Calculating angle with coefficient of static friction value

In summary: Some people call that the y axis, some people call the vertical the y-axis :confused:That's why I say "the normal direction", so there's no confusion.... do you mean on the y axis?Some people call that the y axis, some people call the vertical the y-axis :confused:That's why I say "the normal direction", so there's no confusion.In summary, the conversation discusses finding the angle at which a box will begin to slide on a plank tilted at a certain angle, given the coefficient of static friction. The equation for finding the angle involves taking the components of forces in the normal and slope directions and setting the sum to zero. The final result
  • #1
Fireant
22
0
There's a box sitting on a plank. The coefficient of static friction is 0.33. I want to figure out at which angle the box will begin to slide if the plank is tilted. I know the factors that come into play are FN, Fg or Fgx, and Fs.
So far I have

Fnet= FN-Fs-Fg
=FN- μs-sin∅
=cos-0.33-sin∅

This doesn't look right... can anyone help me out?
 
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  • #2
Hi Fireant! :smile:
Fireant said:
Fnet= FN-Fs-Fg
=FN- μs-sin∅
=cos-0.33-sin∅

I don't understand that at all. :confused:

First, use components in the normal direction to find N.

Then you know the friction is 0.33*N, and you can use components in the slope direction to find θ.

Show us what you get. :smile:
 
  • #3
Hi thank you for responding!

The mass of the box is 10kg. So the normal force here would have to be mg...?

normal force=98N

Ff=μs(FN)
=0.33(98)
=32.34 N


and then
tanθ= 97.5/98
θ=44°

I got 97.5 (for the x component) from pythagorean theory...

I'm pretty lost!
 
  • #4
Hi Fireant! :smile:
Fireant said:
Hi thank you for responding!

The mass of the box is 10kg. So the normal force here would have to be mg...?

No, the weight, W, is mg (98 N).

The normal force, N, is the normal component of the reaction force.

You find N by taking components in the normal direction (of N and W), and equating the sum to zero …

what do you get?
 
  • #5
I think I'm getting confused when you say 'the components in the normal direction'... do you mean on the y axis? Isn't that just FN? Which would be -98N ?
 
  • #6
Fireant said:
I think I'm getting confused when you say 'the components in the normal direction'... do you mean on the y axis?

Some people call that the y axis, some people call the vertical the y-axis :confused:

That's why I say "the normal direction", so there's no confusion.
Isn't that just FN? Which would be -98N ?

What is the normal component of W ?
 
  • #7
Sorry I'm at work, that's why it's taking so long to reply.

the normal component of W would be mgcostheta?
 
  • #8
Yes, so N = mgcosθ.

ok, now find the friction force, and then write the equation for components along the plank.
 
  • #9
so Ff=μsFN
=0.33(mgcosθ)
=0.33(98)(cosθ)
=32.34cosθ
 
  • #10
ok

now the equation along the plank
 
  • #11
would be...

Fnet=Fn-Ff-Fg
0=32.34cosθ-98

?
 
  • #12
but I'm missing

sintheta

right?
 
  • #13
yes.
 
  • #14
0=32.34costheta-sintheta-98
 
  • #15
im not sure how i would isolate theta...
 
  • #16
put = in the middle and divide :smile:
 
  • #17
lol I'm hopeless...! thank you so much for bearing with me

ok going to try this out...

32.34costheta=sintheta-98

sintheta=32.34costheta+98
sintheta/32.34=costheta+98

Sorry, what am i dividing by and how?
 
  • #18
or would it be easier if:

costheta=(sintheta-98)/32.34
 
  • #19
sin/cos = tan ? :wink:
 
  • #20
...oh wow...lol..forgot about that.

so

32.34costheta=sintheta-98
tantheta=130.34
theta=89.6 or 90

that seems wrong..
 
  • #21
Fireant said:
32.34costheta=sintheta-98

that's times 98

write it out properly and you won't make mistakes! :rolleyes: :rolleyes:
 
  • #22
Alright so it must be that

tanθ=32.34/98
θ= 18.26° or 18°
 
  • #23
Looks good! :smile:
 
  • #24
thank u tiny tim! where's my prize?...just kidding...:[
 
  • #25
I'd like to make one comment that I was working towards in the other thread.

With these types of problems, we generally want the governing equations of motion rather than a numerical answer. Having the equations allow us to see what happens for a variety of cases and not just the case for the particular set of numbers you have. For this reason, you should solve problems symbolically and only plug in numerical values on the last step.

Here is how I would solve this problem:

We know that the pellet is on the ramp which is tilted at an angle (we'll call it [itex]\theta_*[/itex]) such that motion is impending. We want to know what this angle is. Our system is the pellet, shown in the sketch below:
N4mOb.png


We'll establish a convenient reference frame, which we'll denote by [itex]x[/itex] and [itex]y[/itex], though we could call it anything we want. Drawing a free-body diagram, we have the weight of the pellet, the friction force (which we'll call [itex]f[/itex]), and the normal force.
piQHH.png

We now find the sum of the forces in the [itex]x[/itex] and [itex]y[/itex] directions (ideally, we'd use vectors!):
[tex]
\begin{align*}
F_x &= mg\sin\theta - f \\
F_y &= N - mg\cos\theta
\end{align*}
[/tex]
Because of the condition of impending motion, we know two things:
  • [itex]a_x = 0[/itex] (the pellet is just about to move!)
  • [itex]f = \mu_s N[/itex], the maximum possible value of the static friction force.
Using Newton's second law:
[tex]
\begin{align*}
0 &= mg\sin\theta_* - f \\
0 &= N - mg\cos\theta_*
\end{align*}
[/tex]
From this, [itex]N = mg\cos\theta_*[/itex], and, using [itex]f = \mu_s N[/itex]:
[tex]
\begin{align*}
0 &= mg\sin\theta_* - \mu_s mg\cos\theta_* \\
mg\sin\theta_* &= \mu_s mg\cos\theta_* \\
\tan\theta_* &= \mu_s
\end{align*}
[/tex]
We now have [itex]\theta_*[/itex] as a function of only one parameter, the coefficient of static friction, which is this case is 0.33. Plugging in, we get [itex]\theta_* \approx 18.26^{\circ}[/itex].

This is a rather interesting result, as it shows that the angle at which an object will overcome friction and begin to move is independent of the mass of the object or even the gravitational acceleration (assuming that it's constant), and depends only on the material of the pellet and the surface (where [itex]\mu_s[/itex] comes from).
 
Last edited:
  • #26
thank you for the thorough explanation.

it's true- i don't think of the equations symbolicaly enough and often become restricted in trying to figure out the numbers. i just need a better understanding of the problems...really see what I'm trying to calculate.

one thing I'm having trouble following is your code, jhae2.718. When you write the equations, it looks like they're coming up as html...or some other type of code...maybe it's my browser?
 
  • #27
Fireant said:
When you write the equations, it looks like they're coming up as html...or some other type of code...maybe it's my browser?

they should be coming up as mathjax (latex): it's one of the great features of pf :tongue2:

i think you need javascript enabled

you may be able to get help on this in the feedback sub-forum

(there's lots of threads on it there already … alternatively, search for "mathjax") :wink:
 

What is the coefficient of static friction?

The coefficient of static friction is a value that represents the amount of force required to overcome the static friction between two surfaces. It is a dimensionless number that is specific to the materials and surfaces in contact.

How is the coefficient of static friction determined?

The coefficient of static friction can be determined experimentally by measuring the force needed to move an object across a surface at a constant speed. The coefficient is then calculated by dividing the force by the weight of the object.

What is the relationship between the coefficient of static friction and the angle of inclination?

The coefficient of static friction is directly related to the angle of inclination between two surfaces. As the angle increases, the coefficient of static friction also increases, meaning more force is required to overcome the friction and move an object.

How is the angle of inclination calculated using the coefficient of static friction?

The angle of inclination can be calculated by taking the inverse tangent of the coefficient of static friction. This can be represented as θ = tan^-1(μ), where μ is the coefficient of static friction.

What are some real-world applications of calculating angle with coefficient of static friction value?

The calculation of angles using the coefficient of static friction is important in many engineering and construction fields. It can be used to determine the stability of structures, such as bridges or buildings, and to ensure that objects will not slip or slide on inclined surfaces.

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