Video: Finding the Magnitude of the Sum of the Moments of Four Parallel Forces Acting along a Rod

Video Transcript

𝐴𝐵 is a rod having a length of 105 centimetres and negligible weight. Forces of magnitudes 240 newtons, 67 newtons, 115 newtons, and 176 newtons are acting on the rod as shown in the figure. Given that 𝐶 and 𝐷 are at the trisection of 𝐴𝐵, determine the algebraic sum of the moments of these forces about point 𝐴.

We’re told that the length of the rod is 105 centimetres and that points 𝐶 and 𝐷 are at the trisection. 105 divided by three is equal to 35. Therefore, the distance between 𝐵 and 𝐷, 𝐷 and 𝐶, and 𝐶 and 𝐴 is 35 centimetres. The moment of a force is equal to the force multiplied by the distance from some point. In this question, we’re asked to take moments about the point 𝐴. The moment of force 𝐴 about this point will be equal to 214 multiplied by zero. This is equal to zero.

The moment of force 𝐵 will be equal to 176 multiplied by 105 as it is 105 centimetres away from 𝐴. This is equal to 18480. The force at point 𝐶 is 67 newtons, and this is 35 centimetres from point 𝐴. 67 multiplied by 35 is equal to 2345. Finally, the force at 𝐷 is 115 newtons. This is 70 centimetres away from point 𝐴. 115 multiplied by 70 is equal to 8050.

We’re told that the forces moving in the anticlockwise direction around point 𝐴 are positive, and those moving in a clockwise direction are negative. Any force moving vertically downwards will move in an anticlockwise direction around point 𝐴. Therefore, the moment at 𝐵 is positive. As the forces at 𝐶 and 𝐷 are acting vertically upwards, these are moving in a clockwise direction. Therefore, the moment will be negative.

The sum of the moments about point 𝐴 are therefore equal to 18480 minus 2345 minus 8050. This is equal to 8085. As the forces are measured in newtons and lengths in centimetres, our units will be newton centimetres. The correct answer is 8085 newton centimetres.