# TENSION FORCE ON ROPE

## Physics lesson, the subject of the tension force in rope. Finding of tension force on rope. Calculations of rope tension at various angles and systems.

In a rope that exposed to any force is occur tension.This tension is transmitted along the rope. At all point of a rope the tension force is equal.
When a force of 20 N is applied to end of the a rope, magnitude of the force at other end of the rope will be 20 N too. If we connection a mass to other end of the rope, magnitude of the force that affected the mass will be 20 N too.

In the following examples describe how to find the rope tension on various systems.
The examples are listed right from easy to difficult.

We are divided the subject to two section due it is too long. The link of second section is at the end of this page.

Question -1 In the figure above is seen a rope that connect from its a end to the Wall. A force of 20 N is applied to the other end of the rope at the direction of the arrow.

The tension force of on "A" point is "TA",
The tension force of on "B" point is "TB",
Applied force to the Wall is "FW"
Find the "TA", "TB", "FW" values.

Solution -1:
This is a easy question. A force applied to any end of a rope is transmitted along the rope. At all point of a rope the tension force is equal. The force exerted on the Wall is 20 N too.
According to this,
TA = 20 N
TB = 20 N
FW = 20 N

Question - 2 In the figure above is mass of the "K" object is 5 kg, mass of the "L" object 6 kg, mass of the "M" object 4 kg.
According to this informations,
Find the tension forces of the T1, T2, and T3 ropes.

Solution -2

Let we start to solve from the most bottom rope.
Tension of the T3 rope is equal to weight of "K" object and is found as the following.
T3 = mK•g
T3 = 5 • (9.8)
T3 = 49 N

Tension of the T2 rope is equal to total of weight of objects K and L and is found as the following.
T2 = mK•g + mL•g
T2 = 5• (9.8) + 6• (9.8)
T2 = 107.8 N

Tension of the T2 rope can be found as the following also.
T2 = T3 + mL • g
T2 = 49 N + 6 • 9.8
T2 = 49 + 58.8
T2 = 107.8 N

Tension of the T1 rope is equal to total of weight of the "K", "L", and "M" objects.
T1 = mK • g + mL • g + mM • g
T1 = 5 • 9.8 + 6• 9.8 + 4• 9.8
T1 = 49 + 58.8 + 39.2
T1 = 147 N

Tension of T1 rope can be found as the following also.
T1 = T2 + T3 + 4 . 9,8
T1 = 49 + 58.8 + 39.2
T1 = 147 N

Question -3 System in the figure is in balance with the T1, T2, T3 ropes and "K" object. Mass of the "K" object is 10 kg.
Find the tension forces on T1, T2 and T3 ropes.

(g (acceleration of gravity) = 10 m/s2)
(sin37° = 0.6, cos37° = 0.8)

Solution -3

Tension force of T1 rope is equal to the weight of the "K" object.
T1 = m•g
T1 = 10 • 10 = 100 N
We will carry the T3 and T1 vectors and create a right triangle. The hypotenuse of the triangle will be T1 vector.
as the following. We will calculate the tension of ropes from trigonometric ratios of a right triangle.
T1 = 100 N was.

 cos53° = T2 T1

 0.6 = T2 100

T2 = 60 N

 cos37° = T3 T1

 0.8 = T3 100

T3 = 80 N

Question - 4 System in the figure is balanced with the T1, T2, T3 ropes and the "K" object.
The tension force on T2 rope is 90 N.
This according to datas,
A) What are tension force of the T1 and T2 ropes.
B) How many kg is mass of the K object.
(g (acceleration of gravity) = 10 m/s2)

Solution 4

The T3 rope tension is equal to the weight of the "K" object.
We will carry the T3 and T2 vectors to up and create a triangle. The hypotenuse of the triangle will be the T3 vector.
as below. For T2 rope tension,

 cos37° = T2 T3

 0.8 = 90 T3

T3 = 112.5

The tension force of the T3 rope and the weight of the "K" object is 112.5 N. The weight of an object equals to the multiplication of mass and gravity.

112.5 = mK • 10
mK = 11.25 kg

For T1 rope tension

 cos53° = T1 F'

 0.6 = T1 112,5

T1 = 67,5 N

Question -5 The system in the figure is consists of T1 and T2 ropes and K body.
The T1 and the T2 ropes can withstand maximum 90 N tension.
According to the above information, how many kg could be the maximum mass of object K?
(g = 10 m/s2 , cos25° = 0.9, cos65° = 0.4)

Solution 5
The weight of K object is a vector.
The tensile forces at the ropes T1 and T2 are also a vector.
The sum of the T1 and T2 vectors must equal the weight of the K object.
Let the weight of the K object be F.

Let we find the relation between the T1 rope tension and the F force.

 T1 = cos25° F

T1 = 0.9F
If the T1 rope can be withstand of force 90 N up.
90 = 0.9•F

 F = 90 0.9

F = 100 N
F = m•g
F = 10•m
m = 10 kg.
Because of is larger the angle of between T1 and ceiling the big part of the weight of K object is on the T1 rope.
Calculate is not necessary for T2 rope. When the mass of the K object is greater than 10 kg, the T1 rope will break off

RISE KNOWLEDGE
06/06/2018

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