Video: Finding the Vector of an Unknown Force Parallel to a Given Force given the Resultant Vector and Finding Its Position

Two parallel forces 𝐹₁ and 𝐹₂ are acting at two points 𝐴 and 𝐡 respectively in a perpendicular direction on line segment 𝐴𝐡, 𝐴𝐡 = 10 cm. Their resultant 𝑅 = βˆ’20𝑖 βˆ’ 16𝑗 is acting at the point 𝐢 that belongs to line segment 𝐴𝐡. Given that 𝐹₂ = βˆ’30𝑖 βˆ’ 24𝑗, determine 𝐹₁ and the length of 𝐡𝐢.

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Video Transcript

Two parallel forces, 𝐹 one and 𝐹 two, are acting at two points, 𝐴 and 𝐡, respectively, in a perpendicular direction on line segment 𝐴𝐡. 𝐴𝐡 equals 10 centimeters. Their resultant, 𝑅 equals negative 20𝑖 minus 16𝑗, is acting at the point 𝐢 that belongs to line segment 𝐴𝐡. Given that 𝐹 two equals negative 30𝑖 minus 24𝑗 determine 𝐹 one and the length of 𝐡𝐢.

In this statement, we’re told the length of line segment 𝐴𝐡, 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 𝐡𝐢.

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 𝐴𝐡, 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 𝐡.

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 𝐡𝐢, where 𝐢 is some distance away from 𝐡 such that it’s the location of the line of action of the resultant force 𝑅.

In order to find that distance 𝐡𝐢, 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 𝐴𝐢 must be equal to three times the length of line segment 𝐡𝐢. That’s the condition we must meet in order for 𝐢 to be the line of action of these two forces.

Now we know that 𝐴𝐢 is equal to 𝐴𝐡 plus 𝐡𝐢. And we’ve written above that that must be equal to three times 𝐡𝐢. So now we want to solve this equation for 𝐡𝐢. It simplifies to two 𝐡𝐢 equals 𝐴𝐡 or 𝐡𝐢 equals 𝐴𝐡 divided by two.

We know that 𝐴𝐡 is equal to 10 centimeters. 𝐡𝐢 therefore must be equal to five centimeters. That’s the distance from point 𝐡 to the line of action of these two forces.

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