Video: Finding the Unknown Components of Two Forces given the Components of the Resultant

The forces 𝐅₁ = 2𝐒 + 2𝐣, 𝐅₂ = π‘Žπ’ + 9𝐣, and 𝐅₃ = 9𝐒 + 𝑏𝐣 act on a particle, where 𝐒 and 𝐣 are two perpendicular unit vectors. Given the forces’ resultant 𝐑 = 2𝐒 βˆ’ 6𝐣, determine the values of π‘Ž and 𝑏.

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Video Transcript

The forces 𝐅 one equals two 𝐒 plus two 𝐣, 𝐅 two equals π‘Žπ’ plus nine 𝐣, and 𝐅 three equals nine 𝐒 plus 𝑏𝐣 act on a particle, where 𝐒 and 𝐣 are two perpendicular unit vectors. Given the forces’ resultant 𝐑 equal to two 𝐒 minus six 𝐣, determine the values of π‘Ž and 𝑏.

Let’s begin by recalling what we actually mean by a resultant force. The resultant force is the overall force acting on the object. Here, it’s the sum of 𝐅 one, 𝐅 two, and 𝐅 three. Now, we’re actually given that the resultant of these forces is two 𝐒 minus six 𝐣. Well, since the resultant is the sum of the three forces, we can say that 𝐅 sub one plus 𝐅 sub two plus 𝐅 sub three must be equal to 𝐑. We’re going to replace each force as given in the question. And we see that two 𝐒 plus two 𝐣 plus π‘Žπ’ plus nine 𝐣 plus nine 𝐒 plus 𝑏𝐣 must be equal to two 𝐒 minus six 𝐣.

Next, we’re going to collect together the 𝐒-components and, separately, the 𝐣-components of each force on the left-hand side. The 𝐒-components are two, π‘Ž, and nine. And the 𝐣-components are two, nine, and 𝑏. And so, we can say that two plus π‘Ž plus nine 𝐒 plus two plus nine plus 𝑏 𝐣 must be equal to two 𝐒 minus six 𝐣. Now, of course, for the vector on the left-hand side to be equal to the vector on the right-hand side of our equation, we know that the individual components must themselves be equal. Equating the components in the 𝐒-direction, and we get two plus π‘Ž plus nine equals two.

Similarly in the 𝐣-direction, our equation is two plus nine plus 𝑏 equals negative six. We can simplify our first equation, and we get π‘Ž plus 11 equals two. Then, we solve for π‘Ž by subtracting 11 from both sides. And we find π‘Ž to be equal to negative nine. Similarly, our other equation becomes 11 plus 𝑏 equals negative six. To solve this equation for 𝑏, we’re going to subtract 11 from both sides. And we get 𝑏 equals negative 17. So, given the information about our three forces and their resultant, we found π‘Ž is equal to negative nine and 𝑏 is equal to negative 17.

And in fact, we could check our answers by substituting π‘Ž equals negative nine and 𝑏 equals negative 17 back into our original forces. Doing that and then finding the sum of the three forces, we do indeed find that it’s equal to two 𝐒 minus six 𝐣, which we saw is equal to 𝐑.

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