Question Video: Finding the Unknown Components of a Force Given Its Position Vector and the Moment Components about an Axis in Three Dimensions | Nagwa Question Video: Finding the Unknown Components of a Force Given Its Position Vector and the Moment Components about an Axis in Three Dimensions | Nagwa

# Question Video: Finding the Unknown Components of a Force Given Its Position Vector and the Moment Components about an Axis in Three Dimensions Mathematics • Third Year of Secondary School

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If the force π = ππ’ + ππ£ β π€, is acting at a point whose position vector is π« = 14π’ β π£ + 12π€ and the π₯- and π¦-components of the moment of the force π about the origin are 73 and 242 units of moment, respectively, find the values of π and π.

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### Video Transcript

If the force π, which is equal to ππ’ plus ππ£ minus π€, is acting at a point whose position vector is π« equal to 14π’ minus π£ plus 12π€ and the π₯- and π¦-components of the moment of the force π about the origin are 73 and 242 units of moment, respectively, find the values of π and π.

In this question, we are told that we have a point with position vector π« equal to 14π’ minus π£ plus 12π€. There is a force vector π equal to ππ’ plus ππ£ minus π€ acting at this point. The π₯- and π¦-components of this vector π are represented by the constants π and π. And it is our job in this question to find these values.

In order to do this, we have been given some information about the moment of the force π, that is, the moment produced by π. Specifically, we are told that the π₯- and π¦-components of this moment are 73 and 242 units, respectively. We recall that a force acting at a distance from a point produces a moment about that point. In this question, we have a force which is acting at some distance from the origin, and this is producing a moment.

A moment can be represented with a vector, which we generally call π¦. And we can calculate π¦ by finding the cross product of the vectors π and π, where π is the force vector which produces a moment about some point and π is the position vector at which π acts relative to that point. Now, in this question, we are told some of the components of π¦, specifically the π₯- and π¦-components. The π₯-component is 73, and the π¦-component, 242. We can therefore write the moment vector π¦ as 73π’ plus 242π£ plus π sub π§ π€, where π sub π§ is unknown as we are not given the π§-component.

This moment vector will be equal to the cross product of the position vector where the force acts and the force vector itself. The position vector is denoted by lowercase π« in this question. We can use this vector equation to find the solution to the question using the known components of the moment vector π¦ and the position vector π« to find the unknown π₯- and π¦-components of the force vector π.

It is important to note at this stage that weβ²re using a vector cross product and not just a multiplication. This means that it is not possible to make π the subject of the equation. Instead, we need to compute the cross product. And we do this by evaluating the three-by-three determinant shown. The elements in the first row are the unit vectors π’, π£, and π€. In the second row, we have the π₯-, π¦-, and π§-components, respectively, of the position vector π«. And in the third row, we have the π₯-, π¦-, and π§-components of the force vector π. Note that the order here is important. We must write them in the same order in the determinant.

We can now populate the determinant with the components of vector π«. They are 14, negative one, and 12. Likewise, the components of vector π are π, π, and negative one. We are now in a position to calculate the determinant. This is done in three parts. Firstly, we have the unit vector π’ multiplied by negative one multiplied by negative one minus π multiplied by 12. This simplifies to π’ multiplied by one minus 12π. Next, we subtract the unit vector π£ multiplied by 14 multiplied by negative one minus π multiplied by 12. This simplifies to negative π£ multiplied by negative 14 minus 12π, which in turn can be rewritten as π£ multiplied by 14 plus 12π. Finally, we add the unit vector π€ multiplied by 14 multiplied by π minus π multiplied by negative one. This is equal to π€ multiplied by 14π minus negative π, which in turn equals π€ multiplied by 14π plus π.

We now have an expression for the determinant of the three-by-three matrix π’, π£, π€, 14, negative one, 12, π, π, negative one. As this is the cross product of vector π« and vector π, this is also equal to vector π¦.

We are now in a position to find the values of π and π by equating the π’- and π£-components in our two equations. Equating the π’-components, we have 73 is equal to one minus 12π. We can subtract 73 and add 12π to both sides of this equation, giving us 12π is equal to negative 72. Dividing through by 12 gives us π is equal to negative six. We can now repeat this process for the π£-components. We have 242 is equal to 14 plus 12π. We solve this equation by firstly subtracting 14 from both sides. We can then divide through by 12 such that π is equal to 19.

We can therefore conclude that the two unknown components of force π are π equals 19 and π equals negative six. And this means that the vector force π is equal to 19π’ minus six π£ minus π€.

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