Question Video: Finding the Moment Vector of a Force Acting At a Point and the Perpendicular Between the Moment and the Line of Action of the Force Mathematics

Given that the force 𝐅 = 4𝐒 βˆ’ 3𝐣 acts through point 𝐴(3, 6), determine the moment 𝐌 about the origin 𝑂 of the force 𝐅. Also, calculate the perpendicular distance 𝐿 between 𝑂 and the line of action of the force.

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Video Transcript

Given that the force 𝐅 equals four 𝐒 minus three 𝐣 acts through the point 𝐴 three, six, determine the moment 𝐌 about the origin 𝑂 of the force 𝐅. Also, calculate the perpendicular distance 𝐿 between 𝑂 and the line of action of the force.

Remember, we can calculate the moment about 𝑂 of some force 𝐅 by finding the cross product of the vector πŽπ€, where 𝐴 is the point at which the force acts, with the vector 𝐅. And we can simplify this by using the two-dimensional definition of a cross product.

So since the point 𝐴 has coordinates three, six, the vector πŽπ€ is the vector three, six. Then, the force 𝐅 is four 𝐒 minus three 𝐣. So in component form, that’s the vector four, negative three. The cross product of two-dimensional vectors is defined as shown. π‘Ž, 𝑏 crossed with 𝑐, 𝑑 gives us π‘Žπ‘‘ minus 𝑏𝑐 times the unit vector 𝐀. In this case then, we’re going to multiply three by negative three and then subtract six times four. And we’ll multiply that by the vector 𝐀. This simplifies to negative 33𝐀. So the moment 𝐌 about the origin of our force is negative 33𝐀.

Next, we need to find the perpendicular distance 𝐿 between the origin and the line of action of the force. And the formula we can use to calculate that is to divide the magnitude of the moment by the magnitude of the force. Well, since the moment is vector negative 33𝐀, its magnitude is simply 33. But of course the magnitude of force 𝐅 is the square root of the sum of the squares of its components. That’s the square root of four squared plus negative three squared, which is equal to five.

This means the perpendicular distance 𝐿 is the quotient of these. It’s 33 divided by five, which is equal to 6.6. So the moment 𝐌 is negative 33𝐀, and that distance 𝐿 is 6.6 length units.

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