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

# Question Video: Finding the Vector of an Unknown Force Parallel to a Given Force given the Resultant Vector and Finding Its Position Mathematics • Third Year of Secondary School

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Two parallel forces πΉβ and πΉβ are acting at two points π΄ and π΅ respectively in a perpendicular direction on line segment π΄π΅, π΄π΅ = 10 cm. Their resultant π = β20π β 16π is acting at the point πΆ that belongs to line segment π΄π΅. Given that πΉβ = β30π β 24π, determine πΉβ and the length of π΅πΆ.

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### Video Transcript

Two parallel forces, πΉ one and πΉ two, are acting at two points, π΄ and π΅, respectively, in a perpendicular direction on line segment π΄π΅. π΄π΅ equals 10 centimeters. Their resultant, π equals negative 20π minus 16π, is acting at the point πΆ that belongs to line segment π΄π΅. Given that πΉ two equals negative 30π minus 24π determine πΉ one and the length of π΅πΆ.

In this statement, weβre told the length of line segment π΄π΅, 10 centimeters. Weβre also told the vector π resulting from the addition of πΉ one and πΉ two. And weβre told the components of πΉ two. We want to determine πΉ one as well as the length of line segment π΅πΆ.

To start off, letβs draw a diagram of these two forces, πΉ one and πΉ two, and the line that theyβre acting on. This line weβve drawn is line segment π΄π΅, which weβre told has a length of 10 centimeters. πΉ one acts perpendicular to that line segment at point π΄. And πΉ two acts in the opposite direction at point π΅.

To solve for πΉ one, the components of that force, we can recall that π, the resultant force, is equal to πΉ one plus πΉ two. If we wrote out this equation by the components of π, πΉ one, and πΉ two, it would look like this. The πth component of πΉ one minus 30 the πth component of πΉ two is equal to the πth component of π negative 20. This must mean that πΉ sub one π is equal to positive 10. Thatβs the only value that makes this sum work.

When we look at the π component, πΉ sub one π minus 24 the πth component of πΉ two is equal to negative 16 the πth component of π. This sum implies that πΉ sub one π is equal to positive eight.

Combining these components, we can write that πΉ one equals 10π plus eight π. Those are the components of the force acting at point π΄. Next, we wanna solve for the distance of the line segment π΅πΆ, where πΆ is some distance away from π΅ such that itβs the location of the line of action of the resultant force π.

In order to find that distance π΅πΆ, we can consider the magnitude of πΉ two to πΉ one, that is, the ratio of those two forces. If we look at the πth component of πΉ one and πΉ two, we see that the πth component of πΉ two is three times greater in magnitude than that of πΉ one, and likewise with the πth component. The πth component of πΉ two is three times greater than that of πΉ one in magnitude.

This tells us that the length of line segment π΄πΆ must be equal to three times the length of line segment π΅πΆ. Thatβs the condition we must meet in order for πΆ to be the line of action of these two forces.

Now we know that π΄πΆ is equal to π΄π΅ plus π΅πΆ. And weβve written above that that must be equal to three times π΅πΆ. So now we want to solve this equation for π΅πΆ. It simplifies to two π΅πΆ equals π΄π΅ or π΅πΆ equals π΄π΅ divided by two.

We know that π΄π΅ is equal to 10 centimeters. π΅πΆ therefore must be equal to five centimeters. Thatβs the distance from point π΅ to the line of action of these two forces.

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