### Video Transcript

A body of weight 24 newtons is
placed on a rough plane inclined at 30 degrees to the horizontal. The coefficient of static friction
between the body and the plane is one over six root three. And the coefficient of kinetic
friction is one over 12 root three. A force is pushing the body upward
in the direction of the line of greatest slope. Determine the magnitude of the
force that would make the body about to move.

Sketching in this scenario, it
could look like this where our body is at rest on an inclined plane of 30
degrees. Situated like this, our body is
subject to a static frictional force, which depends on the coefficient of static
friction given as one over six root three. We’re told that there’s a force,
we’ll call it 𝐹, that acts on the body in the direction of the line of greatest
slope. And we want to figure out how big
that force could be, such that the body would be about to move.

Let’s first give ourselves some
space on screen to work. And then let’s include in our
sketch all of the forces that are acting on our body. We’re told that there’s a weight
force of magnitude 24 newtons. And we also know that there’s a
normal force perpendicular to the surface of our slope. Along with all this, there is a
static frictional force. Now, if we had no applied force,
what we’ve called 𝐹, pushing our body up the incline, then our static frictional
force would point in that same direction. This would be so it would resist
the block sliding down the plane. But in our scenario, we want to
solve for the maximum possible value of 𝐹 such that the block is just barely not
moving up the incline. That means that, here, the static
frictional force is actually acting down the incline.

As we begin analyzing these forces
acting on our body, we can orient ourselves with a pair of orthogonal axes. We’ll say that forces in motion
acting up the incline are positive, and those acting the opposite way are
negative. Likewise, forces in motion
perpendicularly away from our surface are positive, and those in the other direction
are negative. Knowing this, we can apply Newton’s
second law of motion to both of what we’ll call our 𝑦-direction and our
𝑥-direction.

Looking first in the 𝑥-direction,
we see that our force 𝐹 acts in the positive 𝑥, while our frictional force acts
opposite this. These, however, aren’t the only
𝑥-direction forces. That’s because our weight force has
a component right here in the negative 𝑥-direction. Now, this angle right here in this
triangle is equal to 30 degrees. And since this intersection is a
right angle, we can see that this 𝑥-component of our weight force is 24 newtons
times the sin of 30 degrees. The sin of 30 degrees we can recall
is one-half, so we’ll substitute that in right away.

So these then are indeed all the
forces acting on what we’ve called our 𝑥-direction. By Newton’s second law, these add
up to our object’s mass times acceleration in this dimension. But remember that we’re solving for
𝐹 such that our object is not yet in motion. In other words, 𝑎 sub 𝑥 is
zero. That means our whole right side is
zero, which means we can rearrange and see that the maximum force we can apply to
our body without it yet being put in motion is equal to the static frictional force
plus 12 newtons. In general, the force of static
friction is equal to the coefficient of static friction times the normal force
acting on the relevant body.

In our scenario, we know 𝜇 sub
𝑠. But what about 𝐹 sub 𝑁? To figure that out, we’ll want to
look at Newton’s second law of motion applied in the 𝑦-direction. Here, we have a positive, normal
force but then a negative component of our weight force here on our sketch. This component is 24 newtons times
the cos of 30 degrees where the cos of that angle equals the square root of three
over two. Just like in the 𝑥-direction,
because there’s no acceleration in the 𝑦-direction, these forces add up to
zero. This tells us that the normal force
equals 12 times the square root of three newtons.

Now that we know 𝐹 sub 𝑁, we can
substitute it as well as 𝜇 sub 𝑠 into our equation for the frictional force. And multiplying these quantities
together, we see that a square root of three cancels out and that the maximum
frictional force, therefore, is two newtons. 𝐹 then is 14 newtons. This is the magnitude of the
largest force that could act directly up the incline without our body being put in
motion.