Two parallel forces 𝐹 one and 𝐹 two have the same direction, and the distance between their lines of action is 90 centimeters. Given that the magnitude of their resultant is 49 newtons and it is 60 centimeters away from 𝐹 two, find the magnitudes of the two forces, rounding your answer to two decimal places.
Okay, so in this example, we have these two forces 𝐹 one and 𝐹 two. And we know that they’re parallel to one another. Here, we’ve drawn 𝐹 one as having a shorter length and therefore a shorter magnitude than 𝐹 two. But strictly speaking, we don’t know that. We don’t know which of these two forces is greater, or they might have the same magnitude, but just that they act in the same direction and their lines of action are separated by 90 centimeters. Along with this, we’re told that their resultant, the sum of these forces, is 49 newtons. And if we were to draw this resultant, we’ll call it 𝑅, in our sketch, we also know that its line of action is 60 centimeters away from the line of action of 𝐹 two.
Given all this, we want to find the magnitudes of 𝐹 one and 𝐹 two. Because there are two unknowns to solve for, we’ll need two independent equations. We have one such equation here, and we’ll need to find one more.
This first equation, we can say, has to do with the linear sum of our forces. It’s possible though to also consider the rotational effects of these forces, in other words, the moments they create. In general, any force can create a moment 𝑀 about some point called the axis of rotation. The magnitude of that moment equals the component of the force that is perpendicular to a distance between where the force is applied and to that axis.
In a given system, we can choose an axis of rotation to be at any point. For the case of these two parallel forces, let’s say that we locate our axis of rotation right here at the base of the resultant force 𝑅. Because 𝑅 is the resultant force, we can say that 𝑅 times the perpendicular distance from this force’s line of action to our axis of rotation is equal to the sum of the moments about that same point created by 𝐹 one and 𝐹 two.
If we set up the convention that a counterclockwise moment is positive and therefore a clockwise one is negative, then we can say that our resultant force 𝑅 multiplied by the perpendicular distance between this force’s line of action and our axis of rotation is equal to 𝐹 two times 60 minus 𝐹 one times 30. The reason this value in the left zero is because the line of action of 𝑅 passes through our axis of rotation. We can imagine that axis coming into and out of the screen at this X.
The whole left-hand side of this equation then is equal to zero. On the right, we have 𝐹 two multiplied by the perpendicular distance between this force’s line of action and our axis — and note that this moment is positive because it is counterclockwise around our axis — minus 𝐹 one times the perpendicular distance from this force’s line of action to our axis of rotation. Because 𝐹 one tends to create a clockwise rotation about our axis, its moment is negative. This then is our second independent equation involving our two unknowns. If we add 𝐹 one times 30 to both sides of the equation, then we have 30𝐹 one equaling 60𝐹 two. Then, dividing both sides of the equation by 30, we find that 𝐹 one equals 60 over 30 or two times 𝐹 two.
Looking back at our original sketch, we can now see that the relative sizes of our two force vectors aren’t correct. 𝐹 one should actually be twice as long as 𝐹 two, like this. In any case, now that we know that 𝐹 one equals two 𝐹 two, we can replace 𝐹 one in this equation with two 𝐹 two. And so we find that three 𝐹 two equals 49, or 𝐹 two equals 49 over three. And then since 𝐹 one is twice this value, we can write that it’s two times 49 over three. And these are the magnitudes of our two forces.
And before we finish, we want to round these answers to two decimal places. Entering these two fractions on our calculator, to two decimal places, 𝐹 two equals 16.33, while 𝐹 one equals 32.67. And both of these forces have units of newtons. Our final answer then is that the magnitude of 𝐹 one is 32.67 newtons and the magnitude of 𝐹 two is 16.33 newtons.