Tension force is developed in a rope when a weight is attached to it. The tension developed in the rope should be equal to the gravitational pull on the weight. But this is true only for a where the rope is vertically suspended. The formula for tension also depends on the angle of suspension. Here in this article, we look at the examples, formulas, and numerical problems for tension when the rope is suspended at an angle to the ceiling.
The weight of mass m is suspended by two ropes with tension T1 and T2. The tension in both the ropes will be different, so we have to derive two separate sets of the formula for tensions in both strings. Since the weight is static, the net forces acting on the weight in the x and y direction should be zero.
We will start by drawing a free-body diagram and resolve the forces in x and y directions.
Resolving the forces in y-direction: The forces acting in the y-direction are a downward gravitational pull and component of tension forces T1 and T2 in an upward direction. Equating the force we get:
T1 sin(a) + T2 sin(b) = m*g ———-(1)
Resolving the forces in x-direction: The forces acting in the x-direction are the components of tension forces T1 and T2 in opposite directions. Equating the forces we get:
T1cos(a) = T2cos(b)———————(2)
Solving equations (1) and (2), we get the formula for tension.
T1 = [T2cos(b)]/cos(a)]
T2 = [T1cos(a)]/cos(b)]
From the above equations, we can also infer that the higher the angle of suspension, the higher will be the tension, with the maximum tension at a suspension of 90 degrees.
Real-life Examples of an object attached to a string at an angle
Image by skeeze from Pixabay
- Ropes used during rock climbing
- Ropes of a suspension bridge
- String attached to the sail of a ship
- Ropes used to hold a tent
- Strings used in Water sports
Problems on tension formula
The weight of a mass of 10 kg is suspended by two ropes at an angle, find the tension in the ropes.
We will follow the same approach as we used above for deriving the formula for tension. We will first resolve the forces in the x and y direction and form two equations.
Resolving the forces in the x-direction we have:
T1 sin30 + T2 sin60 = 98 N ———-(1)
Resolving the forces in the y-direction we have:
T1cos30 = T2cos60———————(2)
Solving the above equations we get the values for tension T1 and T2:
T1 = 65.1 N T2 = 112.3 N
We can see that T2 is higher than T1, which also shows that angle b is greater than a.
Calculate the tension in the string attached to a kite, provided there is no drag and the string has not pulled the boy.
Here in this case we have only a single string, so we can just resolve the forces in the y-direction and get the formula for tension.
T sin 60 = 50
T = 50/0.866
T = 57.7 N
The tension in the string is much more than the lift generated by the wind. This tension is provided by the muscular force in the hand of the boy holding it.
Tension in a horizontal suspension
A weight of 10 Kg is suspended horizontally by two ropes with tension T1 and T2. The length of the ropes is equal.
Now, in this case, the angles formed by the ropes with the ceiling are 0. Let us try resolving the forces and find the value of tension in the ropes. But such problems can be solved more easily logically rather than by equations. Such problems can be solved only under many assumptions like the equal length of ropes, massless ropes, 100% horizontal suspension, etc.
In the x-direction:
T1 and T2 are opposite, and there is no net force acting in the x-direction, hence they must be equal. Let us apply the formula for tension and verify this.
T1cos0 = T2cos0 => T1 = T2
Now, in the Y-direction:
The entire weight of the box has held the ropes, so the sum of the tensions should be equal to the weight. Let us apply the formula for tension to find out.
T1cos0 + T2cos0 = 98 N => T1 + T2 = 98
T1 = T2 = 49 N
