Two parallel forces, 𝐹 one and 𝐹 two, are acting at two points, 𝐴 and 𝐵, respectively, in a perpendicular direction on line segment 𝐴𝐵. 𝐴𝐵 equals 10 centimeters. Their resultant, 𝑅 equals negative 20𝑖 minus 16𝑗, is acting at the point 𝐶 that belongs to line segment 𝐴𝐵. Given that 𝐹 two equals negative 30𝑖 minus 24𝑗 determine 𝐹 one and the length of 𝐵𝐶.
In this statement, we’re told the length of line segment 𝐴𝐵, 10 centimeters. We’re also told the vector 𝑅 resulting from the addition of 𝐹 one and 𝐹 two. And we’re told the components of 𝐹 two. We want to determine 𝐹 one as well as the length of line segment 𝐵𝐶.
To start off, let’s draw a diagram of these two forces, 𝐹 one and 𝐹 two, and the line that they’re acting on. This line we’ve drawn is line segment 𝐴𝐵, which we’re told has a length of 10 centimeters. 𝐹 one acts perpendicular to that line segment at point 𝐴. And 𝐹 two acts in the opposite direction at point 𝐵.
To solve for 𝐹 one, the components of that force, we can recall that 𝑅, the resultant force, is equal to 𝐹 one plus 𝐹 two. If we wrote out this equation by the components of 𝑅, 𝐹 one, and 𝐹 two, it would look like this. The 𝑖th component of 𝐹 one minus 30 the 𝑖th component of 𝐹 two is equal to the 𝑖th component of 𝑅 negative 20. This must mean that 𝐹 sub one 𝑖 is equal to positive 10. That’s the only value that makes this sum work.
When we look at the 𝑗 component, 𝐹 sub one 𝑗 minus 24 the 𝑗th component of 𝐹 two is equal to negative 16 the 𝑗th component of 𝑅. This sum implies that 𝐹 sub one 𝑗 is equal to positive eight.
Combining these components, we can write that 𝐹 one equals 10𝑖 plus eight 𝑗. Those are the components of the force acting at point 𝐴. Next, we wanna solve for the distance of the line segment 𝐵𝐶, where 𝐶 is some distance away from 𝐵 such that it’s the location of the line of action of the resultant force 𝑅.
In order to find that distance 𝐵𝐶, we can consider the magnitude of 𝐹 two to 𝐹 one, that is, the ratio of those two forces. If we look at the 𝑖th component of 𝐹 one and 𝐹 two, we see that the 𝑖th component of 𝐹 two is three times greater in magnitude than that of 𝐹 one, and likewise with the 𝑗th component. The 𝑗th component of 𝐹 two is three times greater than that of 𝐹 one in magnitude.
This tells us that the length of line segment 𝐴𝐶 must be equal to three times the length of line segment 𝐵𝐶. That’s the condition we must meet in order for 𝐶 to be the line of action of these two forces.
Now we know that 𝐴𝐶 is equal to 𝐴𝐵 plus 𝐵𝐶. And we’ve written above that that must be equal to three times 𝐵𝐶. So now we want to solve this equation for 𝐵𝐶. It simplifies to two 𝐵𝐶 equals 𝐴𝐵 or 𝐵𝐶 equals 𝐴𝐵 divided by two.
We know that 𝐴𝐵 is equal to 10 centimeters. 𝐵𝐶 therefore must be equal to five centimeters. That’s the distance from point 𝐵 to the line of action of these two forces.