ANOTHER static coefficient question

In summary, the problem involves a package of mass m on a 30 degree ramp with a coefficient of friction of \mu_s = \tan \theta. The minimum value of the static coefficient of friction for the package to stay at rest is found by setting \Sigma F_x = 0 and solving for the magnitude of the friction force, which can then be divided by the normal force to find the coefficient of friction. In part (b), the coefficient of friction is reduced to .10 due to rain, and the speed of the package at the bottom of the ramp can be found using the equation {V_f}^2 = {V_0}^2 + 2ax. The normal force can be found using \Sigma
  • #1
dura
9
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...
 
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  • #2
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:
 
  • #3
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.)
 
  • #4
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:
 
  • #5
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. :)
 
  • #6
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.
 

What is the static coefficient of friction?

The static coefficient of friction is a measure of the amount of force required to overcome the resistance between two surfaces in contact with each other without any movement.

How is the static coefficient of friction calculated?

The static coefficient of friction is calculated by dividing the maximum amount of force needed to start an object moving by the weight of the object.

What factors can affect the static coefficient of friction?

The type of surfaces in contact, the amount of force applied, and the presence of any lubricants or contaminants can all affect the static coefficient of friction.

What does a high or low static coefficient of friction indicate?

A high static coefficient of friction indicates that more force is needed to overcome the resistance between two surfaces, while a low static coefficient of friction indicates that less force is needed.

How is the static coefficient of friction used in real-world applications?

The static coefficient of friction is used in various industries, such as manufacturing and engineering, to determine the best materials and surfaces to use for optimal movement and stability.

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