Calculating angle with coefficient of static friction value

[Fireant]

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?

 

[tiny-tim]

Hi Fireant!

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

I don't understand that at all.

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. ​

 

[Fireant]

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!

 

[tiny-tim]

Hi Fireant!

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?​

 

[Fireant]

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 ?

 

[tiny-tim]

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

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 ?

 

[Fireant]

Sorry I'm at work, that's why it's taking so long to reply.

the normal component of W would be mgcostheta?

 

[tiny-tim]

Yes, so N = mgcosθ.

ok, now find the friction force, and then write the equation for components along the plank.

 

[Fireant]

so Ff=μsFN
=0.33(mgcosθ)
=0.33(98)(cosθ)
=32.34cosθ

 

[tiny-tim]

ok

now the equation along the plank

 

[Fireant]

would be...

Fnet=Fn-Ff-Fg
0=32.34cosθ-98

?

 

[Fireant]

but I'm missing

sintheta

right?

 

[tiny-tim]

yes.

 

[Fireant]

0=32.34costheta-sintheta-98

 

[Fireant]

im not sure how i would isolate theta...

 

[tiny-tim]

put = in the middle and divide

 

[Fireant]

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?

 

[Fireant]

or would it be easier if:

costheta=(sintheta-98)/32.34

 

[tiny-tim]

sin/cos = tan ?

 

[Fireant]

...oh wow...lol..forgot about that.

so

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

that seems wrong..

 

[tiny-tim]

32.34costheta=sintheta-98

that's times 98

write it out properly and you won't make mistakes!

 

[Fireant]

Alright so it must be that

tanθ=32.34/98
θ= 18.26° or 18°

 

[tiny-tim]

Looks good!

 

[Fireant]

thank u tiny tim! where's my prize?...just kidding...:[

 

[jhae2.718]

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 \theta_*) such that motion is impending. We want to know what this angle is. Our system is the pellet, shown in the sketch below:

We'll establish a convenient reference frame, which we'll denote by x and y, 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 f), and the normal force.

We now find the sum of the forces in the x and y directions (ideally, we'd use vectors!):
<br /> \begin{align*}<br /> F_x &amp;= mg\sin\theta - f \\<br /> F_y &amp;= N - mg\cos\theta<br /> \end{align*}<br />
Because of the condition of impending motion, we know two things:

• a_x = 0 (the pellet is just about to move!)
• f = \mu_s N, the maximum possible value of the static friction force.

Using Newton's second law:
<br /> \begin{align*}<br /> 0 &amp;= mg\sin\theta_* - f \\<br /> 0 &amp;= N - mg\cos\theta_*<br /> \end{align*}<br />
From this, N = mg\cos\theta_*, and, using f = \mu_s N:
<br /> \begin{align*}<br /> 0 &amp;= mg\sin\theta_* - \mu_s mg\cos\theta_* \\<br /> mg\sin\theta_* &amp;= \mu_s mg\cos\theta_* \\<br /> \tan\theta_* &amp;= \mu_s<br /> \end{align*}<br />
We now have \theta_* as a function of only one parameter, the coefficient of static friction, which is this case is 0.33. Plugging in, we get \theta_* \approx 18.26^{\circ}.

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 \mu_s comes from).

 

[Fireant]

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?

 

[tiny-tim]

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

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")