Acceleration of a system due to gravity

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SUMMARY

The acceleration of a system consisting of three equal masses (A, B, and C) each weighing 2.00 kg, under the influence of gravity (g = 9.81 m/s²), is calculated to be 6.54 m/s². This value is derived from Newton's second law of motion (F = ma), where the total force acting on the system is the weight of masses B and C. The discussion also clarifies that without external forces, the system cannot exceed the gravitational acceleration of 9.81 m/s². Additionally, it explores the potential for achieving greater acceleration using mechanical setups like pulleys, emphasizing the importance of mass distribution and mechanical advantage.

PREREQUISITES
  • Understanding of Newton's second law of motion (F = ma)
  • Basic knowledge of gravitational acceleration (g = 9.81 m/s²)
  • Familiarity with mechanical advantage concepts in physics
  • Ability to analyze systems with multiple masses and forces
NEXT STEPS
  • Study the principles of mechanical advantage in pulley systems
  • Learn about the dynamics of multi-body systems in physics
  • Explore advanced applications of Newton's laws in real-world scenarios
  • Investigate the effects of friction and other forces on system acceleration
USEFUL FOR

Students in physics, educators teaching mechanics, and engineers interested in dynamics and mechanical systems will benefit from this discussion.

Sheepwall
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The problem was present in a physics 1 exam, and I'm reasonably sure I know the answer, yet my friend contradicts me in my conclusion.

Homework Statement


Find magnitude of acceleration of system ABC. Masses of A, B and C are all equal and each has mass 2.00 kg. Let gravitational acceleration g = 9.81 m/s2. See figure below.
Problem.png

Disregard friction - castor, air, and surface.

Homework Equations


Newtons 2nd law of motion:
F = m a (1)

The Attempt at a Solution


The force on the whole system is gravity's effect on B and C (weight of B and platform C). Through (1):

F = (2)(2.00 kg)(9.81 m/s2) = 39.24 N

Acceleration on the whole system is then, under the above force: (Through (1) again):

a = (39.24 N)/[(3)(2.00 kg)] = 6.54 m/s2.

Obviously, the acceleration of A is directed horizontally, while B and C are accelerating downwards. Still, the acceleration of each objects will be the same, yet in their respective directions. It is this value which I am after.

Also, can someone less ignorant than I tell me if I am right in this:
Without other external forces (other than gravity), the system will never accelerate faster than 9.81 m/s2 and the less the mass of A (and the larger the mass of B and C, equivalently), the closer the systems acceleration will be to 9.81 m/s2.

Thank you in advance and I apologize for the triviality of the problem :s
 
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Your analysis is complete and correct for the information you're wanting about the system. Any other questions?
 
Your analysis is correct.
 
Thanks for the (very) quick replies!

Bystander said:
Your analysis is complete and correct for the information you're wanting about the system. Any other questions?

If you don't mind, continuing on the theme of triviality; Could I make something accelerate faster than gravity using wheels and cables, like so (A):
Idea.png

Again, disregarding frictions. How should I think when finding acceleration and tensions in these kinds of "problems?"
 
If "A" is the larger mass, the unlabeled mass can be accelerated at greater than g; you've got the double (one and a halfle?) advantage with the pulley set-up. Think of an asymmetric see-saw, or catapult, with a large mass on the short end and a small mass on the long. Trebuchet. Might be a variety of other gravity driven "siege engines."
 
Okay, so comparing my previous image to the one below:
Idea_2.png

Here, gravity's pull on A (again, like you did, assuming that A's mass is greater than that of the unlabeled), displacing A the distance s, will do the same to the unlabeled weight. In the previous image, displacing A by s, would displace the weight by 2 times s (?). I assume it is harder to displace A in the previous image, then, seeing as the work done by gravity is double there?
 
Sheepwall said:
work done by gravity
Is the same. "mgh" is "mgh" regardless of how the pulley, ropes, are rigged. The length of rope you've got to move past a particular point to lift/lower "A" a distance "h" can be changed, and the force you have to apply to pull the rope can be many times greater or less depending on how much mechanical advantage you "rig" into the block(s) and tackle.
 

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