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.
