D'Alembert's Principle:- A Falling Box

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SUMMARY

D'Alembert's Principle simplifies the application of Newton's second law by rearranging it into a dynamic equilibrium equation, specifically F_net - ma = 0. In the case of a 25kg box dropped from a height of 3.5m, the gravitational force (mg) acts downward while the pseudo force (ma) acts upward. The calculated velocity just before impact, using both conservation of energy and kinematic equations, is 8.287 m/s. While D'Alembert's Principle can be applied, it is more beneficial in scenarios involving torques rather than simple falling objects.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with kinematic equations
  • Knowledge of gravitational potential energy (PE) and kinetic energy (KE)
  • Basic concepts of dynamic equilibrium
NEXT STEPS
  • Study the application of D'Alembert's Principle in torque problems
  • Explore advanced kinematic equations for varying acceleration scenarios
  • Research energy conservation principles in mechanical systems
  • Learn about pseudo forces and their applications in non-inertial reference frames
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Students and educators in physics, particularly those focusing on mechanics, as well as engineers dealing with dynamic systems and forces.

Meekin
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I've been asked to research D'Alembert's principle and solve a question. I've looked up quite a lot of different explanations on the internet of D'Alembert's Principle and I'm not quite grasping how to use it. I understand that you rearrange formula so that they equal 0 (e.g. f - ma = 0 or PE - KE = 0 ) so then you can treat them as static problems. I'm just really struggling on how to apply this to a question.

Homework Statement

A 25kg box is dropped from a height of 3.5m. Using D'Alembert's Principle work out the velocity reached immediately before hitting the ground.

(Gravity to be taken as 9.81ms-2)

Homework Equations



KE = (0.5 x (mass x velocity^2))

PE = mass x acceleration due to gravity x height

Final Velocity^2 = Initial Velocity^2 + (2 x Acceleration x Displacement)

The Attempt at a Solution



I already worked this out using the conservation of energy method and got an answer of Velocity = 8.287ms-1

I then tried playing around with it and realized i could treat it as though it was a box on a horizontal place with no friction

S = 3.5
A = 9.81ms-2
U = 0
V = ?

using v^2 = u^2 + 2as i got the same answer of 8.287ms-1.

This was a guess at a method because its the only other method I've managed to come up with but I'm fairly sure it doesn't use D'Alembert's Principle.

Any help would be appreciated. Thanks guys
 
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Meekin said:
I've been asked to research D'Alembert's principle and solve a question. I've looked up quite a lot of different explanations on the internet of D'Alembert's Principle and I'm not quite grasping how to use it. I understand that you rearrange formula so that they equal 0 (e.g. f - ma = 0 or PE - KE = 0 ) so then you can treat them as static problems. I'm just really struggling on how to apply this to a question.


Homework Statement




A 25kg box is dropped from a height of 3.5m. Using D'Alembert's Principle work out the velocity reached immediately before hitting the ground.

(Gravity to be taken as 9.81ms-2)

Homework Equations



KE = (0.5 x (mass x velocity^2))

PE = mass x acceleration due to gravity x height

Final Velocity^2 = Initial Velocity^2 + (2 x Acceleration x Displacement)


The Attempt at a Solution



I already worked this out using the conservation of energy method and got an answer of Velocity = 8.287ms-1

I then tried playing around with it and realized i could treat it as though it was a box on a horizontal place with no friction

S = 3.5
A = 9.81ms-2
U = 0
V = ?

using v^2 = u^2 + 2as i got the same answer of 8.287ms-1.

This was a guess at a method because its the only other method I've managed to come up with but I'm fairly sure it doesn't use D'Alembert's Principle.

Any help would be appreciated. Thanks guys
You are correct about D'Alemberts principle being, in its simplest form, a rearrangement of Newtons 2nd Law into an equation of dynamic equilibrium, that is , a rearrangement of F_net = ma into simply F_net - ma = 0, where the term " -ma " is a so-called pseudo force or inertial force which acts through the objects center of mass equal in magnitude to F_net and applied in the opposite direction. In your falling box example, the net force is mg acting down and the pseudo inertial force is ma acting up. So you write mg - ma = 0, from which a = g, and you solve for v using your kinematic equation to get of course the same result. The use of D'Alembert here is not very useful and can work against you if you are not careful.
The use of D'Alembert is more appropriate when dealing with torques , because you can then sum moments about any point = 0 for accelerating objects, instead of having to sum torques about the center of mass only if D'Alembert inertial forces are not used.
 

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