Video: Finding the Magnitude and the Position of the Resultant of Five Coplanar Parallel Forces

Points 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸 are lying on the same straight line, such that 𝐴𝐡 = 8 cm, 𝐡𝐢 = 18 cm, 𝐢𝐷 = 12 cm, and 𝐷𝐸 = 11 cm. Five forces of magnitudes 40, 25, 20, 45, and 50 newtons are acting as shown in the figure. Determine their resultant 𝑅 and the distance π‘₯ between its line of action and point 𝐴.

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Video Transcript

Points 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸 are lying on the same straight line, such that 𝐴𝐡 equals eight centimeters, 𝐡𝐢 equals 18 centimeters, 𝐢𝐷 equals 12 centimeters, and 𝐷𝐸 equals 11 centimeters. Five forces of magnitudes 40, 25, 20, 45, and 50 newtons are acting as shown in the figure. Determine their resultant 𝑅 and the distance π‘₯ between its line of action and point 𝐴.

In our figure, we see these five points 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸 all along the same line. At each point, there’s a force of given magnitude and direction acting. The first thing we want to do here is determine the resultant 𝑅 of these five forces. This equals the algebraic sum of all five, and we’ll add them together according to this sign convention here; forces pointing upward are positive. So, that means these two forces must be negative. The resultant 𝑅 then equals 40 plus 25 minus 20 minus 45 plus 50, and this equals 50. All of our forces are in units of newtons, so 𝑅 is as well. We’ve got then the answer to the first part of our question: 𝑅 equals 50 newtons.

Now, what we want to do is determine the distance π‘₯ between the line of action of this resultant force and point 𝐴 on our diagram. The idea here is that our resultant force effectively acts somewhere between the points 𝐴 and 𝐸 on our line. We can pick a point effectively at random and say that 𝑅 acts from here. Given this, our problem statement calls the distance between the line of action of 𝑅 and point 𝐴 π‘₯; it’s this value we want to solve for. To start doing that, let’s mark out the distances we’re given between the adjacent points on this line. Leaving out units, 𝐴𝐡 equals eight centimeters, 𝐡𝐢 equals 18, 𝐢𝐷 is 12 centimeters, and 𝐷𝐸 is 11.

Knowing this, let’s clear some space on screen to work, and we can note that the moment about point 𝐴 created by our resultant force acting a distance π‘₯ away from point 𝐴 is equal to the sum of the moments due to all the other forces acting along this line. Now, in general, the moment created by a force is equal to the component of that force that’s perpendicular to the distance between where the force is applied in a given axis of rotation. In our case, our axis of rotation is point 𝐴. As we start to calculate moments about this point, let’s consider moments in the counterclockwise direction to be positive. Therefore, moments in a clockwise direction are negative.

We can write as a governing equation that the moment 𝑀 sub 𝑅 created by our resultant force as it acts a distance π‘₯ from point 𝐴 is equal to the sum of all the other moments about that same point, point 𝐴. Now, the moment due to the resultant force 𝑅 is equal to that force multiplied by π‘₯. This then is equal to the moment due to the force at point 𝐴 plus the moment due to the force at point 𝐡 plus, and so on, all the way up to the moment due to the force at point 𝐸. When we consider the moment due to the force at point 𝐴, because the line of action of this force passes through point 𝐴, there’s no perpendicular distance between the line of action of the force and our axis of rotation.

Therefore, even though there’s a nonzero force acting at point 𝐴, the moment about point 𝐴 that that force creates is zero. On to the moment due to the force at point 𝐡, this equals the force of 25 newtons times the distance of eight centimeters. And this moment will be positive because it creates a counterclockwise moment about point 𝐴. Moving on to 𝑀 sub 𝐢, this moment will be negative, because it’s clockwise about point 𝐴, times the force of 20 newtons multiplied by a distance of 18 centimeters plus eight. The moment due to the force at point 𝐷 is likewise negative, and its magnitude is 45 times eight plus 18 plus 12.

Finally, there’s the moment due to the force at point 𝐸, which is positive. It equals 50 times eight plus 18 plus 12 plus 11. If we add up everything on the right-hand side of this expression, we get positive 420. We can say then that π‘₯, the distance we want to solve for, equals 420 over 𝑅. 𝑅 equals 50 newtons, and 420 divided by 50 equals 8.4. Just as with the rest of the distances in this example, the units of this distance are centimeters. So, the resultant 𝑅 of all these forces is 50 newtons. And it effectively acts a distance of 8.4 centimeters away from point 𝐴.

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