Given that the force 𝐅 equals four
𝐢 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 𝐎𝐀,
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 𝐎𝐀 is the vector three, six. Then, the force 𝐅 is four 𝐢 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. 𝑎, 𝑏 crossed with 𝑐, 𝑑 gives us
𝑎𝑑 minus 𝑏𝑐 times the unit vector 𝐤. 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 𝐤. This simplifies to negative
33𝐤. So the moment 𝐌 about the origin
of our force is negative 33𝐤.
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 33𝐤, 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 33𝐤,
and that distance 𝐿 is 6.6 length units.