B Exploring the Physics of Stacked Blocks: No Normal Force on Block A

AI Thread Summary
In the discussion about stacked blocks, it is clarified that block A does experience a normal force, but this force is not directly from the ground through block C. Instead, the normal force is transmitted through block B, which supports block A. The normal force acts to counteract the gravitational force on each block, maintaining equilibrium. When block C is removed, gravity affects the remaining blocks, causing them to tumble. The conversation emphasizes the importance of free body diagrams to analyze the forces acting on each block accurately.
Shardul Khare
Messages
23
Reaction score
0
If 3 blocks have some mass(a,b,c) are kept on one another (Vertically i mean)...Why block a doesn't experience the normal force exerted by the ground on block c? won't that force get transferred through b to a?
 
Physics news on Phys.org
Shardul Khare said:
If 3 blocks have some mass(a,b,c) are kept on one another (Vertically i mean)...Why block a doesn't experience the normal force exerted by the ground on block c? won't that force get transferred through b to a?

If there were no gravity and the normal force was an active force (someone pushing block c), then some of that force would be transferred through block c to the next block. But, the normal force is simply opposing gravity, so there is no net force on block c.

Of course, some of the force is transferred as the normal force between blocks c and b.

And, there is nothing special about the blocks, the same sort of internal forces are acting within each block. The bottom of each block is supporting all the block and the ones above it, whereas the top of each block is supporting only the blocks above it.
 
Last edited:
  • Like
Likes Shardul Khare
PeroK said:
If there were no gravity and the normal force was an active force (someone pushing block c), then some of that force would be transferred through block c to the next block. But, the normal force is simply opposing gravity, so there is no net force on block c.

Of course, some of the force is transferred as the normal force between blocks c and b.

And, there is nothing special about the blocks, the same sort of internal forces are acting within each block. The bottom of each block is supporting all the block and the ones above it, whereas the top of each block is supporting only the blocks above it.
At a position within a block, the weight supported will be between the two. wHAT DOES THATMEAN?
 
Shardul Khare said:
At a position within a block, the weight supported will be between the two. wHAT DOES THATMEAN?

I meant in the middle of a block, the weight supported will be the top half of that block plus the blocks above it. There is nothing special about the contact points between the blocks. The same balance of forces is at play throughout each block, eventualy down to inter-molecular forces.
 
Shardul Khare said:
hy block a doesn't experience the normal force exerted by the ground on block c?

It does. When you remove block C, don't the rest of the blocks start tumbling down?
 
  • Like
Likes PeroK
Vanadium 50 said:
It does. When you remove block C, don't the rest of the blocks start tumbling down?
Isnt that because as C is removed so gravity acts on them? Morever if the force acts on A,Why don't we consider it while drawing the FBD of block A?
 
Shardul Khare said:
Isnt that because as C is removed so gravity acts on them? Morever if the force acts on A,Why don't we consider it while drawing the FBD of block A?
Have you drawn free body diagrams of each of the three blocks, showing the forces acting on each, or do you feel that you have advanced to the point beyond which you no longer need to use free body diagrams? If you have drawn the free body diagrams, please write down for us the force balance equation that applies to each of the blocks.
 
Assuming the mass of A is M1
Mass of B is m2 and mass of C is M3...As these blocks are kept vertically and are rest

The force acting on the uppermost A would be Gravitaional M1g downwards and the Normal force (N1)exerted by B on A (Equal to m1g) upwardsForces acting on B would be Normal (N1) Exerted by A on B (Third Law) Equal to m1g and gravitational force downwards equal to m2g and the normal (N2) exerted by block C on B ( N2= mig + m2g) upwards

Forces on C would be N2 downwards (Thid law) and m3g as well as N3 upwards(Between ground and C)
 
Chestermiller said:
Have you drawn free body diagrams of each of the three blocks, showing the forces acting on each, or do you feel that you have advanced to the point beyond which you no longer need to use free body diagrams? If you have drawn the free body diagrams, please write down for us the force balance equation that applies to each of the blocks.
 
  • #10
Shardul Khare said:
Assuming the mass of A is M1
Mass of B is m2 and mass of C is M3...As these blocks are kept vertically and are rest

The force acting on the uppermost A would be Gravitaional M1g downwards and the Normal force (N1)exerted by B on A (Equal to m1g) upwardsForces acting on B would be Normal (N1) Exerted by A on B (Third Law) Equal to m1g and gravitational force downwards equal to m2g and the normal (N2) exerted by block C on B ( N2= mig + m2g) upwards

Forces on C would be N2 downwards (Thid law) and m3g as well as N3 upwards(Between ground and C)
Excellent. Now you have 3 linear algebraic equations in the three unknowns N1, N2, and N3. Please solve for these three unknowns.
 
Back
Top