, F, start subscript, g, end subscript, end color blueD, divided by, 0, point, 25, space, k, g, end fraction, space, left parenthesis, p, l, u, g, space, i, n, space, t, h, e, space, v, e, r, t, i, c, a, l, space, a, c, c, e, l, e, r, a, t, i, o, n, comma, space, m, a, s, s, comma, space, a, n, d, space, v, e, r, t, i, c, a, l, space, f, o, r, c, e, s, right parenthesis
Since we're using Newton's second law for the vertical direction, we only include vertical forces.
The force of gravity (of magnitude start color blueD, F, start subscript, g, end subscript, end color blueD) is directed vertically downward so we include it with a negative sign.
Also, since the tension start color purpleD, T, start subscript, 2, end subscript, end color purpleD is directed diagonally we must break it into horizontal and vertical components as seen below.
We can find the vertical component of the tension start color purpleD, T, start subscript, 2, end subscript, end color purpleD by using the definition of sine.
s, i, n, theta, equals, start fraction, o, p, p, o, s, i, t, e, divided by, h, y, p, o, t, e, n, u, s, e, end fraction, equals, start fraction, start color purpleD, T, start subscript, 2, y, end subscript, end color purpleD, divided by, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, end fraction
start color purpleD, T, start subscript, 2, y, end subscript, end color purpleD, equals, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, s, i, n, theta, equals, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, s, i, n, 30, start superscript, o, end superscript
This is the component we include in Newton's second law for the vertical direction. Also, note that this vertical component of the tension start color purpleD, T, start subscript, 2, end subscript, end color purpleD must be equal to the force of gravity so that the vertical forces can cancel, ensuring no vertical acceleration.
start color purpleD, T, start subscript, 2, end subscript, end color purpleD, equals, start fraction, start color blueD, F, start subscript, g, end subscript, end color blueD, divided by, s, i, n, 30, start superscript, o, end superscript, end fraction, space, left parenthesis, s, o, l, v, e, space, f, o, r, space, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, right parenthesis
start color purpleD, T, start subscript, 2, end subscript, end color purpleD, equals, start fraction, start color blueD, m, g, end color blueD, divided by, s, i, n, 30, start superscript, o, end superscript, end fraction, space, left parenthesis, u, s, e, space, t, h, e, space, f, a, c, t, space, t, h, a, t, space, start color blueD, F, start subscript, g, end subscript, end color blueD, equals, start color blueD, m, g, end color blueD, right parenthesis
start color purpleD, T, start subscript, 2, end subscript, end color purpleD, equals, start fraction, left parenthesis, 0, point, 25, space, k, g, right parenthesis, left parenthesis, 9, point, 8, start fraction, m, divided by, space, s, start superscript, 2, end superscript, end fraction, right parenthesis, divided by, s, i, n, 30, start superscript, o, end superscript, end fraction, equals, 4, point, 9, space, N, space, left parenthesis, c, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, right parenthesis
Now that we know start color purpleD, T, start subscript, 2, end subscript, end color purpleD we can solve for the tension start color greenD, T, start subscript, 1, end subscript, end color greenD using Newton's second law for the horizontal direction.
ax=ΣFxm(use Newtonss second law for the horizontal direction)a_x=\dfrac{\Sigma F_x}{m} \quad \text{(use Newtons's second law for the horizontal direction)}
0, equals, start fraction, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, c, o, s, 30, start superscript, o, end superscript, minus, start color greenD, T, start subscript, 1, end subscript, end color greenD, divided by, 0, point, 25, space, k, g, end fraction, space, left parenthesis, p, l, u, g, space, i, n, space, t, h, e, space, h, o, r, i, z, o, n, t, a, l, space, a, c, c, e, l, e, r, a, t, i, o, n, comma, space, m, a, s, s, comma, space, a, n, d, space, h, o, r, i, z, o, n, t, a, l, space, f, o, r, c, e, s, right parenthesis
We have to use the horizontal component of the tension start color purpleD, T, start subscript, 2, end subscript, end color purpleD since we're using Newton's second law in the horizontal direction.
We can do this by using the definition of cosine.
c, o, s, theta, equals, start fraction, a, d, j, a, c, e, n, t, divided by, h, y, p, o, t, e, n, u, s, e, end fraction, equals, start fraction, start color purpleD, T, start subscript, 2, x, end subscript, end color purpleD, divided by, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, end fraction
start color purpleD, T, start subscript, 2, x, end subscript, end color purpleD, equals, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, c, o, s, theta, equals, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, c, o, s, 30, start superscript, o, end superscript
Note that this horizontal component of the tension start color purpleD, T, start subscript, 2, x, end subscript, end color purpleD must equal the tension start color greenD, T, start subscript, 1, end subscript, end color greenD so the horizontal forces can cancel, ensuring that there is no horizontal acceleration.
start color greenD, T, start subscript, 1, end subscript, end color greenD, equals, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, c, o, s, 30, start superscript, o, end superscript, space, left parenthesis, s, o, l, v, e, space, f, o, r, space, start color greenD, T, start subscript, 1, end subscript, end color greenD, right parenthesis
start color greenD, T, start subscript, 1, end subscript, end color greenD, equals, left parenthesis, start color purpleD, 4, point, 9, space, N, end color purpleD, right parenthesis, c, o, s, 30, start superscript, o, end superscript, space, left parenthesis, p, l, u, g, space, i, n, space, t, h, e, space, v, a, l, u, e, space, w, e, space, f, o, u, n, d, space, f, o, r, space, start color purpleD, T, start subscript, 2, end subscript, end color purpleD, equals, start color purpleD, 4, point, 9, space, N, right parenthesis, end color purpleD
start color greenD, T, start subscript, 1, end subscript, end color greenD, equals, 4, point, 2, space, N, space, left parenthesis, c, a, l, c, u, l, a, t, e, space, a, n, d, space, c, e, l, e, b, r, a, t, e, right parenthesis
Parts of this article were adapted form the following article:
  1. "Normal, Tension, and Other Examples of Forces" from Openstax College Physics. Download the original article free at http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@9.4:27/Normal-Tension-and-Other-Examp