Video Transcript
𝐴𝐵 is a rod of length 114
centimeters and negligible weight. Forces of magnitudes 83 newtons,
225 newtons, 163 newtons, and 136 newtons are acting on the rod as shown in the
following figure. 𝐶 and 𝐷 are the points of
trisection of 𝐴𝐵, and point 𝑂 is the midpoint of the rod. Find the algebraic sum of the
moments of these forces about point 𝑂.
We are asked to find the sum of the
moments of the four forces about point 𝑂. In order to do this, we firstly
need to calculate the distances between each point on the rod. We are told that 𝐴𝐵 has length
114 centimeters. 𝐶 and 𝐷 are the points of
trisection of 𝐴𝐵. And this means that 𝐵𝐷 is equal
to 𝐷𝐶, which is equal to 𝐶𝐴. All three of these will be equal to
114 centimeters divided by three. This is equal to 38
centimeters. The lengths 𝐵𝐷, 𝐷𝐶, and 𝐶𝐴
are all equal to 38 centimeters.
We are told that 𝑂 is the midpoint
of the rod, which means that 𝐵𝑂 is equal to 𝑂𝐴. These lengths are equal to 114
divided by two, which is equal to 57. The lengths 𝐵𝑂 and 𝑂𝐴 are equal
to 57 centimeters. As the lines of action of all four
forces are acting perpendicular to the rod, we can calculate the moment of each
force using the equation 𝑚 is equal to 𝐅 multiplied by 𝐷, where 𝐅 is the
magnitude of the force measured in newtons and 𝐷 is the perpendicular distance to
the point at which we are taking moments, in this question, measured in
centimeters. This means that the units for each
moment will be newton-centimeters.
We can take moments about any point
on the rod. However, in this question, we are
asked to find the moment of the forces about point 𝑂. We can find the sum of the moments
by firstly finding the moment of each force at points 𝐴, 𝐵, 𝐶, and 𝐷. We are told that moments acting in
the counterclockwise direction will be positive. From the diagram, it is clear that
this is true for the 83- and 163-newton forces at points 𝐴 and 𝐷,
respectively. As these produce counterclockwise
moments about 𝑂, the moment will be positive. The forces acting at point 𝐵 and
𝐶 produce clockwise moments about 𝑂. This means that the moments of the
136- and 225-newton forces will be negative.
Let’s begin by calculating the
moment of force 𝐴. This is equal to 83 newtons
multiplied by a perpendicular distance of 57 centimeters, which is equal to 4731
newton-centimeters. As already mentioned, the moment of
the force acting at 𝐵 will be negative. This is equal to negative 136
multiplied by 57, which is equal to negative 7752 newton-centimeters. The moment of the force acting at
point 𝐶 is also negative. It is equal to negative 225
multiplied by 19, as the distance 𝑂𝐶 is 19 centimeters. The moment here is equal to
negative 4275 newton-centimeters.
Finally, we have the moment of the
force acting at 𝐷 which is equal to 163 multiplied by 19. This is equal to 3097
newton-centimeters. The sum of the moments is therefore
equal to 4731 minus 7752 minus 4275 plus 3097. This is equal to negative 4199. The algebraic sum of the moments of
the four forces about point 𝑂 is negative 4199 newton-centimeters.
