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If the force πΉ, where πΉ equals negative two π plus πΏ π minus nine π, is acting on the point π΄ β four, five, negative two β and the moment π sub π΅ of the force about the point π΅ β negative four, negative four, three β is negative 91π plus 82π plus two π, determine the value of πΏ.
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Given the force πΉ and also given the coordinates of points π΄ and π΅ and, finally, given the moment π sub π΅ of the force about the point π΅, we want to solve for the value of πΏ given in the equation for πΉ.
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Weβll start off by drawing in the points π΄ and π΅ on a set of coordinate axes.
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With our π-, π-, and π-directions written in, we can locate points π΄ and π΅ on this graph.
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Point π΄ has coordinates π equals four π, equals five, and π equals negative two.
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And point π΅ has coordinates π equals negative four, π equals negative four, and π equals three.
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The force πΉ acts from point π΄.
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Weβre not entirely sure what direction because we donβt know itβs π-component.
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If we connect points π΄ and π΅ by a vector, we can call that vector π
.
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Itβs the distance vector between the point π΄ at which the force acts and the point π΅ at which our moment is measured.
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Based on our diagram, we can say that π
is equal to π΄ minus π΅.
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And when we enter in the coordinates of those two points and calculate π
, we find itβs a vector with components eight π, nine π, and negative five π.
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Now weβll record π
, πΉ, and π sub π΅ off to the side so we can begin to solve for πΏ.
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Now that we know the displacement vector π
, the force πΉ, and the moment about point π΅, π sub π΅, we can recall that this moment is equal to the cross product of π
and πΉ.
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This cross product is equal to the determinant of this matrix, with our unit vectors written in followed by the components of our two vectors.
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When we apply this relationship to our scenario, we see itβs the πΉ sub π-component we want to solve for.
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If we know that, then weβll know πΏ.
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We can figure out πΉ sub π or πΏ by focusing either on the π- or the π-component of π sub π΅.
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Just to choose one of those two ways, letβs focus on the π-component and let that lead us to the value of πΏ.
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The πth component of the moment, which we can call π sub π΅π, is equal to π
sub π sub times πΉ sub π minus π
sub π times πΉ sub π.
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And we can replace πΉ sub π with πΏ because thatβs what it is.
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Now the πth component of π sub π΅ is negative 91.
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Thatβs equal to the πth component of π
, which is positive nine, times the πth component of πΉ, which is negative nine, minus the πth component of π
, negative five, multiplied by πΏ.
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Negative nine times nine is negative 81.
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And adding 81 to both sides gives us the expression negative 10 is equal to five πΏ, or πΏ is equal to negative two.
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Thatβs the value of the πth component of πΉ.