Static Friction Formula with Solved Examples

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Static friction is mainly concerned with the frictional force applied on a body or an object and acts as a resistive force that is responsible for keeping the object at rest. It is known as a self-regulating force. The value of the static friction ranges between zero and the smallest value of force that is needed by the body to start the motion. The static friction formula is the product of the coefficient of static friction and the normal force. In this article, we will read about the static friction formula, and look at some related solved examples.

Read More: Types of Friction

Table of Contents

What is Static Friction?
Static Friction Formula?
What is the Coefficient of the Friction?
Solved Examples
Things to Remember
Sample Questions

Key Takeaways: Static Friction, Coefficient of static friction, Friction, Formula, frictional force, resistive force, normal force.

What is Static Friction?

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A type of frictional force that is resistive in nature, which after being applied on a body or an object at rest, is characterized as static friction.

Static friction

Static friction

Read More: Coefficient of Static friction

Static Friction Formula

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The formula of the static friction mainly helps in calculating the frictional force and its coefficient. The formula for the same can be written as the following.

F_s = μ_s × F_n

Here,

F_s refers to the static friction

μ_s refers to the coefficient of static friction

F_n refers to the normal force

Work done by static friction

Work done by static friction

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Static friction

What is the Coefficient of the Friction?

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The amount of friction generated is mainly based on the nature of the objects or the bodies in contact with one another. For instance, the denser and the concrete objects are known for generating a greater amount of frictional force.

Therefore, the coefficient of friction majorly measures that how easily one object or body moves with respect to another. This shall be understood via the value of the coefficient of the friction, such as, if the value of the coefficient of friction is high, then the amount of friction between the objects is greater. However, the coefficient of friction does not represent any absolute unitary value.

Also Read:

Kinetic Friction	Electric Flux
Sliding friction	Rolling friction

Solved Examples

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Let us now look at some solved examples related to the static friction formula:

Example 1: Calculate the static friction if the actual force acting upon some car of 100kg is 200 N. However, the car remains in the static position. Given is the coefficient of friction, that is, 0.2.

Solution. Given: Normal force, F_n = 200N,

Coefficient of friction, μ_s = 0.2,

Therefore, Static friction shall be

F_s = μ_s × F_n

= 0. 2 × 200N

= 40 N

So, the static friction will be 40 N.

Example 2: Calculate the static friction if the coefficient of friction is given as 0.2. And the force of 300 N is acting on a TV box of 2 kg lying still on the floor.

Solution. Given: F_n (Normal force) = 300 N,

μ_s (Coefficient of friction) = 0.2,

Static friction is given by F_s = μ_sF_n

= 0.2 × 300 N

F_s = 60 N.

Things to Remember

The frictional force works completely against the movement of the body.
In general, force is responsible for the movement of any body in rest, but static friction is a kind of resistive force that prevents movement.
The formula of the static friction mainly helps in calculating the frictional force and its coefficient.
The formula for the static friction can be written as F_s= μ_s × F_n
The coefficient of friction measures that how easily one object or body moves with respect to another.

Sample Questions

Ques. Calculate the static friction if the actual force acting upon some box of 5kg is 300 N. However, the box remains in the static position. Given is the coefficient of friction, that is, 0.3. (3 marks)

Ans. Given: Normal force, F_n = 300N,

Coefficient of friction, μ_s = 0.3,

Therefore, Static friction shall be

F_s = μ_s × F_n

= 0.3 × 300N

= 90 N

So, the static friction will be 90 N.

Ques. Calculate the friction coefficient if 100 N is the value of the static frictional force. And a toy car of mass 4 kg is being pushed by a person but it remains at the static position. (5 marks)

Ans. Given: Mass, m = 4 kg,

Static frictional force, F_s = 100N,

Normal force, F_n = m × g

F_n = 5 × 10, now for acceleration due to gravity,

G = 10 ms⁻²

Hence, F_n = 50N

Now, Formula is,

F_s = μ_s × F_n

After rearranging it,

μ_s = F_s x F_n

On Putting the values,

μ_s =100/50

μ_s =2

Therefore, the friction coefficient is 2.

Ques. Calculate the static friction if the coefficient of friction is given as 0.3. And the force of 200 N is acting on a snack box of 5 kg lying still on the floor. (3 marks)

Ans. Given: F_n (Normal force) = 200 N,

μs (Coefficient of friction) = 0.3,

Static friction, F_s = μ_sF_n

= 0.3 × 200 N

F_s = 60 N

Ques. Calculate the friction coefficient when Amy is hauling a toy car of mass 4 kg which was at rest earlier on the floor. And if the given value of the static frictional force is 50 N. (2 marks)

Ans. Given: m (Mass) = 4 kg,

F_s (static frictional force) = 50 N,

F_n (Normal force) = mg

= 4 Kg × 9.8 m/s2

F_n = 39 N

μ_s = F_s/F_n

μ_s = 50/39

μ_s = 1.282

Ques. If a conveyor is dumping sand onto a cone-shaped pile. Find the maximum angle θ if the given coefficient of static friction between the sand grains is μ_s. (3 marks)

Ans. Given, the maximum sliding angle = θ

The acting force of gravity on the block is F₁ = Mgsinθ

The maximum friction force that is opposing the sliding force is F₂ = Mgcosθμ_s.

At some angle θ, the block will eventually be on the verge of sliding off.

This is the maximum angle θ and this will happen when F₁ = F₂.

Therefore, the θ_max = a tan(μ_s)

Ques. If the shear strength of each bolt is 5000 lbs of the two boards that are bolted together with two bolts. And the squeeze force between the boards is 500 lbs. The coefficient of static friction between the boards is μ_s = 0.5, what is the maximum force F that can be applied to the boards and not pull them apart? Find the static friction. (3 marks)

Ans. In order to pull the boards apart, the shear strength the bolts have and the friction force between the boards must exceed. Therefore, the maximum pull force must be of value below the force needed to do this.

Hence, F_max = 2 × 5000 + 500 × 0. 5.

F_max = 10250 lbs

Also Read:

Mechanical Properties of Fluids	Bernoulli’s principle
Hydraulic Machines	Friction Force Formula

Static Friction Formula with Solved Examples

What is Static Friction?

Static Friction Formula

What is the Coefficient of the Friction?

Solved Examples

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

2.
Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

3.
In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.

4.
Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

5.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

6.
The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

Similar Physics Concepts

Comments

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Static Friction Formula with Solved Examples

What is Static Friction?

Static Friction Formula

What is the Coefficient of the Friction?

Solved Examples

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

2.Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

3.In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.

4.Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

5.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

6.The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at:

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

2.
Assertion : Induced emf produced in a coil will be more when the magnetic flux linked with the coil is more. Reason (R): Induced emf produced is directly proportional to the magnetic flux.

3.
In a Young's double-slit experiment, two waves each of intensity I superpose each other and produce an interference pattern. Prove that the resultant intensities at maxima and minima are 4I and zero respectively.

4.
Assertion : In Young’s double-slit experiment, the fringe width for dark and bright fringes is the same. Reason (R): Fringe width is given by \( \beta = \frac{\lambda D}{d} \), where symbols have their usual meanings.

5.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

6.
The figure represents the variation of the electric potential \( V \) at a point in a region of space as a function of its position along the x-axis. A charged particle will experience the maximum force at: