Finding the time it takes for two stacked blocks to travel

In summary: But what i wish i could show you is the diagram. I should probably mention that the top block has a rope attached to it and this is where that question i asked you about calculating static friction on its own came from. Knowing this, would it affect the answer in any way?
  • #1
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Homework Statement


The coefficient of static friction is 0.604 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.104. Force F causes both blocks to cross a distance of 3.44m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.13kg and the mass of the upper block is 2.25kg?

Homework Equations

The Attempt at a Solution


I looked at the attached file that my teacher gave us
I found the normal of the top block which was 22.0 N
Since it equals one of the normals of the bottom block, I substituted in the same value in for calculating the other normal. This "newfound" normal was used to calculate the acceleration in the x direction using Fnet=ma ---> ma = friction static - friction kinetic

I ended up getting 6.86 x 10^-2 s which is wrong, please help!
 

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  • #2
alexi_b said:

Homework Statement


The coefficient of static friction is 0.604 between the two blocks shown. The coefficient of kinetic friction between the lower block and the floor is 0.104. Force F causes both blocks to cross a distance of 3.44m, starting from rest. What is the least amount of time in which the motion can be completed without the top block sliding on the lower block, if the mass of the lower block is 1.13kg and the mass of the upper block is 2.25kg?

Homework Equations

The Attempt at a Solution


I looked at the attached file that my teacher gave us
I found the normal of the top block which was 22.0 N
Since it equals one of the normals of the bottom block, I substituted in the same value in for calculating the other normal. This "newfound" normal was used to calculate the acceleration in the x direction using Fnet=ma ---> ma = friction static - friction kinetic

I ended up getting 6.86 x 10^-2 s which is wrong, please help!
Let's call the block under "A" et the block on top "B". The applied force is only on A? And it is horizontal?

What you have to do is to determine the maximum horizontal force on block B (top) such that it does not slide. This is given by the max static friction force. Calculate this first, and then set that equal to ##m_B a##. That will give you the acceleration, which will be the acceleration of both blocks since they move as one.
 
  • #3
nrqed said:
Let's call the block under "A" et the block on top "B". The applied force is only on A? And it is horizontal?

What you have to do is to determine the maximum horizontal force on block B (top) such that it does not slide. This is given by the max static friction force. Calculate this first, and then set that equal to ##m_B a##. That will give you the acceleration, which will be the acceleration of both blocks since they move as one.
does this maximum static friction force equal the force being applied to pull the boxes? In the FBD it indicates so but do i just calculate the mac static friction force alone?
 
  • #4
alexi_b said:
does this maximum static friction force equal the force being applied to pull the boxes? In the FBD it indicates so but do i just calculate the mac static friction force alone?
No, it is not equal to the force applied. You need to use the equation for the maximum static friction force on an object (the one that contains ##\mu_s##).
 
  • #5
nrqed said:
No, it is not equal to the force applied. You need to use the equation for the maximum static friction force on an object (the one that contains ##\mu_s##).
So i get:

Fs = (coefficient of static friction) x (normal force of top block)
Fs = 13.3

13.3 = mb(a)
13.3= 2.25kg (a)
a = 5.91 m/s^2

d = Vit + 1/2at^2 (Vit vanishes because there's no initial velocity)
d = 1/2at^2
3.44 = 1/2 (5.91)t^2
3.44/2.955 = t^2 //sqrt both sides
1.08 s = t

this is apparently wrong :(
 
  • #6
alexi_b said:
So i get:

Fs = (coefficient of static friction) x (normal force of top block)
Fs = 13.3

13.3 = mb(a)
13.3= 2.25kg (a)
a = 5.91 m/s^2

d = Vit + 1/2at^2 (Vit vanishes because there's no initial velocity)
d = 1/2at^2
3.44 = 1/2 (5.91)t^2
3.44/2.955 = t^2 //sqrt both sides
1.08 s = t

this is apparently wrong :(
I don't see any mistake. The force is only acting on block A, right?
 
  • #7
nrqed said:
I don't see any mistake. The force is only acting on block A, right?
yeah it is.

But what i wish i could show you is the diagram. I should probably mention that the top block has a rope attached to it and this is where that question i asked you about calculating static friction on its own came from. Knowing this, would it affect the outcome?
 
  • #8
alexi_b said:
yeah it is.

But what i wish i could show you is the diagram. I should probably mention that the top block has a rope attached to it and this is where that question i asked you about calculating static friction on its own came from. Knowing this, would it affect the outcome?
AHH! It changes everything. I was assuming there were no other force than friction on the top block
 
  • #9
nrqed said:
AHH! It changes everything. I was assuming there were no other force than friction on the top block
yes i am so sorry ;(! that's why I was getting confused because I thought that the tension in the rope = static friction of the top block. I am back to square one now, please help!
 
  • #10
alexi_b said:
So i get:

Fs = (coefficient of static friction) x (normal force of top block)
Fs = 13.3

13.3 = mb(a)
13.3= 2.25kg (a)
a = 5.91 m/s^2

d = Vit + 1/2at^2 (Vit vanishes because there's no initial velocity)
d = 1/2at^2
3.44 = 1/2 (5.91)t^2
3.44/2.955 = t^2 //sqrt both sides
1.08 s = t

this is apparently wrong :(
Your method would be right if F were applied to the lower block, but it is applied to the top block. Clearly, the friction between the lower block and ground will be important, but you have not considered it.
 
  • #11
alexi_b said:
yes i am so sorry ;(! that's why I was getting confused because I thought that the tension in the rope = static friction of the top block. I am back to square one now, please help!
No, *I* am sorry, I opened the file you had attached, but I did not notice there were two pages! I just saw the first one, so I did not realize there was a rope there. I am sorry
 
  • #12
haruspex said:
Your method would be right if F were applied to the lower block, but it is applied to the top block. Clearly, the friction between the lower block and ground will be important, but you have not considered it.
It is my fault, I had not seen the FBD and I gave instructions thinking that there were no horizontal forces on the top block. My mistake.
 
  • #13
nrqed said:
It is my fault, I had not seen the FBD and I gave instructions thinking that there were no horizontal forces on the top block. My mistake.

So if i calculate the force of the normal from the top block, according to the diagram it is the same magnitude for the bottom block? Or do i reverse its signs?
 
  • #14
alexi_b said:
So if i calculate the force of the normal from the top block, according to the diagram it is the same magnitude for the bottom block? Or do i reverse its signs?
Watch out, there are two normal forces on the bottom block, one due to the contact with the top block and one with the floor. Calculate these two. Then calculate the net horizontal force on the bottom block, using for the static friction force the maximum value, which you had calculated before, ## f_s = \mu_s m_{top} g##. Then the net horizontal force on the bottom block is ##f_s - f_k##. Set this to ##m_{bot} a## and then do as before.
 
  • #15
nrqed said:
Watch out, there are two normal forces on the bottom block, one due to the contact with the top block and one with the floor. Calculate these two. Then calculate the net horizontal force on the bottom block, using for the static friction force the maximum value, which you had calculated before, ## f_s = \mu_s m_{top} g##. Then the net horizontal force on the bottom block is ##f_s - f_k##. Set this to ##m_{bot} a## and then do as before.
so if i calculate the normal force of the top block to be 22.0 N, since there are two normal forces acting on the bottom block, the normal force of the top block will act as one of the normal forces of the bottom block. Should it be 22.0 N or -22.0 N?
 
  • #16
alexi_b said:
so if i calculate the normal force of the top block to be 22.0 N, since there are two normal forces acting on the bottom block, the normal force of the top block will act as one of the normal forces of the bottom block. Should it be 22.0 N or -22.0 N?
The magnitude is 22 N while the y component is -22 N. But for the calculation of the static friction force, ##f_s##, you need to use the magnitude. You will of course get the same result as before.
 
  • #17
nrqed said:
The magnitude is 22 N while the y component is -22 N. But for the calculation of the static friction force, ##f_s##, you need to use the magnitude. You will of course get the same result as before.
im still lost, I don't know what to do :(, any more hints?
 
  • #18
alexi_b said:
im still lost, I don't know what to do :(, any more hints?
You need to analyse the forces and acceleration of the lower block.
 
  • #19
alexi_b said:
im still lost, I don't know what to do :(, any more hints?
First step: calculate the maximum static friction force ##f_s## (it is the same static friction force on th stop block an don the bottom block by the action-reaction principle ). Second step: calculate the force of kinetic friction on the bottom block (due to the friction with the ground). Do you see how to do those two steps?
 

1. How do you calculate the time it takes for two stacked blocks to travel?

The time it takes for two stacked blocks to travel is calculated using the equation t = √(2h/g), where t is the time in seconds, h is the height in meters, and g is the acceleration due to gravity (9.8 m/s²).

2. What factors affect the time it takes for two stacked blocks to travel?

The factors that affect the time it takes for two stacked blocks to travel include the height of the stack, the acceleration due to gravity, and any external forces acting on the blocks, such as friction or air resistance.

3. Can you use this equation for any two stacked blocks, regardless of their weight?

Yes, this equation can be used for any two stacked blocks, regardless of their weight. The time it takes for two stacked blocks to travel is determined by the height of the stack and the acceleration due to gravity, which are constants that do not depend on the weight of the blocks.

4. Is the time it takes for two stacked blocks to travel affected by the surface they are traveling on?

Yes, the surface the blocks are traveling on can affect the time it takes for them to travel. If the surface is rough, it will create more friction, which can slow down the blocks and result in a longer travel time. If the surface is smooth, there will be less friction and the blocks will travel faster.

5. How accurate is this equation in determining the time it takes for two stacked blocks to travel?

This equation is an ideal model and does not take into account factors such as air resistance or the shape of the blocks, so it may not always accurately predict the time it takes for two stacked blocks to travel. However, it can provide a good estimate in simpler scenarios.

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