Question Video: Finding the Moment and Perpendicular Distance of a Resultant Force about a Pivot Point Mathematics

Given that πβ = β2π’ + 2π£, πβ = β3π’ β π£, and πβ = π’ β 4π£ are acting at the point π΄(2, 3), determine the moment π of the resultant of the forces about the point π΅(β2, β1), and calculate the length of the perpendicular line πΏ joining the point π΅ to the resultantβs line of action.

03:25

Video Transcript

Given that π one equals negative two π’ plus two π£, π two equals negative three π’ minus π£, and π three equals π’ minus four π£ are acting at the point π΄ two, three, determine the moment π of the resultant of the forces about the point π΅ negative two, negative one and calculate the length of the perpendicular line πΏ joining the point π΅ to the resultantβs line of action.

Recall that the moment π of a force π acting from a point π about a pivot point π is given by π« cross π, where π« is the vector π to π. To determine the moment π in this question, we first need to find the resultant of the three forces π one, π two, and π three. We can do this through simple vector addition. So the resultant force, which we will call just π, is equal to π one plus π two plus π three. So we have negative two π’ plus two π£ minus three π’ minus π£ plus π’ minus four π£. This all comes to negative four π’ minus three π£.

Now that we have π, we need to determine the vector π« from the pivot point π, which in this case is the point π΅ negative two, negative one to the point of action π, which in this case is the point π΄ two, three. So π« is given by the position vector of π΄ two, three minus the position vector of π΅ negative two, negative one, which comes to the vector four, four. The moment π is then given by π« cross π, which is the determinant of the three-by-three matrix π’, π£, π€, four, four, zero, negative four, negative three, zero. Both π« and π are in the π₯π¦-plane and have a π€-component of zero. Therefore, only the π€-component of their cross product will be nonzero.

Taking this determinant by expanding along the top row gives us negative 12 minus negative 16π€. So this gives us the first part of our answer. The moment π of the force π about the point π is equal to four π€.

Now we are asked to calculate the length of the perpendicular line πΏ joining the point π΅ to the resultantβs line of action. This is equivalent to finding the perpendicular distance between the pivot point π΅ and the line of action of the force. And recall that these quantities are related by the equation the magnitude of the moment π equals the magnitude of the moment π multiplied by the perpendicular distance between the pivot point and the line of action πΏ. Rearranging for πΏ gives us the magnitude of π over the magnitude of π. π is equal to four π€. So its magnitude is just four.

And the magnitude of π can be found using the Pythagorean theorem on its components. So we have the square root of negative four all squared plus negative three all squared. This comes to the square root of 25, which is equal to five. And we ignore the negative square root since we are dealing with a length, which is strictly positive, so πΏ is equal to four over five. In decimal form then, the second part of our answer, the length of the perpendicular line πΏ joining the point π΅ to the resultantβs line of action, is equal to 0.8 length units.

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