Does a top pan balance show a greater value when a person jumps on it?

  • Context: Undergrad 
  • Thread starter Thread starter adjacent
  • Start date Start date
  • Tags Tags
    Balance
Click For Summary
SUMMARY

A top pan balance shows a greater value when a person jumps on it due to the principles of momentum and force. When a person jumps, their velocity increases, and upon landing, the change in momentum results in an additional force being exerted on the balance. This force of impact, combined with the gravitational force acting on the person, leads to a higher reading on the scale. The relationship between momentum (defined as mass times velocity) and force (the rate of change of momentum) is crucial in understanding this phenomenon.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Basic knowledge of momentum and its formula (M*V)
  • Familiarity with the concept of force as the rate of change of momentum
  • Awareness of the conservation of momentum principle
NEXT STEPS
  • Study Newton's Second Law of Motion in detail
  • Learn about the conservation of momentum and its applications
  • Explore the relationship between force and momentum in various physical scenarios
  • Investigate the implications of momentum in both classical and relativistic physics
USEFUL FOR

Physics students, educators, and anyone interested in the principles of mechanics and dynamics, particularly those studying the effects of force and momentum in real-world applications.

adjacent
Gold Member
Messages
1,552
Reaction score
62
A top pan balance measures the force.So if I jump(or fall from a height) on it,Why does it show a greater value?(Acceleration and mass does not change,so the force is same)??
 
Physics news on Phys.org
When you jump and approach the balance you gain a velocity.When you fall on the pan all of your velocity becomes zero which means your momentum has changed.Applying conservation of momentum the momentum is transferred to the pan and as we know change in momentum over an interval of time causes a force.So then the pan would experience the gravitational force as well as the force caused due to change in your momentum
 
It shows a greater value because the force of impact is in addition to the weight of the object. The weight is the force it takes to keep you standing there motionless. The force of impact is the force it takes to slow you down from your falling speed to a stop on top of the force needed to keep you there.

The larger your momentum before impact, the larger your force of impact, and the greater the value the scale will show.
 
Oh,Momentum.Maybe I'll learn it later. :)
 
adjacent said:
Oh,Momentum.Maybe I'll learn it later. :)

momentum is nothing but M*V(mass times velocity).The second law of motion itself states that force is the rate of change of momentum.How can you know about force but not about momentum?
 
A lot of the time the net force on an object is taught as equal to its mass times the derivative of its velocity instead of derivative of mass times velocity, especially at freshman level physics.

Momentum is more than just total of M*V. It is, "The thing which is conserved because of Newton's third law". Light has momentum, but we wouldn't describe its momentum as M times V.
 
jfizzix said:
A lot of the time the net force on an object is taught as equal to its mass times the derivative of its velocity instead of derivative of mass times velocity, especially at freshman level physics.

Momentum is more than just total of M*V. It is, "The thing which is conserved because of Newton's third law". Light has momentum, but we wouldn't describe its momentum as M times V.

Yeah momentum is M*V/√(1-V^2/c^2) .So you expect a freshman level physics student to understand that and about how photon has momentum but can't expect him to know that force is the rate of change of momentum.ok :rolleyes:
 
quawa99 said:
Yeah momentum is M*V/√(1-V^2/c^2) .So you expect a freshman level physics student to understand that and about how photon has momentum but can't expect him to know that force is the rate of change of momentum.ok :rolleyes:

Momentum is not just \frac{M\vec{v}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} either. Certainly not for a (massless) photon in any case, but that's beside the point.

The point is that momentum is a thing that is not defined by its experessions, but by its total value being conserved via Newton's third law. That where the expressions come from.

Also, for goodness sake, it's barely two weeks into the semester. When I was a freshman in physics I couldn't say I understood much of anything only two weeks into the class. Everyone learns it for the first time somewhere.
 
Last edited:
In the freshman physics curriculum, usually momentum conservation is taught (just) after Newton's second law, which is why \vec{F}=m\vec{a} is more ubiquitous than \vec{F}=\frac{d\vec{p}}{dt} (among other reasons).
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
461
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 29 ·
Replies
29
Views
3K