Video Transcript
A body weighing 195 newtons is on a
rough plane that is inclined at an angle of 45 degrees to the horizontal. If the coefficient of friction
between the body and the plane is root three over three, which of the following is
true about the body? Is it (A) the body is in a stable
state of equilibrium, (B) the body is sliding down the plane, or (C) the body is on
the point of sliding down the plane?
And let’s begin by simply sketching
a diagram of all of our forces. Here is our plane, which makes an
angle of 45 degrees to the horizontal. A body rests on the plane that
weighs 195 newtons. That means the force that acts
vertically downwards on the plane is 195 newtons. We know this means that there will
be a reaction force. The reaction force acts
perpendicular to the plane in the opposite direction to the weight of the body.
Now, we’re told that the body rests
on a rough plane. This means there is a frictional
force that’s trying to stop the body from moving down the plane; let’s call that
𝐹r. But of course, we know that
friction is equal to 𝜇𝑅, where 𝜇 is the coefficient of friction and 𝑅 is the
reaction force of the plane on the body perpendicular to the plane. In this case, we know 𝜇 is equal
to root three over three. Our job is going to be to find
𝑅. And to do so, we add in this
right-angled triangle. By including this triangle, we’ll
be able to calculate the components for the weight that act parallel to the plane
and perpendicular to the plane.
To decide whether the body is
sliding down the plane or on the point of sliding down the plane or, in fact, in
equilibrium, we need to establish whether the force of the weight that acts parallel
to the plane is greater than or equal to the force of friction which acts in the
opposite direction. So, let’s resolve perpendicular to
the plane to find 𝑅.
We know that the body isn’t jumping
off the plane. And so, this means that the sum
total of the forces acting perpendicular to the plane is zero. Alternatively, we can say that the
reaction force is equal to the force of the weight that acts perpendicular to the
plane. But what is that force? Well, let’s label the part of the
right-angled triangle that we’re interested in; let’s label it 𝑥. Then, we can see we can use right
angle trigonometry to find the value of 𝑥. 𝑥 is the side adjacent to the
included angle of 45 degrees, whereas 𝐻 is the hypotenuse. We can, therefore, use the cos
ratio.
Remember, cos 𝜃 is adjacent over
hypotenuse. And we can say that cos of 45 is 𝑥
over 195. We rearrange by multiplying both
sides by 195. And we find 𝑥 is 195 cos of
45. This acts in the opposite direction
to the reaction force. So, we can say that 𝑅 minus 195
cos of 45 equals zero. We can add 195 cos of 45 to both
sides of our equation. So, 𝑅 is 195 cos of 45 or 195 root
two over two. And of course, this is newtons.
We now know what 𝑅 is. So, next, we resolve parallel to
the plane. Let’s begin by calculating the size
of the frictional force. Remember that’s 𝜇𝑅. We know that 𝜇 is the coefficient
of friction. It’s root three over three, and 𝑅
is 195 root two over two. Correct to one decimal place, that
gives us 79.6 newtons.
Our next job is to consider the
component of the weight that acts parallel to the slope. If we go back to our right-angled
triangle, that’s 𝑦. This time, that’s the side opposite
the included angle. By using the sine ratio, we see
that sin of 45 equals 𝑦 over 195. 𝑦 is equal to 195 sin 45 then. This time, correct to one decimal
place, we get 137.9 newtons. It should be quite clear to us that
137.9 newtons is greater than 79.6 newtons. This means the force acting down
and parallel to the plane is greater than the force acting up and parallel to the
plane.
Since this is the case, the body
can’t be in equilibrium or even at the point of sliding down the plane. It must be moving in the direction
of the greater force; it must be sliding down the plane. So, the answer is (B) the body is
sliding down the plane.
