Stacked blocks on an accelerating elevator

Click For Summary
The problem involves four stacked blocks in a downward-accelerating elevator, requiring the calculation of the contact force between the 30kg and 40kg blocks. Initially, a force of 650.5N was calculated based on the combined weight and acceleration of the blocks above the 40kg block. However, upon further consideration, it was determined that the correct contact force should account for the difference in forces acting on the blocks, resulting in a value of 330.5N. The discussion also highlights that when an elevator accelerates downward, objects feel lighter, and in free fall, they experience weightlessness. Ultimately, the correct magnitude of the contact force between the 30kg and 40kg blocks is confirmed to be 330.5N.
Chris0101
Messages
4
Reaction score
0

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.
 
Physics news on Phys.org
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
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 17 ·
Replies
17
Views
2K
Replies
13
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
42
Views
2K
Replies
4
Views
821
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
639