Video Transcript
A uniform ladder having a weight of
140 newtons and a length of seven meters rests with one end 𝐴 on smooth horizontal
ground and its other end 𝐵 on a smooth vertical wall. The ladder is kept in equilibrium
by attaching its end 𝐴 to a string fixed to the junction of the wall and the ground
vertically below 𝐵. The ladder is inclined to the
horizontal at an angle of 45 degrees. If a man whose mass is 85 kilograms
climbs the ladder to the point 4.9 meters away from end 𝐴, find the tension in the
string rounded to two decimal place. Use the value 𝑔 equals 9.8 meters
per square second for gravitational acceleration.
Before we perform any calculations
here, let’s begin by drawing a free body diagram. Remember, this is a very simple
sketch of the scenario showing all relevant forces. Here is our ladder inclined to the
horizontal at an angle of 45 degrees. The ladder is uniform, and it has a
weight of 140 newtons. This means the weight force can be
considered to be acting exactly halfway along the ladder. Since the ladder is seven meters,
the force of the weight must act at 3.5 meters from either end. We’re then told that the ladder is
kept in place by attaching end 𝐴 to a piece of string. And so we add a tensional force
acting to the right. It keeps the ladder in place.
Now Newton’s third law of motion
tells us that since the ladder will exert a force on both the ground and the wall,
there will be normal reaction forces of the ground and the wall on the ladder. We’re going to call them 𝑅 sub 𝐴
and 𝑅 sub 𝐵. Note that there are no frictional
forces involved here. And that’s because we’re told both
the ground and the wall are smooth. There’s one final force we need to
consider. A man whose mass is 85 kilograms
stands on the ladder at point 4.9 meters away from the base 𝐴. The downward force of his weight by
Newton’s second law of motion is mass times gravity; that’s 85𝑔. And now we have all of our
forces. And we see that we’re looking to
find the value of tension in the string. So let’s clear some space so we can
perform some calculations.
Now we were told the body was in
equilibrium. And so there are two things that we
can consider. Firstly, for a rigid body to be in
equilibrium, the sum of all forces acting on the body must be equal to zero. We often think about these in terms
of the vertical forces and the horizontal forces. Similarly, the sum of all moments
that act on the body must also be equal to zero. We think about moments in terms of
a counterclockwise and a clockwise direction. And of course, we calculate a
moment by multiplying the force by the perpendicular distance of the point of action
of the force to the point about which the body is trying to turn. And so we’re going to begin by
using the fact that the sum of the forces must be equal to zero.
We could do this both horizontally
and vertically, but actually we’re trying to find the value of 𝑇. And we’re really not interested in
𝑅 sub 𝐴. So we’re only going to perform this
horizontally. We’re going to take the direction
to the right, the direction in which the tension force is acting to be positive. And so the sum of the forces acting
in this direction must be 𝑇 minus 𝑅 sub 𝐵. We said the sum of these forces is
zero. So our first equation is 𝑇 minus
𝑅 sub 𝐵 equals zero, which we could alternatively write as 𝑇 equals 𝑅 sub
𝐵. Now that we’ve dealt with the
forces in the horizontal direction, we’re going to deal with the moments of the
forces about a point.
Now we can take moments about any
point, but there are a lot more forces acting at point 𝐴 than there are at point
𝐵. So we’re going to take moments
about 𝐴 with counterclockwise being a positive direction. There are three forces that we’re
interested in. Those are the 140-newton force, the
85𝑔 force, and 𝑅 sub 𝐵. But remember, moments are
calculated by multiplying the force by the perpendicular distance of the line of
action of this force to the point about which the body is trying to rotate. And so we’re going to need to
resolve each of these forces to find the component of them that acts perpendicular
to the body, perpendicular to the ladder.
So we draw a right-angled triangle
to each of these forces, with an included angle of 45 degrees. We want to find the values of 𝑥,
𝑦, and 𝑧. These are the components of these
forces that act perpendicular to the ladder. Let’s begin with 𝑥. That’s the component of the weight
of the ladder. We have the hypotenuse in this
triangle, and we’re trying to find the adjacent. So we’re going to use the cosine
ratio. When we do, we find that cos of 45
is 𝑥 over 140, and so 𝑥 must be equal to 140 times cos of 45 degrees. Now, since cos of 45 is actually
root two over two, this becomes a 140 times root two over two which is 70 root
two. And so the component of the weight
of the ladder that acts perpendicular to the ladder is 70 root two newtons.
This force is trying to turn the
ladder in a clockwise direction, and so its moment is going to be negative. It’s force times distance, which is
70 root two times 3.5. And so the moment is negative 70
root two times 3.5. We’re going to repeat this process
to find the value of 𝑦. Once again, we use the cosine
ratio. But this time, our equation is cos
45 equals 𝑦 over 85𝑔. And so 𝑦 is 85 𝑔 times cos of
45. We’re going to use the fact that 𝑔
is 9.8. And we find this is equal to 416.5
root two. Once again, this force is trying to
rotate the ladder in a clockwise direction, and so its moment is going to be
negative. It’s acting at a point 4.9 meters
away from 𝐴, and so its moment is negative 416.5 root two times 4.9.
There’s one more moment that we’re
interested in. And to find it, we’re going to need
to calculate the component of 𝑅 sub 𝐵 that acts perpendicular to the ladder. This time, we’re going to use the
sine ratio. We know that the hypotenuse of this
is 𝑅 sub 𝐵. And we’re trying to find its
opposite. sin of 45 then is 𝑧 over 𝑅 sub
𝐵, meaning that 𝑧 is 𝑅 sub 𝐵 times sin of 45 degrees. Now in fact remember, we said that
𝑇 was equal to 𝑅 sub 𝐵. We also know that sin of 45 is root
two over two, so we can write 𝑅 sub 𝐵 times sin of 45 as root two over two 𝑇. The moment this time is going to be
positive, since this force is acting in the counterclockwise direction. And so we get root two over two 𝑇
times seven since that’s the length of the ladder.
We know that the sum of these
moments is equal to zero. So we simplify this expression and
then set it equal to zero. We get negative 2285.85 root two
plus 3.5 root two 𝑇 equals zero. We want to solve for 𝑇, but before
we do, we’re going to divide through by root two. We then add 2285.85 to both sides,
and our last job is to divide through by 3.5. That’s 653.1 or, correct to two
decimal places, 653.10. And so we see the tension is 653.10
newtons.
