ANOTHER static coefficient question

Click For Summary

Homework Help Overview

The problem involves a package case resting on a ramp and requires determining the minimum static coefficient of friction for the case to remain at rest, as well as analyzing the motion of the case when the coefficient of friction is reduced. The subject area includes concepts from static and kinetic friction, forces on inclined planes, and motion under constant acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of mass in solving part (a) and explore the relationship between normal force and friction. There are attempts to derive equations based on force diagrams and the conditions for equilibrium.

Discussion Status

Some participants have provided guidance on how to approach the problem by suggesting the use of free body diagrams and the equations of motion. There is an ongoing exploration of how to express forces in terms of mass and angle, with some participants questioning their understanding of the relationships involved.

Contextual Notes

Participants note the challenge of working with unknown variables, specifically the mass of the object, and how this impacts their ability to derive the necessary equations. There is a recognition that certain data may not be required to solve the problem.

dura
Messages
8
Reaction score
0
Problem reads: A package case of mass m is left 10 m up on a 30 degree ramp overnight where it is at rest. a) what is the minimum value of the static coefficient of friction for this to be ture? Prove in general that the coefficient of static friction is [tex]\mu_s = \tan \theta[/tex]. b) during the night itrains and water seeps under the case reducing the coefficient of friction to .10 (both static and kinetic). With what speed does the case reach the bottom of the ramp?

Ok... can I move on with part a without mass of the object? I am trying to move forward with [tex]F_{s} max = \mu_s FN[/tex]

Oh man, physics...
 
Physics news on Phys.org
Did you draw a diagram of the forces? You have mg, N, and the friction. If you want the object to stay still [tex]\Sigma F[/tex] on it needs to be 0, in both X and Y axis. I suggest (and so will your teacher) that you lay the X axis on the ramp's plane.

From [tex]\Sigma F_y = 0[/tex] you will be able to find the normal force (N) as a function of the object's mass m. Once you have that, write down [tex]\Sigma F_x = 0[/tex] to find the magnitude of the friction force. Divide that by N and you will have the coefficient of friction. :smile:
 
As for (b), once the object is moving the friction is constant. You can solve the question by finding [tex]\Sigma F_x[/tex] (it's no longer zero; X is the only axis in which there is movement). Of course that depends on the object's mass, which you don't have, but you only need the constant acceleration so you divide by m and get a real number. From there it's just using this equation:

[tex]{V_f}^2 = {V_0}^2 + 2ax[/tex]

(You can also use energies and work to find the final velocity, but I'm not sure if you already studied that.)
 
Thank you Chen for your reply! I still have stupid questions however.
I did draw a FBD and see that the sum of all forces must be 0. I am still confused. When I take [tex]\Sigma F_y = 0[/tex], as FN + (-mg cos 30) = 0... not knowing the normal force or the mass of the object, I am left with gravity and the angle only. What am I missing? (other than brains)


Originally posted by Chen
Did you draw a diagram of the forces? You have mg, N, and the friction. If you want the object to stay still [tex]\Sigma F[/tex] on it needs to be 0, in both X and Y axis. I suggest (and so will your teacher) that you lay the X axis on the ramp's plane.

From [tex]\Sigma F_y = 0[/tex] you will be able to find the normal force (N) as a function of the object's mass m. Once you have that, write down [tex]\Sigma F_x = 0[/tex] to find the magnitude of the friction force. Divide that by N and you will have the coefficient of friction. :smile:
 
Originally posted by dura
Thank you Chen for your reply! I still have stupid questions however.
I did draw a FBD and see that the sum of all forces must be 0. I am still confused. When I take [tex]\Sigma F_y = 0[/tex], as FN + (-mg cos 30) = 0... not knowing the normal force or the mass of the object, I am left with gravity and the angle only. What am I missing? (other than brains)

You can leave N as a function of m, and as you noticed:

[tex]F_N = (g\cos 30) m[/tex]

When you write down [tex]\Sigma F_x = 0[/tex] and replace [tex]f_s[/tex] with [tex]F_N\mu[/tex] you will find that the mass cancels. :)
 
Originally posted by dura
...When I take [tex]\Sigma F_y = 0[/tex], as FN + (-mg cos 30) = 0... not knowing the normal force or the mass of the object, I am left with gravity and the angle only. What am I missing?
You are doing fine. Just keep going.

When it looks like you need data that is not given, chances are you don't really need it. That's the case here.

The equation for y, gives FN = mg cos 30. Now write the equation for the x-components. Then combine them, as Chen advised.
 

Similar threads

Replies
6
Views
3K
Replies
18
Views
4K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 28 ·
Replies
28
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K