Tension formula: Tension in a vertically suspended wire with a weight

Tension is a force that acts opposite to the pulling force applied on a rope, wire, string, or any one-dimensional material. You might have noticed many objects being hanged using a rope or cables. It might be big bridges called suspension bridges or small ones like decorative hangings on a Christmas tree. In this article, we will look at more such examples and calculate the formula for tension in a rope used for hanging objects.

The tension in such vertically suspended wires cannot be calculated simply by a single formula. We have to derive the formula for tension for each and every case. The tension formula will vary depending on the number of weights involved, the number of ropes, the angle of attachment, and more such factors.

The formula for tension for a rope with a single weight attached to it

When a weight is attached to a rope, there are three scenarios. This image shows the formula for tension for all the three scenarios.
When weight is attached to a rope, there are three scenarios. This image shows the formula for tension for all three scenarios.

When a weight is attached to a rope there can be three scenarios possible:

  1. The object is at rest.
  2. The object is moving downwards
  3. The object is moving upwards

In the first case, the object is in a state of rest. So the tension in the string will be equal to the gravitational force mg. So, the formula for tension: T = m*g

In the second case, the object is moving downwards with acceleration(a). Examples of such cases include the pulley and elevator. You can read this article to find out about the formula for tension in the cables of an elevator in detail. Here, the net force(m*a) should be equal to the sum of the forces:

m*a = m*g – T

T = m*g – m*a = m*(g-a)

In the third case, the object is moving upwards with an acceleration(a). This case is again seen in elevators and pulley systems. Here the net force(m*a) should be again equal to the sum of the forces. In this equation, tension will be positive because it moves in the direction of motion.

m*a = T – m*g

T = m(g+a)

From the above formulas for tension, it can be seen that the tension in a rope is high when the rope is pulled upwards compared to being lowered. It is obvious, as we require more force to pull up an object than to drop it down.

The formula for tension for a rope with two weights attached to it

Now, let us see the formula for tension when two weights m1 and m2 are attached to the ceiling with the help of two ropes.

Tension in a vertically suspended wire with two weights attached to it.
Tension in a vertically suspended wire with two weights attached to it.

Since the ropes are different let us assign the tension as T1 and T2. There will be a gravitation force m1*g and m2*g acting on blocks 1 and 2 respectively. First, let us draw the free-body diagram of both blocks and find the equations for tension in the ropes.

Free-body diagram for tension formula in block2
The free-body diagram for tension in block2

In block 2, since the blocks are at rest, the tension T2 is equal to the gravitational force on block2.

T2 = m2*g —–(1)

Free-body diagram for tension formula in block1
The Free-body diagram of tension in block1

In block1, the tension of both ropes will be acting on the block along with the gravitational force(m1g). The net force on the block is zero.

T1-T2-m1g = 0 —–(2)

From equations (1) and (2) we can find the solution for tensions T1 and T2.

Numerical problems on tension

  1. Find the mass of the block. A tension of 40 N is acting on the rope which is in a state of rest.
Number problem on tension in a rope
Number problem on tension in a rope

Since the system is at rest we can directly apply the tension formula we derived.

T = m*g

40 = m*9.8

m = 40/9.8 = 4.08 Kg

The mass of the block will 4.08 Kg. The unit is a kilogram as the tension was given in the SI unit of force.

2. The tension in a rope is 40N with a block of mass m attached to it at one end. The whole system is moving downwards with an acceleration of 4 m/s2. Find the mass of the block.

Numerical problem of tension in a rope
The numerical problem of tension in a rope

We will again apply the formula for tension in the case of downward acceleration.

T = m(g-a)

40 = m(9.8-4)

m =40/5.8 = 6.9 kg

3. Two blocks of weight 2Kg and 3Kg are attached by two ropes. Find the tension T1 and T2 in the ropes.

Tension in a vertically suspended wire with weight.
Tension in a vertically suspended wire with weight.

From the above derivations for tension formula for a similar case, we arrived at two equations:

T2 = m2*g —–(1)

T2 = 3*9.8 = 29.4 N

T1-T2-m1g = 0 —–(2)

T1-29.4-(2*9.8) = 0

T1 = 9.8 N

Hence the values for the tension of the ropes are 9.8 N and 29.4 N.

For cases where the number of blocks is more than two the same derivation can be used to arrive at a formula for tension and then solve the equations for tension value as shown.

Real-life examples of tension:

Tension formula
Tension in a string suspended vertically with a weight
Image by Rondell Melling from Pixabay
  • A suspension bridge consists of a network of cables able to lift the weight of an entire bridge. The golden gate bridge in California is one of the most famous suspension bridges in the world.
  • The decorations on a Christmas tree have a small string attached to them, which bears the entire weight of the object.
  • The chandelier in our homes and churches have strong chains to bear the weight of such a heavy object.

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