Stacked blocks on an accelerating elevator

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SUMMARY

The contact force between the 30kg and 40kg blocks in a downward-accelerating elevator (3.2 m/s²) is calculated to be 330.5N. Initially, the total force acting on the 40kg block was miscalculated as 650.5N, which included the effects of both the 50kg and 30kg blocks. Upon reevaluation, the correct approach involves recognizing that the apparent weight decreases when the elevator accelerates downward, leading to a revised contact force calculation. The final conclusion confirms that the contact force is indeed 330.5N.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of forces in a non-inertial reference frame
  • Basic principles of weight and apparent weight in accelerating systems
  • Ability to perform calculations involving mass, acceleration, and force
NEXT STEPS
  • Study the effects of acceleration on apparent weight in different scenarios
  • Learn about free-fall conditions and their implications on contact forces
  • Explore advanced applications of Newton's laws in non-inertial frames
  • Investigate the relationship between mass, weight, and gravitational acceleration
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding forces in accelerating systems.

Chris0101
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Homework Statement


Four blocks of masses 20kg, 30kg, 40kg, and 50kg are stacked on top of one another in an elevator in order of decreasing mass with the lightest mass on the top of the stack. The elevator moves downward with an acceleration of ##3.2m/s^2## . Find the contact force between the 30kg and 40kg masses.

Homework Equations



Newton's second law:
##F = ma##

In the case of the question at hand:
##F = mg + ma##

The Attempt at a Solution



Due to the fact that the elevator is moving downward with a positive acceleration, the total force that the top two blocks apply onto the 40kg block will be:

##F = mg + ma##
##F = (50kg)(9.81m/s^2) + (50kg)(3.2m/s^2)##
##F = 490.5N + 160.0N##
##F = 650.5N##


This shows that the force acting on the 40 kg block from the top two blocks (20kg & 30kg) is 650.5N. This downward magnitude of force is the amount of force that the 40kg block has to support which is also the normal force that the 40kg block. With that, the contact force between the 30kg and 40kg block is 650.5N

I'm not sure if this is correct. If someone can correct me on this question that would be great. Thanks for the help.
 
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Suppose that the downward acceleration was 9.81 m/s (that is, g). The elevator's acceleration would then match that due to gravity alone. That would mean the contents would essentially be in free-fall. Would the force then be 2g*50 kg ? In other words, do things feel heavier when the elevator accelerates downwards? Or do they feel lighter?
 
When the elevator is in free fall, one will experience weightlessness. With that said, if the the elevator is accelerating downward at a fraction of g, one's apparent weight will be less than his/her true weight. With your tip, I see where I went wrong and contact force between the 30kg and 40kg block should be the difference of the the two forces which is a magnitude of 330.5N.

Unfortunately, I do not have a reference to look up this answer, but is 330.5N the correct magnitude of the contact force between the 30kg and 40kg block, or am I forgetting something?
 
Chris0101 said:
is 330.5N the correct magnitude of the contact force between the 30kg and 40kg
Yes.
 
Thanks for the help
 

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