How Does a Falling Chain Affect Scale Readings?

In summary, the question is about the reading on a scale when a chain of mass M and length L is released and falls onto the scale. The normal force of the scale and the gravity force of the chain are the only forces acting. The chain's linear mass density is denoted by r = M/L. Using Newton's second law and conservation of energy, an expression for the normal force is derived, but it does not match the expected answer. The question asks for help in figuring out if there is an error in the application of Newton's second law, energy conservation, or the algebraic calculations.
  • #1
Sam_Goldberg
46
1

Homework Statement



Hi guys, I have a question on problem 4.11 in Kleppner and Kolenkow's mechanics book. A chain of mass M and length L is suspended vertically with its lowest end touching a scale. The chain is released and falls onto the scale. What is the reading of the scale when a length of chain, x, has fallen?

Homework Equations



I will denote the chain's linear mass density by r = M/L.

The Attempt at a Solution



Okay, there are only two forces acting on the chain: the normal force of the scale and the gravity force due to its mass. The normal force N is what we're interested in; we want it as a function of position. I will write Newton's second law for the chain:

r (d/dt)[(L - x)(-x')] = N - rLg

The left hand side is the change in vertical momentum of the chain; the right side is the normal force and weight. Note that x' = dx/dt. This rewrites as:

N/r = Lg - Lx'' + (x')^2 + xx'' (equation 1)

Now, let's use conservation of energy on the chain. The normal force at the bottom does no work, but the conservative gravitational force does. We get (equating the energy at chain fallen x to the energy at the beginning of the fall):

(1/2)r(L - x)(x')^2 + r(L - x)g(L - x)/2 = rLg*L/2

Simplifying,

(x')^2 + g(L - x) = g(L^2)/(L - x) (equation 2)

I will now take the time derivative of each side to get:

x'' = (g/2) + (g/2)(L^2)/[(L - x)^2] (equation 3)

Now I substitute equations 2 and 3 into equation 1 to come up with the expression for the normal force on the chain. After some algebra, I get:

N = rg[(-L/2) + (3/2)x + (L^2)/(2L - 2x)]

which is not the answer. Am I messing up Newton's second law, the energy conservation, or my algebra after I plug things in? Thanks in advance for the help.
 
Physics news on Phys.org
  • #2
x''=g should be true, shouldn't it?
 
Last edited:
  • #3


I would suggest double-checking your calculations and equations to ensure that they are correct. It is possible that you may have made a mistake in one of your equations or algebraic manipulations. I would also recommend reviewing the problem statement and making sure that you are interpreting it correctly. If you are still having trouble, I would suggest seeking assistance from a professor or fellow student for clarification or guidance. It is important to approach scientific problems with a critical and careful mindset in order to arrive at the correct solution.
 

FAQ: How Does a Falling Chain Affect Scale Readings?

What is a chain falling on a scale?

A chain falling on a scale is a physics experiment where a chain is dropped onto a scale from a certain height, and the weight of the chain is measured at different points during its descent.

What is the purpose of this experiment?

The purpose of this experiment is to demonstrate the principles of conservation of energy and the relationship between potential energy and kinetic energy.

How does the weight of the chain change as it falls?

As the chain falls, its weight will decrease due to the conversion of potential energy into kinetic energy. This is because the chain's potential energy decreases as it moves closer to the ground and its kinetic energy increases.

What factors can affect the results of this experiment?

The results of this experiment can be affected by the height from which the chain is dropped, the length and weight of the chain, and any external factors such as air resistance.

What can we learn from the results of this experiment?

The results of this experiment can help us understand the relationship between potential energy and kinetic energy, as well as the effects of gravity on falling objects. It can also serve as a practical demonstration of the principles of physics and how they can be applied in real-world scenarios.

Back
Top