Video Transcript
The length of a rod 𝐴𝐵 is 111
centimeters, and its weight is 95 newtons, which is acting at its midpoint. The rod is resting horizontally on
two supports, where one of them is at end 𝐴 and the other is at a point 𝐶, which
is 30 centimeters away from 𝐵. A weight of 71 newtons is suspended
from the rod at a point that is nine centimeters away from 𝐵. Find the magnitude of weight 𝑤
that should be suspended from end 𝐵 so that the rod is about to rotate, and
determine the value of the pressure 𝑃 exerted on 𝐶 in that situation.
Let’s begin by drawing a free body
diagram showing this scenario. Here is our rod. Now, the downwards force of its
weight acts at its midpoint. So that’s 55.5 centimeters from
either end. Now, we also have this 71-newton
force. Now, that’s acting downwards at a
point nine centimeters away from 𝐵. We want to add a weight 𝑤 at point
𝐵. So we add that to the diagram and
two further forces. Those are the reaction force of the
support on the rod. We’ll call them 𝑅 sub 𝐴 and 𝑅
sub 𝐵, respectively.
Now, the rod is about to rotate, so
it’s essentially in limiting equilibrium. So we can say two things. Firstly, the sum of all the forces
acting on the body is zero. And secondly, the sum of the
moments is also equal to zero. So we’ll begin by considering the
sum of the forces. We’re going to take the direction
in which the reaction forces are acting as being positive.
And so this means that the
95-newton force, the 71-newton, and the weight force 𝑤 must be acting in the
opposite direction. And so we can therefore say that
the sum of the forces, which we know is equal to zero, is 𝑅 sub 𝐴 plus 𝑅 sub 𝐶
minus 95 minus 71 minus 𝑤 equals zero. And if we add 95 and 71 to both
sides of this equation, we get 𝑅 sub 𝐴 plus 𝑅 sub 𝐶 minus 𝑤 equals 166.
There’s not a lot more we can do
with this. And so we move on to the next bit
of information. And this is the sum of the moments
of our forces is also equal to zero. Let’s clear some space and find a
point about which we want to take moments.
Now, since we have this unknown
force 𝑤 at 𝐵, let’s take moments about 𝐵. And of course we define a direction
to be positive. This time, let’s choose
counterclockwise. We begin by thinking about the
reaction force at 𝐴. We know that this is trying to move
the body in a clockwise direction. And so its moment is going to be
negative. It’s negative 𝑅 sub 𝐴 times
111.
The moment of the weight force is
acting in a counterclockwise direction, so it’s positive. It’s 95 times 55.5. We then have negative 𝑅 sub 𝐶
times 30. That’s the moment of the reaction
force at 𝐶. And then we have 71 times nine. That’s the moment of this 71-newton
force. The sum of these moments, which
we’re of course measuring in newton centimeters here, is equal to zero. And this simplifies as shown.
Now, there’s one bit of information
that we haven’t yet used. And that is that the rod is on the
point of rotating. Now, since we’re adding a weight at
𝐵, we know the object is going to be rotating about 𝐶. And so this means that, at this
exact moment, there must be no downwards force of the rod on point 𝐴. And so, similarly, the reaction
force at 𝐴 must also be equal to zero. So we can rewrite our equation by
letting 𝑅 sub 𝐴 be equal to zero. And we’ll solve by adding 30𝑅 sub
𝐶 to both sides. So 30𝑅 sub 𝐶 is 5911.5. And when we divide both sides of
this by 30, we get 𝑅 sub 𝐶 equals 197.05. But how does this help?
We were trying to find the value of
the pressure 𝑃. But of course pressure is the
amount of force acting on an area. So since this force is acting on a
point, we can say that 𝑅 sub 𝐶 must actually be equal to 𝑃. And so 𝑃 must be equal to 197.05
newtons. Let’s clear some space and use this
information in our first equation here. Using 𝑅 sub 𝐴 equals zero and 𝑅
sub 𝐶 as 197.05, our equation is as shown. We add 𝑤 to both sides and then
subtract 166. And so we get that 𝑤 is 31.05 or
31.05 newtons.
