Calculating Mass using F=ma and Friction

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Homework Help Overview

The discussion revolves around calculating mass using the equation F=ma in the context of a friction problem. The original poster has a force of 500N and a friction force of 125N but lacks a given acceleration, leading to questions about whether to use 9.81 m/s² as the acceleration.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between applied force, friction, and mass. The original poster attempts to calculate mass using the gravitational acceleration, while others question the appropriateness of this assumption. There are discussions about the normal force and its relation to the friction force.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on how to approach the problem. Some participants suggest using additional data from a chart to determine the coefficient of friction, while others clarify the relationship between forces involved.

Contextual Notes

There is a lack of explicit information regarding the acceleration and the coefficient of friction, which are crucial for solving the problem. The original poster is working with a friction simulator and is encouraged to provide more context or visual aids to facilitate understanding.

Drew552
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Homework Statement


Well i have the force which equals 500N and I don't have a given acceleration so should I be substituting 9.81 as the acceleration? or is their another way to find it given that I also have the friction which equals 125N. So which way is the right way of going?

Homework Equations


F=Ma

The Attempt at a Solution


Well what I've done is 500=m x 9.81, which goes to 500/9.81=m, which then gives me 51.02kg, so is that correct because it doesn't seem right to me.

Any help would be much appreciated,

Thanks.
 
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Can you please state the exact problem you are asked to solve, word for word.
 
Its a fill in the chart question so its not really a word for word one, i have all the values but this one. If i can provide anything else please tell me.
 
You need to describe the situation a bit more. You have the force on what? What is it doing?
 
Well its on a friction simulator online. I'm pushing on a unknown mass with 500N of applied force. The friction force comes out as 125N and the sum of the forces is 375N to the right direction. I am asked to find the mass and the weight of the object. I want to know the mass of the object which in turn will let me work out the weight with W=mg.

Thanks.
 
Drew552 said:
I'm pushing on a unknown mass with 500N of applied force.
OK. Across a horizontal surface?

Drew552 said:
The friction force comes out as 125N and the sum of the forces is 375N to the right direction.
So the only horizontal forces involved are the applied force and the friction?

Drew552 said:
I am asked to find the mass and the weight of the object.
Are you given any additional information? Such as the acceleration or the coefficient of friction?
 
Yes it is being pushed on an horizontal surface. And yes the only horizontal forces are the force and friction. No i am not given the acceleration but I am assuming it could possibly be 9.81 and it doesn't give me the coefficient of friction.
 
Drew552 said:
No i am not given the acceleration but I am assuming it could possibly be 9.81
That's the acceleration due to gravity. No reason to think the mass has that acceleration.

I do not see that enough information has been given to solve the problem. Does the diagram or chart have any other info in it? Or a previous problem?
 
Am i allowed to attach a picture of the chart and also the picture of the simulator??
 
  • #10
Drew552 said:
Am i allowed to attach a picture of the chart and also the picture of the simulator??
Go ahead.
 
  • #11
.
 
  • #12
Simulator picture.PNG
Chart.PNG
 
  • #13
So you have plenty of other info! Hint: Use the data for the other masses to figure out the coefficient of friction. (In fact, if you understand the chart, you don't even need to calculate the coefficient. You can just read off the mass that you need.)
 
  • #14
I do? Well point taken then, thanks for the hint, ill try to to do my best at getting the answer but still unsure how to get the friction coefficient
 
  • #15
Hint: What's the basic relationship for kinetic friction that uses the coefficient of friction?
 
  • #16
F(kinetic)= u(coefficient) x F(normal)??
 
  • #17
Drew552 said:
F(kinetic)= u(coefficient) x F(normal)??
Exactly! And in these examples, what creates the normal force?
 
  • #18
a entity pushing against something thus making the kinetic energy?
 
  • #19
so if i had a 20N force pushing a against a box the normal force would equal 20N right?
 
  • #20
The normal force is the force pushing each mass against the surface. What is that force?
 
  • #21
The force is the 2 entities rubbing together?
 
  • #22
Drew552 said:
The force is the 2 entities rubbing together?
Sure, but what supplies that force? It's not the applied force, since that force is horizontal. Hint: What's the only vertical force acting on the masses?
 
  • #23
gravity?
 
  • #24
Drew552 said:
gravity?
Of course!

Now pick anyone of the masses and use the given data to figure out the coefficient of friction (which is a constant for this problem).
 
  • #25
so it would be 40=u+9.8 Assuming 40 is the mass
 
  • #26
so is the coefficient of friction constant over one horizontal surface no matter the mass?
 
  • #27
Drew552 said:
so is the coefficient of friction constant over one horizontal surface no matter the mass?
Yes, assume for this problem that the coefficient of friction is constant regardless of mass. (Meaning: Assume the masses are all the same material.)

Pick one of the masses that are given. Find the normal force, which is just the weight. You are given the friction force. Use those values to calculate the coefficient.

Once you have the coefficient, you can use it with the friction force to calculate the unknown mass.
 
  • #28
Ok, thank you for all your help, I think I've got it down now. ill post again if i come to any complications.

Thanks a lot :)
 

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