**Tension force** (F_{T}) 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: **F _{T}**

**= mg ± ma**, the value of tension force acting on an object can be calculated.

When an object is moving upward, **F _{T} = mg + ma**

When an object is moving downward,

**F**

_{T}= mg – maWhen an object is at rest,

**F**

_{T}= mgLet’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/s^{2} 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/s^{2})

Solution:

Given data:

Mass of a wooden block, m = 2 kg

Acceleration of a wooden block, a = 25 m/s^{2}

Tension force acting on a wooden block, F_{T} = ?

Using the formula of tension force, (when an object is moving upward)

F_{T} = mg + ma

F_{T} = (2 × 9.81) + (2 × 25)

F_{T} = 19.62 + 50

F_{T} = 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/s^{2}. 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/s^{2}

Tension force acting on a block, F_{T} = ?

Using the formula of tension force, (when an object is moving downward)

F_{T} = mg – ma

F_{T} = (8 × 9.81) – (8 × 5)

F_{T} = 78.48 – 40

F_{T} = 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, F_{T} = ?

Using the formula of tension force, (when an object is at rest)

F_{T} = mg

F_{T} = 2 × 9.81

F_{T} = 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/s^{2} (b) When a box is moving downward with the acceleration of 6 m/s^{2} (c) When a box is at rest

Solution:

Given data:

Mass of a box, m = 4 kg

Tension force acting on a box, F_{T} = ?**(a)** When a box is moving upward with the acceleration of 8 m/s^{2}

Acceleration of a box, a = 8 m/s^{2}

Using the formula of tension force, (when an object is moving upward)

F_{T} = mg + ma

F_{T} = (4 × 9.81) + (4 × 8)

F_{T} = 39.24 – 32

F_{T} = 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/s^{2}

Acceleration of a box, a = 6 m/s^{2}

Using the formula of tension force, (when an object is moving downward)

F_{T} = mg – ma

F_{T} = (4 × 9.81) – (4 × 6)

F_{T} = 39.24 – 24

F_{T} = 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)

F_{T} = mg

F_{T} = 4 × 9.81

F_{T} = 39.24 N

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

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**Related:**

- Force Equation
- Normal Force Equation
- Net Force Formula
- Applied Force Formula
- Magnetic Force Equation
- Centripetal Force Equation
- Centrifugal Force Equation
- Spring Force Equation
- Electric Force Equation

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