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The forces π
one equals two π’ plus two π£, π
two equals ππ’ plus nine π£, and π
three equals nine π’ plus ππ£ act on a particle, where π’ and π£ are two perpendicular unit vectors.
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Given the forcesβ resultant π equal to two π’ minus six π£, determine the values of π and π.
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Letβs begin by recalling what we actually mean by a resultant force.
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The resultant force is the overall force acting on the object.
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Here, itβs the sum of π
one, π
two, and π
three.
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Now, weβre actually given that the resultant of these forces is two π’ minus six π£.
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Well, since the resultant is the sum of the three forces, we can say that π
sub one plus π
sub two plus π
sub three must be equal to π.
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Weβre going to replace each force as given in the question.
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And we see that two π’ plus two π£ plus ππ’ plus nine π£ plus nine π’ plus ππ£ must be equal to two π’ minus six π£.
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Next, weβre going to collect together the π’-components and, separately, the π£-components of each force on the left-hand side.
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The π’-components are two, π, and nine.
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And the π£-components are two, nine, and π.
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And so, we can say that two plus π plus nine π’ plus two plus nine plus π π£ must be equal to two π’ minus six π£.
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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.
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Equating the components in the π’-direction, and we get two plus π plus nine equals two.
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Similarly in the π£-direction, our equation is two plus nine plus π equals negative six.
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We can simplify our first equation, and we get π plus 11 equals two.
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Then, we solve for π by subtracting 11 from both sides.
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And we find π to be equal to negative nine.
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Similarly, our other equation becomes 11 plus π equals negative six.
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To solve this equation for π, weβre going to subtract 11 from both sides.
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And we get π equals negative 17.
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So, given the information about our three forces and their resultant, we found π is equal to negative nine and π is equal to negative 17.
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And in fact, we could check our answers by substituting π equals negative nine and π equals negative 17 back into our original forces.
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Doing that and then finding the sum of the three forces, we do indeed find that itβs equal to two π’ minus six π£, which we saw is equal to π.