Calculate Pressure of Sand on Paper: D=19.6 Pa

In summary, the problem is to calculate the pressure exerted by sand grains falling on a 10cm X 10cm sheet of paper at a rate of 100 grains per second. The grains have a mass of 1 mg and are falling from a height of 1m under gravity. The sheet of paper is sticky, so the grains stick after they fall. The possible answers are A) 5.0 Pa, B) -5.0 Pa, C) -19.6 Pa, D) 19.6 Pa. To solve this problem, you need to calculate the change in momentum of each grain when it hits the sheet and the rate of change of momentum for the stream of sand grains. The velocity of each
  • #1
Bgerst103
32
0

Homework Statement



Calculate the pressure of sand grains falling on a 10cm X 10cm sheet of paper. Assume the grains are each 1 mg in mass, the ate pf the sand as it falls is 100 grains per second. The sane is falling from a height of 1m under gravity, and the sheet it falls to is sticky, so that the grains stick after they fall. Calculate the PRESSURE the sand exerts on the sheet of paper each second.

A) 5.0 Pa
B) -5.0 Pa
C) -19.6 Pa
D) 19.6 Pa

Homework Equations



P=F/A

The Attempt at a Solution



I tried finding F by doing mg of 100 grains and dividing that by .01 cm^2 but didn't come up with any of these answers.
 
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  • #2
What is the velocity of each grain of sand when it hits the sheet of paper? What is the change in momentum of each grain of sand when it hits the sheet?

Chet
 
  • #3
Chestermiller said:
What is the velocity of each grain of sand when it hits the sheet of paper? What is the change in momentum of each grain of sand when it hits the sheet?

Chet

Using Vf^2=Vi^2+2a(DeltaX) i got that the velocity of each grain is 19.6 m/s when it hits. How do I solve for change in momentum and how are these implemented into the solution?
 
  • #4
Bgerst103 said:
Using Vf^2=Vi^2+2a(DeltaX) i got that the velocity of each grain is 19.6 m/s when it hits. How do I solve for change in momentum and how are these implemented into the solution?
If a 1mg mass changes speed from 19.6m/s to 0, what is the magnitude of its change in momentum?
 
  • #5
Bgerst103 said:
Using Vf^2=Vi^2+2a(DeltaX) i got that the velocity of each grain is 19.6 m/s when it hits. How do I solve for change in momentum and how are these implemented into the solution?
The velocity drops to zero when the grain hits and sticks to the paper. So, what is its change in momentum? (Incidentally, I get a different velocity after it falls 1 meter. I get ~ 4.5 m/s. Did you remember to take the square root?) You are going to be trying to find the rate of change of momentum of the sand grains that hit the paper.

Chet
 
  • #6
Chestermiller said:
The velocity drops to zero when the grain hits and sticks to the paper. So, what is its change in momentum? (Incidentally, I get a different velocity after it falls 1 meter. I get ~ 4.5 m/s. Did you remember to take the square root?) You are going to be trying to find the rate of change of momentum of the sand grains that hit the paper.

Chet

Oops, you're right. I forgot to take the square root. My professor gave us a bank of equations such as p=mv, pf=pi, deltaP=F(deltaT), where p is momentum. So, from the looks of it, I need to find the time using vf=vi + at to find the change in momentum? If that's right I got ~.46 s. Then to calculate F I did mg and plugged that back into deltaP=F(deltaT) which ends up looking like .000001 (9.8) x .46 = .000005. However, I'm rather confused why calculating mg for each grain of sand and dividing by the area doesn't work. How is momentum incorporated into the equation P= F/A.
 
  • #7
Bgerst103 said:
Then to calculate F I did mg
Then don't. As pointed out already, you need to calculate the rate of change in momentum of the sand as it strikes the sheet. If a 1mg of sand moving at 4.5 m/s comes to rest, what is its change in momentum? How often does that happen each second? What therefore is the rate of change of momentum of the stream of sand due to its impact with the sheet?
 
  • #8
Chestermiller said:
The velocity drops to zero when the grain hits and sticks to the paper. So, what is its change in momentum? (Incidentally, I get a different velocity after it falls 1 meter. I get ~ 4.5 m/s. Did you remember to take the square root?) You are going to be trying to find the rate of change of momentum of the sand grains that hit the paper.

Chet

So I calculated the change in momentum to be 0.0443 which I then plugged into P=F/A to get 4.43 Pa. I'm hoping the right answer is 5.0 Pa and that they just rounded up.
 
  • #9
I get a change in momentum per grain of 0.000001 kg x 4.43 m/s = 0.00000443 kg m /s. If there are 100 grains/second hitting the paper, the rate of change of momentum of the grains is 100/s x 0.00000443 kg m/s = 0.000443 kg m/s2=0.000443 Newtons. This force is spread over an area of 0.01 m2, so the pressure from the impact is 0.0443 N/m2= 0.0443 Pa. This doesn't match any of the answers.

In addition to the change in momentum of the grains, their weight is also building up on the paper. For 100 grains/s, the rate of mass buildup is 0.0001 kg/sec. The weight of this buildup is 0.00098 N/sec. The rate of increase of pressure resulting from this weight buildup is 0.00098/0.01 = 0.098 Pa/sec. So, the total pressure after 1 second is 0.143 Pa, and after 2 seconds, it is 0.24 Pa.

These results also do not match the answers provided.

Haruspex, what am I doing wrong?

Chet
 
  • #10
Chestermiller said:
I get a change in momentum per grain of 0.000001 kg x 4.43 m/s = 0.00000443 kg m /s. If there are 100 grains/second hitting the paper, the rate of change of momentum of the grains is 100/s x 0.00000443 kg m/s = 0.000443 kg m/s2=0.000443 Newtons. This force is spread over an area of 0.01 m2, so the pressure from the impact is 0.0443 N/m2= 0.0443 Pa. This doesn't match any of the answers.

In addition to the change in momentum of the grains, their weight is also building up on the paper. For 100 grains/s, the rate of mass buildup is 0.0001 kg/sec. The weight of this buildup is 0.00098 N/sec. The rate of increase of pressure resulting from this weight buildup is 0.00098/0.01 = 0.098 Pa/sec. So, the total pressure after 1 second is 0.143 Pa, and after 2 seconds, it is 0.24 Pa.

These results also do not match the answers provided.

Haruspex, what am I doing wrong?

Chet
No idea, Chet. Everything in your post looks right to me.
 
  • #11
Chestermiller said:
I get a change in momentum per grain of 0.000001 kg x 4.43 m/s = 0.00000443 kg m /s. If there are 100 grains/second hitting the paper, the rate of change of momentum of the grains is 100/s x 0.00000443 kg m/s = 0.000443 kg m/s2=0.000443 Newtons. This force is spread over an area of 0.01 m2, so the pressure from the impact is 0.0443 N/m2= 0.0443 Pa. This doesn't match any of the answers.

In addition to the change in momentum of the grains, their weight is also building up on the paper. For 100 grains/s, the rate of mass buildup is 0.0001 kg/sec. The weight of this buildup is 0.00098 N/sec. The rate of increase of pressure resulting from this weight buildup is 0.00098/0.01 = 0.098 Pa/sec. So, the total pressure after 1 second is 0.143 Pa, and after 2 seconds, it is 0.24 Pa.

These results also do not match the answers provided.

Haruspex, what am I doing wrong?

Chet

You did everything right. I emailed the professor who said they used 100m rather than 1m for the answers so everything is off by a factor of 100.
 
  • #12
Bgerst103 said:
You did everything right. I emailed the professor who said they used 100m rather than 1m for the answers so everything is off by a factor of 100.
It seems to me this would only account for a factor of 10.

Chet
 

1. How do you calculate the pressure of sand on paper?

To calculate the pressure of sand on paper, you will need to know the density of sand and the area of the paper. First, multiply the density of sand, which is typically given in kilograms per cubic meter, by the acceleration due to gravity, which is approximately 9.8 meters per second squared. Then, divide this result by the area of the paper in square meters. This will give you the pressure of the sand on the paper in Pascals (Pa).

2. What is the unit of measurement for pressure?

The unit of measurement for pressure is Pascals (Pa). It is named after French mathematician and physicist Blaise Pascal and is defined as one newton per square meter. Other commonly used units for pressure include atmospheres (atm), pounds per square inch (psi), and bar (bar).

3. How does the pressure of sand on paper affect the paper's strength?

The pressure of sand on paper can significantly affect the paper's strength. As the pressure increases, the paper may start to deform and ultimately tear if the pressure is too high. The type and weight of the paper also play a role in its strength, as thicker and stronger paper can withstand higher pressures compared to thinner and weaker paper.

4. What is the role of sand in determining the pressure on paper?

The presence of sand on paper can increase the pressure on the paper due to its weight. Sand particles have a higher density compared to air, and when placed on top of the paper, they exert a downward force, resulting in an increase in pressure. The amount of sand and its distribution on the paper's surface will also affect the pressure exerted.

5. How can the pressure of sand on paper be reduced?

The pressure of sand on paper can be reduced by either decreasing the amount of sand on the paper or increasing the surface area of the paper. The pressure can also be reduced by using a thicker and stronger paper that can withstand higher pressures. Additionally, placing a supporting material, such as a board or tray, underneath the paper can help distribute the weight of the sand and reduce the pressure on the paper's surface.

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