Adding mass when a blocks rests on another block?

In summary, the conversation discusses the relationship between the amplitude of motion and the maximum amplitude for which the upper block does not slip on the lower block in a system of a 1.0 kg mass riding on top of a 5.0 kg mass attached to a spring with a spring constant of 50 N/m and a coefficient of static friction of 0.50. The maximum demand placed on the frictional force is determined by the equation for static friction, and the upper limit is reached when the force of the spring is greater than the frictional force. The equations of ΣF=ma for each block are used to explain this relationship.
  • #1
dantechiesa

Homework Statement


The figure shows a 1.0 kg mass riding on top of a 5.0 kg mass which is attached
to a spring. The 5.0 kg mass oscillates on a frictionless surface, and the spring constant
is k = 50 N/m. There is friction between the upper block and the lower block, and the
coefficient of static friction is 0.50.

a) As the amplitude of the motion is increased, there is a maximum amplitude for
which the upper block does not slip on the lower block. Explain why, and calculate
this amplitude.

Homework Equations


kx = Ffr

The Attempt at a Solution


Can someone explain to me - possibly through a FBD, why, when figuring out question a), the Ffr between the 2 blocks is 6.0kg * 9.8 * .50?

Why would it not be the normal force of the top box? Simply 1 * 9.8 * .5?

How does the larger bottom block exert a normal force up and why does this not happen with, let's say an object lying on the Earth (we don't include the mass of the earth)

Thanks.
 
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  • #2
The coefficient of static friction allows you to compute the maximum force that is available from static friction. If something is sitting still, the frictional force is zero, but if you push on the weight, or try to pull the rug out from under it, it will supply the necessary frictional force to oppose movement. There is an upper limit to this frictional force, given by the equation for static friction with the coefficient of static friction.
 
  • #3
Charles Link said:
The coefficient of static friction allows you to compute the maximum force that is available from static friction. If something is sitting still, the frictional force is zero, but if you push on the weight, or try to pull the rug out from under it, it will supply the necessary frictional force to oppose movement. There is an upper limit to this frictional force, given by the equation for static friction with the coefficient of static friction.
Thanks for the reply,
how come I use the mass of the lower block when computing this friction force?

Wouldn't I simply use the normal force from the top block only?
 
  • #4
dantechiesa said:
Thanks for the reply,
how come I use the mass of the lower block when computing this friction force?

Wouldn't I simply use the normal force from the top block only?
Yes, from the top block only. ## F_N=W=mg ##.
 
  • #5
Charles Link said:
Yes, from the top block only. ## F_N=W=mg ##.
So computing the question a),
finding the distance at which the box begins to slip.
Would be at the point where force of the spring, kx = Ffr of the box

50 * x = .5 1.0 * 9.8

x = .098

However, the answer key states: as the amplitude increases, more force is required to accelerate the upper block in its SHM – eventually, the force required is larger than what (static) friction can provide, and the upper block will slip on the lower block. It does so when the amplitude of the SHM is A = 0.59

.59 is only found when the Ffr is found using: .5 * (1.0 + 5.0) * 9.8
adding the two boxes mass. Why?
 
  • #6
dantechiesa said:
Would be at the point where force of the spring, kx = Ffr of the box
Why?
 
  • #7
haruspex said:
Why?
Because once the force of static friction of the top box is overcome by the force of the spring, the box will begin to slide.

no?
 
  • #8
dantechiesa said:
Because once the force of static friction of the top box is overcome by the force of the spring, the box will begin to slide.

no?
Different boxes. Draw a separate FBD for box.
 
  • #9
haruspex said:
Different boxes. Draw a separate FBD for box.
After thinking about it some more, I am starting to really get lost.

Does the attached diagram look right?

Turns out i missed fg and fn of lower block. So pretend I have those in
 

Attachments

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  • #10
dantechiesa said:
After thinking about it some more, I am starting to really get lost.

Does the attached diagram look right?

Turns out i missed fg and fn of lower block. So pretend I have those in
The diagram is right. Don't worry about vertical forces on the lower block - they will just balance out.
What is the maximum demand placed on the frictional force?
 
  • #11
haruspex said:
The diagram is right. Don't worry about vertical forces on the lower block - they will just balance out.
What is the maximum demand placed on the frictional force?

I don't understand why it would not be equal to kx.

If the bottom box is 'shoved' forward by the spring by a force of kx, then as x gets larger and largr, the total force will.

The frictional force is only going to be able to withstand so much, and at a certain x, will be overcome.

What is flawed in my reasoning?

Thanks.
 
  • #12
dantechiesa said:
I don't understand why it would not be equal to kx.
If the spring force and the frictional force on the lower box were in balance then the lower box would not accelerate. It does accelerate. The frictional force is therefore allowed to be less than the spring force.
The equation of relevance is ΣF=ma. Write out that equation for each box, on the assumption that there is no slipping.
 
  • #13
haruspex said:
If the spring force and the frictional force on the lower box were in balance then the lower box would not accelerate. It does accelerate. The frictional force is therefore allowed to be less than the spring force.
The equation of relevance is ΣF=ma. Write out that equation for each box, on the assumption that there is no slipping.

Top box = 1.0 * a = Ffr
Bottom box = 5.0 * a = kx
Sorry but I am not seeing the relationship between the two (+ i don't think my fnet are right..)
 
  • #14
dantechiesa said:
Bottom box = 5.0 * a = kx
You are missing a horizontal force. Remember - bodies exert equal and opposite forces on each other.
 
  • #15
haruspex said:
You are missing a horizontal force. Remember - bodies exert equal and opposite forces on each other.
Ok, so since there is no friction on the bottom, the other surface the lower block is in contact with is the upper block.
So the Ffr between the 2 blocks also applies to the bottom box?
5.0 * a = kx +(?) ffr?
 
  • #16
dantechiesa said:
+(?) ffr?
Equal and opposite
 
  • #17
haruspex said:
Equal and opposite
Presuming you mean kx - Ffr = ma.

DOes that mean we now need to find a in order to find x?

Secondly, can you explain why its like that?

Thank you!
 
  • #18
dantechiesa said:
DOes that mean we now need to find a in order to find x?
Yes, and your equation for the top box tells you the value of a.
The coefficient of friction sets a limit on the acceleration that can be achieved by the top box. If the lower box exceeds that then slipping will occur.
 
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  • #19
haruspex said:
Yes, and your equation for the top box tells you the value of a.
The coefficient of friction sets a limit on the acceleration that can be achieved by the top box. If the lower box exceeds that then slipping will occur.
It all worked,
Thanks for sticking it out!
 

Related to Adding mass when a blocks rests on another block?

1. How does adding mass affect the stability of a block resting on another block?

Adding mass to a block that is resting on another block can increase its stability by increasing its weight and therefore its resistance to being toppled over.

2. Can adding mass to a block resting on another block change its center of gravity?

Yes, adding mass to a block can change its center of gravity, which can affect its stability. The center of gravity is the point at which the weight of an object is evenly distributed, and adding mass can shift this point.

3. How does the position of the added mass affect the stability of the block?

The position of the added mass can greatly affect the stability of the block. Placing the mass closer to the bottom of the block can increase stability, while placing it at the top can decrease stability.

4. Is there a limit to how much mass can be added to a block resting on another block?

There is no specific limit to how much mass can be added to a block, but adding too much mass can cause the block to become too heavy and unstable. It is important to consider the weight and size of the base block when adding mass.

5. How can adding mass to a block resting on another block affect the friction between the two blocks?

Adding mass to a block can increase the force between the two blocks, which can increase the friction between them. This can help to keep the blocks in place and prevent them from slipping or sliding.

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