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 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. [How can a rope exert a force?]
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
=
ΣF
) for a direction in which the tension is directed.
m
3. Solve for the tension using the Newton's second law equation a
=
ΣF
.
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 2.0 kg box of cucumber extract is being pulled across a frictionless table by a rope at an angle
θ = 60o as seen below. The tension in the rope causes the box to slide across the table to the right with
m
an acceleration of 3.0 2 .
s
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.
[What if we accidentally chose to use the second law in the vertical direction?]
ax =
3.0
ΣFx
m
(use Newtons's second law for the horizontal direction)
m
T cos60o
=
s2
2.0 kg
(plug in the horizontal acceleration, mass, and horizontal forces)
[Where did the Tcos60 term come from?]
T cos60o = (3.0
T =
(3.0
m
)(2.0 kg) (get the T isolated on one side)
s2
m
)(2.0 kg)
s2
cos60o
(solve algebraically for T )
T = 12 N (calculate and celebrate)
Example 2: Box hanging from two ropes
A 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 T2 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 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 =
0=
ΣFy
m
(use Newtons's second law for the vertical direction)
T2 sin30o − Fg
0.25 kg
(plug in the vertical acceleration, mass, and vertical forces)
[Where did the Tsin30 come from?]
T2 =
Fg
sin30o
(solve for T2 )
T2 =
mg
sin30o
(use the fact that Fg = mg)
T2 =
(0.25 kg)(9.8
sin30o
m
)
s2 = 4.9 N (calculate and celebrate)
Now that we know T2 we can solve for the tension T1 using Newton's second law for the horizontal
direction.
ax =
0=
ΣFx
m
(use Newtons's second law for the horizontal direction)
T2 cos30o − T1
0.25 kg
(plug in the horizontal acceleration, mass, and horizontal forces)
[Where did the Tcos30 term come from?]
T1 = T2 cos30o
(solve for T1 )
T1 = (4.9 N)cos30o
(plug in the value we found for T2 = 4.9 N)
T1 = 4.2 N (calculate and celebrate)
[Attributions and References]
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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.
2 Votes
• Comment • Flag
3 months ago by
Chris Gorman