𝐴𝐵 is a rod with 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 the point 𝑂.
Okay, so looking at our figure, we see this rod 𝐴𝐵 with point 𝐴 on the right end and 𝐵 on the left. We also see these four forces acting either vertically upward or vertically downward on this rod, and as well as this, the points of trisection of the rod, 𝐷 and 𝐶 — these are points that divide the rod into three even lengths — and the midpoint of the rod point 𝑂.
Putting our focus on this midpoint, we want to calculate the total moment about point 𝑂 due to these four forces. As we begin to do that, let’s recall the fact that the length of our rod, the length of line segment 𝐴𝐵, is 114 centimeters. And we’ll also say that we want to calculate the moment about point 𝑂. We’ll call this 𝑀 sub 𝑂.
We can now clear away our problem statement. And let’s recall the fact that when it comes to moments of force, the moment 𝑀 created by some force around a given point is equal to the component of that force perpendicular to the line between where the force is applied and the point of interest times the distance from that point of interest to the applied force. To calculate 𝑀 sub 𝑂 then, the net moment about point 𝑂 due to these four forces, we’ll add together the moment created by each one individually to get that total.
And notice that we have a sign convention for the moments that are generated by these forces. If a force tends to cause rotation in a counterclockwise direction about point 𝑂, then that moment is considered positive. So, for example, let’s consider the force applied at point 𝐴. This force will create a moment of 83 newtons multiplied by this distance 𝐴𝑂. And we see, because this will tend to create a counterclockwise rotation about point 𝑂, this individual moment will be positive.
To keep track of the moments created by these four forces, let’s solve for them individually. The moment created by this force — we’ll call it 𝑀 sub 83 — is equal to the force of 83 newtons times the length of this line segment 𝐴𝑂. We’re told that 𝑂 is the midpoint of line segment 𝐴𝐵, which has a length of 114 centimeters. 114 divided by two is 57, and so this is 𝑀 83, the moment due to the 83-newton force about point 𝑂.
Next, we’ll consider 𝑀 225 the moment due to the 225-newton force. This is equal to that force times the distance of the line segment 𝐶𝑂. If we recall that point 𝐶 is at one of the trisection points of our line segment 𝐴𝐵, then that indicates that this distance here is 114 centimeters divided by three. That’s 38 centimeters. And if we subtract that distance from the distance from point 𝐴 to point 𝑂, half the length of the rod, then we find that line segment 𝐶𝑂 equals half the length of the rod, 57 centimeters, minus 38 centimeters. That equals 19 centimeters.
We’re just about done with this moment. But notice that this force of 225 newtons would tend to create a clockwise rotation about point 𝑂. Therefore, we’ll add a negative sign to this moment. So while the 83-newton force created a positive moment, the 225-newton force creates a negative moment about point 𝑂.
Moving on to our next force, 163 newtons, this force creates a moment which equals that force magnitude times the length of the line segment 𝑂𝐷. And we can note that the length of this line segment is equal to that of 𝐶𝑂. That’s because point 𝐷, just like point 𝐶, is a trisection point of the line segment 𝐴𝐵. So line segment 𝑂𝐷 is 19 centimeters as well. And we see that this force tends to create a counterclockwise moment about point 𝑂 and therefore is positive. And then, lastly, we consider the moment created by our 136-newton force. This equals 136 newtons multiplied by the length of our line segment 𝑂𝐵. And that distance, we can recall, is equal to the length of 𝐴𝑂, 57 centimeters. Lastly, we consider what direction of rotation this force would tend to create around point 𝑂. We see that its moment would rotate in a clockwise direction and therefore is negative by our sign convention.
Alright, now that we’ve calculated each moment individually, let’s combine them to solve for the net moment about point 𝑂. 𝑀 sub 𝑂 is equal to 𝑀 83 plus 𝑀 225 plus 𝑀 163 plus 𝑀 136. And if we multiply and then combine all of these terms on our calculator, the answer we get is negative 4199 newton centimeters. This is the cumulative or net moment around point 𝑂.