Please enter the values to find the tension (force) in a rope, string, or similar object.
This calculator helps you to calculate the tension (force) in a rope, cable, or string that is used to lift (stretch) or pull an object. With this tool, you can make tension calculations in physics for:
Tension is a stretching or pulling force that is transmitted axially along an object such as a rope, cable, chain, etc. to pull an object. It is a special kind of force that acts in a direction opposite to the compression and acts on opposite ends of the rope.
Tension in a cable (due to object hanging) can be classified in two cases:
Suppose an object is lifted with a string, as shown in the above picture. In this condition, the tension in the string is equal to the weight of the object due to gravity (approximately 9.8 m/s²). Now if we consider the upward force as positive and the downward force as negative, they cancel each other’s effect, and the overall sum of both forces equates to zero.
ΣF↑ = 0 = T + (-W) T = W
where;
This is a bit more complex case, where the tension is distributed along two ropes used to suspend an object of hanging mass ‘m’. This force has influence along the horizontal and vertical components of the force.
As the gravitational force acts vertically downwards, we will only consider the vertical components of the pulley tension force, such that:
\(\sum{F}↑=0\) \(T_{1y} + T_{2y} + \left(-W\right) = 0\)
Moving ‘-W’ to the other side of the equation
\(W = T_{1y} + T_{2y}\)
The components of the angle along
\(T_{1y}\) and \(T_{2y}\) can be expressed in terms of \(T_{1}\) and \(T_{2}\),
such that:
\(T_{1y} = T_{1} × sin\left(α\right)\)
\(T_{2y} = T_{2} × sin\left(β\right)\)
\(W = T_{1} × sin\left(α\right) + T_{2} × sin\left(β\right)\) — (1)
Now coming to the horizontal components, there is no movement in this direction because the whole system is in a static equilibrium state. It shows that both of the x components are equal to each other.
\(T_{1}x = T_{2}x\) or \(T_{1} × cos\left(α\right) = T_{2} × cos\left(β\right)\) Moving \(cos\left(α\right)\)
to the other side; \(T_{1} = \dfrac{T_{2} × cos\left(β\right)}{cos\left(α\right)}\)
Putting the value of \(T_{1}\) in equation (1);
\(W = T_{1} × sin\left(α\right) + T_{2} × sin\left(β\right)\)
\(W = T_{2} * [\dfrac{cos\left(β\right)}{cos\left(α\right)}] × sin\left(α\right) + T_{2} × sin\left(β\right)\)
\(W = T_{2} × [\dfrac{cos\left(β\right) × sin\left(α\right)}{cos\left(α\right) + sin\left(β\right)}]\)
\(T_{2} = \dfrac{W}{[\dfrac{cos\left(β\right) × sin\left(α\right)}{cos\left(α\right) + sin\left(β\right)}]}\)
Now we have;
\(T_{1} = \dfrac{W}{[\dfrac{cos\left(β\right) × sin\left(α\right)}{cos\left(α\right) + sin\left(β\right)}]} × [\dfrac{cos\left(β\right)}{cos\left(α\right)}]\)
The given formula is considered to calculate tension (force) considering tension formula with angle of its orientation.
\(T_{1} = \dfrac{W}{[\dfrac{cos\left(α\right) × sin\left(β\right)}{cos\left(β\right) + sin\left(α\right)}]}\)
In dynamic equilibrium, the value of acceleration (a) is not zero. In this condition, tension in a string has variable cases, including:
| Motion of the Object | Rope Tension (T) |
|---|---|
| Moving Upward with Acceleration (a) | T = W + ma |
| Moving Downward with Acceleration (a) | T = W - ma |
| Suspended (Not Moving) | T = W |
| Moving Upward or Downward at Uniform Speed | T = W |
Our tension calculator also considers these cases to help you find tension in a string (cable) under dynamic equilibrium state.
Steps to find the tension force applied through a string while pulling an object:
Considering the terms of physics, the tension force calculator is capable of determining the pulling force acting either with rope, wire, cable, etc.
Let’s solve a couple of examples to better understand the concept of tension!
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A mass of 10kg is attached to a string and pulled against a frictionless surface at an angle of \(35^\text{o}\). What is the tension in the string?
Solution:
Step 1: As there is no frictional force, so tension will be equal to gravitational force, such that:
\(T = f_{g}\)
Step 2: Write the expression for tension in the string.
\(\displaystyle T = mgsin(\theta)\)
Step 3: Input given values to solve for the result.
\(\displaystyle T = 10 * 9.8 sin(35^\text{o})\)
\(\displaystyle T = 56.154 N\)
Find the tension in a string that is used to hand a tire of 30kg at a height of 15m?
Solution:
Step 1: Find the total of all forces.
\(F_{g} = m * g\) \(F = T + F_{g}\)
\(F = T + \left(30kg\right) * 9.8ms^{-2}\)
\(F = T - 294N\)
Step 2: Identify acceleration. Since the tire is not moving, there is no acceleration.
Step 3: Determine the tension force:
\(F = T - 294N\) \(30*0 = T - 294N\)
\(T = 294N\)
Instead of manual calculations, you can solve your examples by simply adding the values into the tension calculator.
Yes, it is. When you tie an object of a certain hanging mass with a string, an internal pulling force is generated in the string that helps connect the object and the reference point. This is why, tension is regarded as contact force.
The tension force is negative along the opposite side of the direction of motion.
As we know the work equation:
W = F*S
Now as it is known that tension in a string does not cause any displacement, so ‘S=0’
W = F*0
W = 0
Hence proved, the work done by tension is always zero.
The tension and gravitational forces act in the opposite directions of each other. Now if the hanging object is not balanced by tension, it will accelerate towards the ground due to the fore of gravity.
From the source of Wikipedia: Tension (physics), System in equilibrium, System under net force
From the source of Khan Academy: The force of tension, Super hot tension, Tension in an accelerating system and pie in the face, Mild and medium tension
From the source of Lumen Learning: Normal Force, Tension, and Other Examples of Forces, Normal Force, Tension, Real Forces and Inertial Frames, Problem-Solving Strategies
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