Now let's take our slope example and one more time apply it but this time with friction.
Let's say that you have a block that rests on a slope
with the coefficient of static friction of 0.5,
the coefficient of kinetic friction of 0.3 and you wanna get the block to move.
The question is to get the block right to the threshold of sliding,
what angle does the ramp need to be lifted to?
Try solving this yourself and remember your trig identities as you do.
And see if you can solve for the angle that the block has to be lifted to on the slope.
The way to solve a problem like this is again first is to draw a diagram
and you can see what we have here is the forces that we typically included
where you have your gravitational force directly downwards,
we've shaded it because we've actually broken it already into its two components
as usual we have the gravitational force down the slope fg sine of theta.
We have the gravitational force into the slope fg cosine of theta.
And we have the normal force pushing directly out of the slope as well as a coordinate system.
We're going to add one more force now which is the frictional force
and again the frictional force is going to try to oppose the motion of our object
so if the object would want to slide down the slope,
the frictional force will be trying to oppose the motion, pushing up the slope.
The question is, can we get try to get this block to the threshold of sliding downwards?
In other words, we want the acceleration down the slope to be right at that threshold of zero
so that it can be anything more and be a positive acceleration.
So let's see if we can solve to get that.
All we wanna do is first write out Newton's second law.
Let's go ahead and write it in the x direction.
The forces in the x direction you see are fg sine of theta acting downwards
and f of friction f-sub-f acting upwards.
Keeping with our sine convention, what this will look like is fg sine of theta minus force of friction.
This will be equal to, by Newton's second law, mass times this acceleration in the x direction.
In the y direction, we'll have something similar,
when we have a normal force acting upwards
and then in a negative y direction, so minus, we have fg times the cosine of theta.
And this is equal to the mass times the acceleration in the y direction.
The first thing you wanna do in any problem with a slope like this
where you still have a y equation and especially if when you're going to consider friction
is to recognize that the acceleration into in and out of the slope is zero
because again, the object is not gonna fly off the slope
because of forces is not gonna go into the surface because of the forces
and so we know it's not going to move into in or out of the slope.
So in the y direction we can say that the acceleration of the object is zero.
So now let's figure out the forces in the x direction.
We have fg times the sine of theta minus,
and now we have our frictional force which we plug in for our frictional force.
We know first of all that the object isn't moving yet.
We're trying to get it to the threshold of motion but it has not actually started sliding yet.
So what we need to do is plug in a static friction
but what can we put in for static friction,
because again static friction could be less than or equal to mu-sub-s times the normal force.
For this problem since we're trying to get it right to the threshold of motion,
we already know that we maxed out the static friction,
we pushed it right to its limit so that the gravitational force pulling the object down
and the friction trying to fight that force and keep the object where it is
are right at their threshold, so we know that we can instead of using the less than or equal to,
use the equal to mu-sub-s times the normal force
because again, it's right at that limit where the static friction is about to give away. So let's use that.
We have the frictional force which is mu-sub-s times the normal force.
And this is equal to mass times acceleration in the x direction.
Now what I'm gonna do in the x direction, is similar to what I just did in the y direction,
except now it will be for a very different reason. So it's important to keep this distinct.
In the y direction, what I said is the object isn't moving into in or out of the slope.
For the x direction, I would also say that the acceleration is zero but it's for a very different reason.
This y direction argument that I just made with the acceleration into in
or out of the slope being zero
is a very general argument that you can use again and again
but the acceleration in the x direction that I'm about to call zero
is because of something specific in this problem.
In this problem we are asked about the threshold.
The box is right at the threshold of sliding downwards.
If that's true then the acceleration in the x direction, along the slope is zero,
because it's about to start moving, but we just care about what force must I be to get it right to zero
because any more force than that, and the object starts sliding
so we just wanna get it to that threshold.
And what do we need that, that threshold to be?
So that's what's we're going to do say the acceleration in the x direction is zero,
but only because in this problem we're considering a threshold.
When we do this we can see that fg times the sine of theta minus mu-sub-s times fn is zero
which means that I can add the mu-sub-s fn to both sides and equate this forces and get something like this.
That fg times the sine of theta, is equal to mu-sub-s times fn.
And this is a fancy way of saying, that the gravity trying pull the object down the slope,
is in perfect equilibrium with the coefficient or the static friction trying to keep it from moving down the slope.
We're asking about that threshold and what angle must this be,
what angle must theta be so that this become equal?
If you look at this equation, you notice that we know f-sub-g
because we can write mg for the gravitational force.
We know mu-sub-s because that is given in the problem to be 0.5.
We want to solve for theta so this is something to find,
but then what do we do with this normal force?
We have to somehow find the normal force and this is why in the friction problem,
the y direction into in and out of the slope, actually becomes important.
So let's use the y equation to figure out what the normal force has to be.
Going to our y equations, since we've said the accelerations are zero,
we can solve for the normal force by adding fg cosine of theta to both sides
and now we have an expression for the normal force
which we can plug in to the x direction equation.
Doing that we now have fg times the sine of theta is equal to mu-sub-s times the normal force
which we just solved from the y equation
so notice we're just using this equation to plug in f-sub-n, giving us fg times the cosine of theta.
So this is a great place to pause because it's always easy when you're doing a problem like this,
to get lost in the variables and the equations as we rearrange things.
So at this point let's reassess what it is we're trying to find in this problem.
We're asked about an angle specifically what angle do we need to lift this slope
so that the object starts sliding. So we're looking for theta,
so what we would like to do is rearrange this equation
in a way that we can solve for theta, the angle.
What I'm gonna do next is do exactly that is so try to rearrange equation so solve for theta.
So doing that, I'm going to divide both sides by fg times the cosine of theta
and very conveniently what this will do for us is get rid of this fgs
telling us that mu-sub-s is equal to sine of theta divided by cosine of theta
which from our trigonometry is just equal to the tangent of theta.
And this is in fact, the famous result that's quoted many times in many physics classes
telling you that the only thing the angle of the slope depends on
is the coefficient of static friction so if you know the coefficient of static friction,
you know exactly what angle you need to lift your slope to
so that an object on that slope will start sliding.
In this particular problem we can actually solve for what that angle that would be.
Since we know that mu-sub-s is equal to 0.5.
So now we know that the tangent of theta is equal to 0.5.
And since this is not one of our special angles,
this is something that you plug into a calculator or be given on an exam.
What you would do is set up the angle theta is equal to what we call the arc tangent
where the inverse tangent of 0.5
which is just a fancy way of saying what angle if I took the tangent of it
would give me 0.5 and the calculator or something will tell you
that this angle happens to be approximately 26.570 degrees.
In general, this equation here, is the key result.
Where mu-sub-s the coefficient of static friction is equal to the tangent of theta.
This result that we have of 26.570 degrees is just a particular result for this problem.