Tension Force Formula | Problems (With Solutions)

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Tension Force Formula

Tension force (FT) is equal to the weight (mg) plus/minus the product of mass (m) and acceleration (a) of an object. Using the formula of tension force: FT = mg ± ma, the value of tension force acting on an object can be calculated.

When an object is moving upward, FT = mg + ma
When an object is moving downward, FT = mg – ma
When an object is at rest, FT = mg

Let’s solve some problems based on this formula, so you’ll get a clear idea.

Tension Force Practice Problems

Problem 1: A wooden block of mass 2 kg is hung on a rope. If a wooden block is accelerating at the rate of 25 m/s2 in the upward direction, then calculate the value of tension force acting on a wooden block. (Take the value of gravitational acceleration, g = 9.81 m/s2)

Solution:

Given data:
Mass of a wooden block, m = 2 kg
Acceleration of a wooden block, a = 25 m/s2
Tension force acting on a wooden block, FT = ?

Using the formula of tension force, (when an object is moving upward)
FT = mg + ma
FT = (2 × 9.81) + (2 × 25)
FT = 19.62 + 50
FT = 69.62 N

Therefore, the tension force acting on a wooden block is 69.62 N.

Problem 2: One block of mass 8 kg is attached with a string and it is accelerating downward at the rate of 5 m/s2. Find the value of tension force acting on a block.

Solution:

Given data:
Mass of a block, m = 8 kg
Acceleration of a block, a = 5 m/s2
Tension force acting on a block, FT = ?

Using the formula of tension force, (when an object is moving downward)
FT = mg – ma
FT = (8 × 9.81) – (8 × 5)
FT = 78.48 – 40
FT = 38.48 N

Therefore, the tension force acting on a block is 38.48 N.

Problem 3: What is the value of tension force acting on a rubber tyre of mass 2 kg hanging from a rope? (Note that a rubber tyre is at rest)

Solution:

Given data:
Mass of a rubber tyre, m = 2 kg
Tension force acting on a rubber tyre, FT = ?

Using the formula of tension force, (when an object is at rest)
FT = mg
FT = 2 × 9.81
FT = 19.62 N

Therefore, the tension force acting on a rubber tyre is 19.62 N.

Problem 4: Calculate the tension force acting on a box of mass 4 kg for the following: (a) When a box is moving upward with the acceleration of 8 m/s2 (b) When a box is moving downward with the acceleration of 6 m/s2 (c) When a box is at rest

Solution:

Given data:
Mass of a box, m = 4 kg
Tension force acting on a box, FT = ?

(a) When a box is moving upward with the acceleration of 8 m/s2

Acceleration of a box, a = 8 m/s2

Using the formula of tension force, (when an object is moving upward)
FT = mg + ma
FT = (4 × 9.81) + (4 × 8)
FT = 39.24 – 32
FT = 7.24 N

Therefore, the tension force acting on a box is 7.24 N.

(b) When a box is moving downward with the acceleration of 6 m/s2

Acceleration of a box, a = 6 m/s2

Using the formula of tension force, (when an object is moving downward)
FT = mg – ma
FT = (4 × 9.81) – (4 × 6)
FT = 39.24 – 24
FT = 15.24 N

Therefore, the tension force acting on a box is 15.24 N.

(c) When a box is at rest

Using the formula of tension force, (when an object is at rest)
FT = mg
FT = 4 × 9.81
FT = 39.24 N

Therefore, the tension force acting on a box is 39.24 N.

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