Mechanical Advantage Formula: Definition, Unit and Important Questions

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Mechanical Advantage is the ratio of input force to output force. Mechanical advantage is used to assess the performance of a machine and to calculate the gain in force by the mechanical system. The term mechanical advantage is frequently used in reference to simple machines. It is used while studying the mechanism and working of a lever. Let’s have a look at the formula of mechanical advantage and some important questions related to mechanical advantage.

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Table of Content

Introduction to Mechanical Advantage
Formula of Mechanical Advantage
Unit of Mechanical Advantage
Things to Remember
Sample Questions

Introduction to Mechanical Advantage

Mechanical Advantage refers to the ratio of input force to output force.

Work refers to the amount of energy that transfers when a body is moved by an external force propagated in the direction of displacement. So, work done is the product of displacement or distance and force.

A large force applied at a small distance and a light force applied at a long distance can perform the same amount of work. So, the mechanical advantage is the ratio of input force to output force.

The extent to which friction and other factors are decreasing the actual work output of the machine determines the efficiency of the machine. A frictionless machine will have 100% efficiency. The ratio of the actual mechanical advantage and theoretical mechanical advantage is equal to the efficiency of the machine.

In an ideal case, the force of the object and the force applied to overcome the applied force are the same.

Also Check:

Central Force

Moment of Force

Balanced Force

Formula of Mechanical Advantage

Mechanical Advantage is the ratio of force applied to the object (input force) to the effort required to overcome the applied force (output force). So, the mechanical advantage formula is given by:

\(MA = \frac{F_{B}}{F_{A}}\)

In this formula,

MA = Mechanical Advantage

F_B = Force of the object

F_A = Effort to overcome the applied force

Mechanical Advantage

Mechanical Advantage

Check Important Formula for Linear Momentum

Unit of Mechanical Advantage

Mechanical Advantage does not have a Unit. Both factors in the mechanical advantage formula are the same (force) and have the same unit, so mechanical advantage is a dimensionless parameter.

Things of Remember Based on Mechanical Advantage Formula

Mechanical Advantage refers to the ratio of input force to output force.
Mechanical Advantage is used to assess the performance of a machine. It is used for the calculation of gain in force by the mechanical system.
The term mechanical advantage is frequently used in reference to simple machines. It is used while studying the mechanism and working of a lever.
The ratio of the actual mechanical advantage and theoretical mechanical advantage is equal to the efficiency of the machine.
The formula of mechanical advantage is \(MA = \frac{F_{B}}{F_{A}}\). In the formula, MA stands for Mechanical Advantage. F_B stands for Force of the object and F_A stands for Effort to overcome the applied force. Mechanical Advantage does not have a Unit.
In an ideal case of mechanical advantage, the force of the object and the force applied to overcome the applied force are the same.

Also Check Further:

Bulk Modulus

Velocity

Poisson's Ratio

Sample Questions Based on Mechanical Advantage Formula

Ques: Describe an ideal case of Mechanical Advantage. (2 Marks)

Ans: In an ideal case of mechanical advantage, the force of the object and the force applied to overcome the applied force are the same.

Ques: Mechanical Advantage is the ratio of input force to output force. Explain. (2 Marks)

Ans: Work is the amount of energy that transfers when a body is moved by an external force propagated in the direction of displacement. So, work done is the product of displacement or distance and force. A large force applied at a small distance and a light force applied at a long distance can perform the same amount of work. So, mechanical advantage is the ratio of input force to output force.

Ques: What is the value of Mechanical Advantage when the force is 500 Newton force is needed to overcome a load of 1000 Newton? (3 Marks)

Ans: In the above case, we need to find the mechanical advantage of the machine.

Effort to overcome the applied force = 1000 N = F_B

Force of the object = 500 N = F_A

Mechanical Advantage = F_B / F_A

Mechanical Advantage = 1000 / 500

Mechanical Advantage = 2

The value of Mechanical Advantage is 2.

Also Read More:

Law of Inertia

Friction Force Formula

Frictional Force

Ques: What is the effort needed to overcome the applied force if the force of the object is 200 N and Mechanical advantage of the machine is 5? (3 Marks)

Ans: In the above case, we need to find the effort needed to overcome the applied force.

Mechanical Advantage = 5 = MA

Force of the object = 200 N = F_A

So to find the effort to overcome the applied force (F_B):

\(MA = \frac{F_{B}}{F_{A}}\)

F_B = MA × F_A

F_B = 5 × 200

F_B = 1000 N

So the effort to overcome the applied force is 1000 N.

Ques: What is the force required to overcome a load of 720 Newton if Mechanical advantage of the lever is 9? (3 Marks)

Ans: In the above case, we need to find the force of the object.

Mechanical Advantage = 9 = MA

Effort to overcome the applied force = 720 N= F_B

So to find the force of the object (F_A):

\(MA = \frac{F_{B}}{F_{A}}\)

9 = 720 / F_A

F_A = F_B / MA

F_A = 720/ 9

F_A = 80 N

So, the force of the object is 80 Newton.

Ques: What is the value of Mechanical Advantage when the force is 150 Newton force is needed to overcome a load of 900 Newton? (3 Marks)

Ans: In the above case, we need to find the mechanical advantage of the machine.

Effort to overcome the applied force = 900 N = F_B

Force of the object = 150 N = F_A

Mechanical Advantage = F_B / F_A

Mechanical Advantage = 900 / 150

Mechanical Advantage = 6

The value of Mechanical Advantage is 6.

Also Read:

Gravitational Potential Energy

Newton’s Second Law of Motion

Newton’s Third Law of Motion

Ques: What is the effort needed to overcome the applied force if the force of the object is 300 N and Mechanical advantage of the machine is 6? (3 Marks)

Ans: In the above case, we need to find the effort needed to overcome the applied force.

Mechanical Advantage = 6 = MA

Force of the object = 300 N = F_A

So to find the effort to overcome the applied force (F_B):

\(MA = \frac{F_{B}}{F_{A}}\)

F_B = MA ×F_A

F_B = 6 × 300

F_B = 1800 N

So,the effort to overcome the applied force is 1800 N.

Ques: What is the force required to overcome a load of 500 Newton if Mechanical advantage of the lever is 10? (3 Marks)

Ans: In the above case, we need to find the force of the object.

Mechanical Advantage = 10 = MA

Effort to overcome the applied force = 500 N= F_B

So to find the force of the object (F_A):

\(MA = \frac{F_{B}}{F_{A}}\)

10 = 500 / F_A

F_A = F_B / MA

F_A = 500 / 10

F_A = 50 N

So, the force of the object is 50 Newton.

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Mechanical Advantage Formula: Definition, Unit and Important Questions

Introduction to Mechanical Advantage

Formula of Mechanical Advantage

Unit of Mechanical Advantage

Things of Remember Based on Mechanical Advantage Formula

Sample Questions Based on Mechanical Advantage Formula

CBSE CLASS XII Related Questions

1.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

2.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

3.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

5.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

6.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

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Mechanical Advantage Formula: Definition, Unit and Important Questions

Introduction to Mechanical Advantage

Formula of Mechanical Advantage

Unit of Mechanical Advantage

Things of Remember Based on Mechanical Advantage Formula

Sample Questions Based on Mechanical Advantage Formula

CBSE CLASS XII Related Questions

1.Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

2.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

3. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

5.Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

6.Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

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SUBSCRIBE TO OUR NEWS LETTER

1.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

2.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

3.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

5.
Suppose a pure Si crystal has \( 5 \times 10^{28} \) atoms per \( \text{m}^3 \). It is doped with \( 5 \times 10^{22} \) atoms per \( \text{m}^3 \) of Arsenic. Calculate majority and minority carrier concentration in the doped silicon. (Given: \( n_i = 1.5 \times 10^{16} \, \text{m}^{-3} \))

6.
Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now: