Video Transcript
A body of mass 16 kilograms was
placed on a smooth plane that was inclined at 45 degrees to the horizontal. A horizontal force of 48
kilogram-weight was acting on the body towards the plane. Given that the line of action of
the force, the body, and the line of greatest slope all lie in the same vertical
plane, determine the acceleration of the body. Consider the acceleration due to
gravity to be 𝑔 equals 9.8 meters per square second.
To answer this question, we’re
simply going to begin by sketching a diagram. This diagram doesn’t need to be to
scale, but it should be roughly in proportion so we can accurately model what’s
going on. We have a smooth plane inclined at
45 degrees to the horizontal. Now, the fact that this plane is
smooth tells us simply that it exerts no frictional force on the body placed on
it. We have a body whose mass is 16
kilograms placed on this plane. This means it exerts a downward
force on the plane. And that force is equal to mass
times acceleration due to gravity. We’ll call acceleration due to
gravity 𝑔, although we are told it’s 9.8 meters per square second.
The downward force the body exerts
on the plane is 16𝑔. Now, we also have a horizontal
force of 48 kilogram-weight acting on the body towards the plane. And so we can model this as the
body exerting a horizontal force on the plane itself. The problem is we’re given
information about this force in kilogram-weight. And currently, we’ve modeled the
weight of the body to be in newtons. But really a kilogram-weight is
simply another way of measuring a force. And we can convert from
kilogram-weight to newtons by recognizing that one kilogram-weight is equal to 9.8
newtons. And so our horizontal force of 48
kilogram-weight is roughly equal to 48 times 9.8 newtons. That’s 470.4. So our horizontal force is 470.4
newtons.
We’re looking to calculate the
acceleration of the body. And so we need to make an
assumption about the direction in which we think the body is going to move. Let’s assume it’s going to move up
the plane. And so the acceleration is positive
in this direction. Really, it doesn’t matter
though. If we assumed the acceleration to
be acting in the opposite direction, the sign would tell us the direction in which
it really acts. Now, there’s one further force that
we can model on our diagram. That force is the normal reaction
of the plane on the body. Let’s call that 𝑅.
Now, we don’t really need to
consider 𝑅 in this question. But it’s always worth drawing on
the diagram so we have everything modeled. And then the formula we’re going to
use to help us to solve this problem is 𝐅 equals 𝑚𝑎: force is equal to mass times
acceleration. We have modeled the acceleration as
acting parallel to the plane. So we’re going to consider the
components of our forces that act in this direction. Let’s begin by looking at our
horizontal force. We’re going to add a right-angled
triangle as shown.
We know the included angle in this
triangle is 45 degrees. And that’s because corresponding
angles are equal. We want to find the component of
this force that is parallel to the plane. So let’s label that side of our
triangle 𝑥 or 𝑥 newtons. 𝑥 is the adjacent side in this
triangle, and the hypotenuse is 470.4 newtons. So let’s use the cosine ratio to
form an equation. We know that cos 𝜃 is adjacent
over hypotenuse. So in this case, cos of 45 degrees
equals 𝑥 over 470.4. We can work out 𝑥 by multiplying
by 470.4. So 𝑥 is 470.4 times cos of 45, but
cos of 45 is root two over two. And so we find 𝑥 to be equal to
235.2 root two. That is the component of this force
that acts parallel to the plane.
We’ll now repeat this process for
the weight force. We add a right-angled triangle
in. And the included angle once again
is 45 degrees. I’ve called the component of this
force that’s parallel to the plane 𝑦 or 𝑦 newtons. Once again, we have the adjacent
and hypotenuse. So we’re going to use the cosine
ratio. cos of 45 is 𝑦 divided by 16𝑔. And if we multiply by 16𝑔, we get
16𝑔 times cos of 45.
We’re now going to use the fact
that 𝑔 is 9.8. Using this along the side the fact
that cos of 45 is root two over two and we find 𝑦 is 78.4 root two. We now have forces acting parallel
to the plane. We need to find the sum of these
forces. This one acts in the same direction
as the acceleration force, so we’re going to assume it’s positive. This one however acts in the
opposite direction. It’s acting parallel to and down
the plane. So its force is negative, meaning
the sum of our forces is 235.2 root two minus 78.4 root two. This of course is equal to mass
times acceleration. Remember, we’re using this
formula. The mass of the body is 16
kilograms, and the acceleration we said was equal to 𝑎.
So we have an equation: 235.2 root
two minus 78.4 root two equals 16𝑎. This left-hand side simplifies to
156.8 root two. So to solve our equation, our last
step is to divide through by 16. 156.8 divided by 16 is 49 over
five. So we find our acceleration 𝑎 is
equal to 49 root two over five or 49 root two over five meters per square
second. Note, now, that had we modeled the
acceleration to be acting in the opposite direction — in other words, down the
plane — we would have received a result of negative 49 root two over five. The magnitude of the acceleration
though would have still been 49 root two over five. This just tells us that it’s
decelerating in this direction.
