How To Find Tension Force With Acceleration: Steps, Problem Examples

In physics we deal with many kinds of forces like weight, collision, normal force, etc. This article deals with how to find tension force with acceleration.

Tension force is a contact force that comes into action whenever there is any object that is pulled using a cable, string, rope or a cord. It is a reactive force that counters the pulling force along a string. To determine tension force with acceleration, we use the results from Newton’s second and third laws of motion.

The most common example on how to find tension force with acceleration is the case where bodies are hung from a point through a chain or string.

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Steps on how to find tension force with acceleration:

Consider an object O of mass m kg suspended from a point. Assume that the object is accelerating up or down with an acceleration of a m/s2.

Our initial step is to construct a free body diagram. This diagram illustrates the different types of forces along with the direction in which these forces acts.

Since the body is suspended, the force of gravity is the primary force acting on it. The expression for  the gravitational force acting on any object of mass M kg is given by

F = Mg

where g is the acceleration due to gravity whose value is 9.8 m/s2. This expression is a result of the Netwon’s second law of motion which says that net force is given by the product of a body’s mass and its acceleration.

So here, the weight acting on the body W is given by

W = Mg

and is directed downwards.

Let T be the tension force acting along the string. Now we use Newton’s third law of motion to find the Tension force in the system. The third law states that

“To every action, there is an equal and opposite reaction”.

The summation of all upward forces must be equal to the summation of all downward forces if the system was in equilibrium.

Now, Newton’s second law of motion is applied since the body is accelerating. Hence, the Tension force T is given by

T = Wpm ma

There are a few cases:

  • If W and a are along the same direction, then tension force is given by T = W + ma
  • If W and a are along opposite directions, then tension force is given by T = W – ma
  • If there is no acceleration, then tension force is equal to the body’s weight, i.e., T = W

The basic assumptions made here are: the rope or string is massless and the motion is frictionless.

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How To Find Tension Force With Acceleration: Problem Examples

Problem 1:

Consider two objects of masses m1 and m2 respectively that are attached to the two ends of a massless rope passing over a frictionless and massless pulley. As shown in the figure, m2 is placed on an inclined plane. Neglecting friction, find the acceleration of the masses and tension in the string if m1 is 20 kg and m2 is 10 kg and θ is 45°.

how to find tension force with acceleration
Figure 1


For the mass m1 :

For the first object of mass m1, free body diagram is constructed as shown.

Free body diagram of m1

The forces acting on m1 are its weight m1g (directed downwards) and tension in the string T (acting upwards). From the figure, it is understood that m1 accelerates upwards. Hence apply second law of Newton,

T – m1g = m1a……………………………(1)

For the mass m2 :

Free body diagram for m2 is as shown in the figure.

Free body diagram of m2

The forces acting on this mass is due to its weight and normal reaction force N. These forces are resolved into components to determine the tension force. The components of m2g are m2g cosθ and m2g sin θ .  Of these m2g sin θ is parallel to the plane and opposite to the tension T. And m2g cosθ balances N. Hence we have,m2g cosθ

m2g sin θ – T =m2a………………………………………(2)

m2g cosθ = N…………………………………………………….(3)

Adding 1 and 2 equations,

T – m1g + m2g sin θ – T = m1a + m2a

a=m2g sin θ – m1g/m1+m2………(4)

Substituting the value of a in equation (1),


T=m1/m2 g sin θ – m12g/m1+m2 + m1g

T =m1/m2 g(1+sinθ )/m1+m2

It is given, m1 = 20 kg; m2 = 10 kg; θ= 45°

We have to find acceleration a and Tension T;

From equation (4), we have,

a = 10* 9.8sin 45-20*9.8/20+10

a =- 4.22 m/s2

Hence acceleration of the masses is -4.22 m/s2.

Now, substitute this value of a and m1 in equation (1),

T = m1g + m1a

T = m1(a+g)

T= 20 (-4.22 +9.8)

T = 111.6 N

Hence Tension in the string is 111.6 N.

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Deeksha Dinesh

Hello, I am Deeksha Dinesh, currently pursuing post-graduation in Physics with a specialization in the field of Astrophysics. I like to deliver concepts in a simpler way for the readers.

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