How to Find Coefficient of Friction on Sandpaper Using Applied Force and Mass?

Click For Summary
To find the coefficient of friction on sandpaper, the applied force and mass of the block are used to determine the frictional force. By moving the block slowly to maintain constant speed, the applied force equals the frictional force, allowing for the calculation of friction without acceleration. The normal force is calculated using the mass and gravitational acceleration. The frictional force can then be plotted against the normal force on a graph to determine the coefficient of friction from the slope. This method effectively utilizes experimental data to derive the necessary values for the lab project.
Havenater23
Messages
38
Reaction score
0

Homework Statement


I have a lab project in physics and I'm having a bit of trouble. We measured the Applied force of a moving block of wood using a spring scale. The first trials were don't with just the table , knowing the mass and applied force I was able to find the friction force; simply it was the same as F app. My problem now is that we did other trials on sandpaper, and I have the Applied force and mass and I don't know how to find the mass. We are suppose to find the Co efficent of friction, but not through math ; we have to use a graph where : y = mx aka Ff=U*Fn

Which isn't too bad, I just can't figure out how to get Force friction with only the mass and applied force when it is using the sand paper.

Mass = 0.27952 kg Force applied = 2.0N
Fn = Fg
mg= 0.27952(9.8)= 2.739296 N

Help please


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
I am assuming that the block or mass is moving slowly or at constant speed when the force is applied...in the 2nd example, you know the applied force so you then know the kinetic friction force, and since you have calculated the normal force, plot that point on the graph with Ffriction on the y-axis and the normal force on the x axis. Then try another mass, and do the same, moving it slowly; and yet another...your graph should approximate a straight line, the slope of which is____?
 
Can't find friction, yes I have the Force normal , but no coefficient. Even if I know the applied force I don't know how that helps me find friction example


Fapp-Ffrict = MA


I only know F app and mass



Ffrict = U * Fn

Don't see how you solve unless Fapp in the 2nd problem also equals Ffrict.
 
Havenater23 said:
Can't find friction, yes I have the Force normal , but no coefficient. Even if I know the applied force I don't know how that helps me find friction example


Fapp-Ffrict = MA


I only know F app and mass



Ffrict = U * Fn

Don't see how you solve unless Fapp in the 2nd problem also equals Ffrict.
That's right, and that is why i mentioned that in your experiment, you move the object slowly so it doesn't accelerate (a =0). Then Fapp-Ffrict = 0, or Fapp = ffrict. If you use too much of a force and the object accelerates, now you need a means of determining the acceleration. Instead, move it s-l-o-w-l-y, without accelerating. Per Newton 1, an object at rest or moving in a straight line at constant speed, will remain at rest or moving in a straight line at constant speed, unless the net force acting on it is greater than 0. Keep it at constant speed, and F_net = 0.
 
Oh true, didn't even think about it. Ha.

Thank you , appreciate it.
 
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
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
17
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K