D’Alembert’s Principle: Statement, Derivation and Mathematical Representation

D’Alembert’s Principle states that: for a unit of mass of debris, sum of difference of pressure acting at the machine and the time derivatives of the momenta is 0 when it is projected onto any digital displacement. D’Alembert's Principle is an alternative form of Newton’s second law of Motion. The principle indicates that any system of forces is in equilibrium if the exerted forces are added to fictitious forces.  

Key takeaways: D’Alembert’s principle. Newton’s Second Law, newton, work, force, displacement, Atwood’s machine

Read More: Physical Significance of Electric Field

---

What is D’Alembert’s Principle?

[Click Here for Sample Questions]

According to D’Alembert’s Principle,

“The sum of the differences between the forces exerted on a system of particles (with non-zero rest mass) and the time derivatives of the momenta of the system is zero when the system is projected into virtual displacement.”

D’Alembert’s principle is another interpretation of Newton’s law but the former is a more generalized version as it changes a problem in kinetics to statics. It is used in solving problems in statics, a branch of mechanics that analyses systems that have no acceleration and are in equilibrium with respect to their environment. And also is used to deal with problems in dynamics, consistent with constraint forces.

---

Mathematical Representation of D’Alembert’s Principle 

[Click Here for Sample Questions]

The mathematical representation of the principle is done using Newton's dot notation. Thus in general form that is in a system of particles with variable mass, it is written as:

\(\sum_{i} (F_i - m_iv_i - M_iv_i) . \sigma r_i =0\)

Where F_i is the net force exerted on the i-th particle 

m_i is the mass of the i-th particle

v_i is the velocity of the i-th particle

r_i is the virtual displacement of the i-th particle

Or, it can be also written as:

\(\sigma W = \sum_{i} (F_i - m_ia_i).\sigma r_i = 0\)

Where ai is the acceleration of the i-th particle.

Also check:

---

D’Alembert’s Principle Derivation

[Click Here for Sample Questions]

D’Alembert’s Principle has been derived through two cases: Equation for variable mass system and Equation for constant mass system.

Equation for Variable Mass System (General Case)

From the statement of D’Alembert’s principle, we have to find the time derivative of momentum for an i-th particle with variable mass and sum that to the force exerted on the particle. From newton’s second law the momentum of an i-th particle is given as: P_i = m_iv_i

Differentiating momentum of an i-th particle with respect to time,

\(\dot {P_i} = \dot {m_iv_i} + v_i \dot{m_i}\)

But we know acceleration is a = dv/dt. Using this in the above equation we get,

\(\dot {P_i} = m_i \dot{v_i} + m_ia_i\)

Summing up all the particles in the system we get the equation for d’Alembert’s principle for the variable mass system.

\(\sum_{i} (F_i - m_ia_i - M_iv_i) . \sigma r_i =0\)

Equation for Constant Mass System

Using newton’s second law that the acceleration of a particle is a virtue of the total force, 

\(F^{T}_{i} = m_ia_i\)

\(F^{T}_{i} = m_ia_i = 0\)

Now, for a virtual displacement r_i, the virtual work W done by the total and inertial forces on each particle is added to zero since the system is in equilibrium.

So we get virtual work as:

\(\sigma W =\sum_{i} F^{T}_{i} . \sigma r_i - \Sigma m_ia_i . \sigma r_i =0\)[[

But, the above equation does not express work for arbitrary displacements, so now including constraint forces along with exerted force we get:

\(\sigma W =\sum_{i} F_{i} . \sigma r_i - \Sigma m_ia_i . \sigma r_i =0\)

Now if the virtual displacements are in the direction orthogonal to constraint forces the second term sum to zero, and we get the final equation of the d’Alembert’s principle for the constant mass system as:

\(\sigma W =\sum_{i} (F_{i} - m_ia_i) . \sigma r_i =0\)

The equation tells us that, for a dynamic system, no work is done by the difference between exerted forces and inertial forces. 

---

Applications of D’Alembert’s Principle

[Click Here for Sample Questions]

Here we use D’Alembert’s Principle to solve some trivial problems in mechanics and understand that the principle makes the results easy to analyze. 

Also check:

Mass falling under Gravitational Field

We now use D’Alembert’s principle to describe the equilibrium of a mass of m falling under the effect of a constant gravitational field g. Consider the z-axis as positive upward, then the force on the mass m is −mg and the work due to this force under vertical displacement δz is δWG = −mgδz. The inertial force on the mass m is given by −m (d² z/dt²) and the work due to this force under vertical displacement δz is, δWI = −m (d² z/dt²)δz. 

Using the d’Alembert’s Principle we set the sum of the two to zero gives us:

We arrive conversely to the expected result from Newton's law F= ma here a = (d² z/dt²) = −g. 

Although in this case, D’Alembert’s principle gives no advantages over normal procedures, it becomes more useful in problems with constraints.

Also check:

Bead on Frictionless Vertical Hoop

In the case of a bead sliding on a vertical loop, the force of gravity is not in the direction of motion. Here the component of gravity orthogonal to the hoop and the force of the hoop on the bead that constrains the bead to move in a circle does no work on the bead.

The work on the bead due to gravity for a small displacement δs = RδΦ (R is the radius of the loop) along the wire is δWG = −mgRsinΦ δΦ. The acceleration of the bead has two components: a radial component and a tangential component.

For radial component of acceleration with velocity v is ar = −v²/R, 

Whereas the tangential velocity is v = ds/dt

Or, ds/dt = R(dΦ/dt) …..(i)

Using a = dv/dt we get,

• Or, at = dv/dt = (d/dt)(ds/dt)= d² s/dt²
• Or, at = R(d²Φ/dt²) [Using (i)]

The work done by the radial component of the inertial force mv²/R is zero. The work done by tangential component −mat = −mR(d ²Φ/dt²) is given by,

δWI = −mR (d²Φ/dt²) δs

Or, δWI = −m R2(d²Φ/dt²) δΦ. [Using δs = RδΦ ]

Now, from D’Alembert’s principle we get,

Or, δWI + δWG = 0

• Or, −mgRsinΦ δΦ.−mR² (d²Φ/dt²) δs = 0
• Or, −mR [ gsinΦ + R (d²Φ/dt²) ]  δs= 0
• Or, gsinΦ + R (d²Φ/dt²) = 0
• Or, d²Φ/dt² + (g/R) sinΦ= 0

Which is similar to the differential equation of linear S.H.M. is d²x/dt² + Φx = 0 with x = Φ and 

Φ =  g/R (angular frequency)

Figure: A bead slides frictionlessly on a vertical hoop of radius R under the influence of gravity

Also check:

Atwood’s Machine

By using D'Alembert's principle Atwood's machine becomes easy to solve. The state of the machine is determined by the positions of the two masses along with the U-shaped coordinate s looping over the frictionless pulley with the string. The work done by gravity g on the left-hand mass m under the displacement δs is given by:

δWL = −mgδs, 

The work done by gravity acting on the right-hand mass M is δWR = Mgδs. 

The total work done by gravity on the system of two masses m and M is δWG = (M − m) gδs

The work done by the inertial forces on the two masses is,

δWI = −(M + m)(d² s/dt²)δs

Now, from D’Alembert’s principle, we get,

δWI + δWG = 0

• Or, δW = [ (M − m)g − (M + m)((d² s/dt²)) ] δs = 0
• Or, d² s/dt² = g (M − m)/(M + m)

Figure: Atwood’s machine

---

Things to Remember

• The principle of virtual work states that for a system in equilibrium, there will be no virtual work done by the forces acting on the system.
• D’Alembert’s principle also known as Lagrange–d'Alembert principle, states that the sum of the differences between the forces exerted on a system of particles(with non-zero rest mass) and the time derivatives of the momenta of the system is zero when the system is projected into virtual displacement.
• Where Fi is the net force exerted on the i-th particle; mi is the mass of the i-th particle; vi is the velocity of the i-th particle; ri is the virtual displacement of the i-th particle
• The applications of d’Alembert’s principle are: mass falling under gravity, parallel axis theorem, and frictionless vertical hoop with a bead.

Also check:

---

---

Previous Year Questions

1. From the top of a tower a body AA is projected vertically up… [JKCET 2007]
2. Assuming earth to be an inertial frame… [JKCET 2009]
3. A body is sliding on a smooth inclined plane requires 4 second to reach the… [JIPMER 1999]
4. A body of mass 5kg is suspended by a spring balance on… [BITSAT 2006]
5. A body of mass MM hits normally a rigid wall with velocity… [BITSAT 2018]
6. A man of weight 80kg is standing in an elevator which is moving with… [DUET 2003]
7. A light inextensible string that goes over a smooth fixed pulley… [BITSAT 2013]
8. A ball is released from the top of a tower… [JKCET 2007]
9. There are two identical springs each of spring constant k… [UPSEE 2017]
10. A 10gm bullet moving directly upward at 1000 m/s strikes and passes through… [UPSEE 2017]
11. A body slides down a frictionless inclined plane starting from… [UPSEE]
12. A force F=75N is applied on a block of mass 5kg along the fixed smooth incline… [UPSEE 2016]
13. A wooden box lying at rest on an inclined surface of a wet wood is held at static… [TS EAMCET 2017]
14. Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle θ… [JEE Main 2019]
15. Given in the figure are two blocks A and B of weight 20N and 100N, respectively… [JEE Main 2015]
16. Consider a cylinder of mass M resting on a rough horizontal rug that is pulled… [JEE Main 2014]
17. A uniform sphere of weight W and radius 5cm is being held by a… [JEE Main 2013]
18. A particle is released on a vertical smooth semicircular track from point… [JEE Main 2014]
19. A particle is released on a vertical smooth semicircular track from point… [JEE Main 2014]
20. A conical pendulum of length 1m makes an angle… [JEE Main 2017]

---

Also check:

Check out: