If the force 𝐅 equals negative three 𝐢 plus 𝑚𝐣 is acting on the point 𝐴 five, one, where its moment vector about the point 𝐵 three, four is 𝐤, determine the value of 𝑚 and the perpendicular distance 𝐿 between 𝐵 and the line of action of the force.
Recall that the moment 𝐌 of a force 𝐅 acting from a point 𝑃 about a point 𝑂 is given by 𝐫 cross 𝐅, where 𝐫 is the vector 𝑂 to 𝑃. Recall also that the magnitude of 𝐌 is equal to the magnitude of 𝐅 multiplied by the perpendicular distance 𝐿 from the point of action of 𝐅 to the pivot point. In this case, we have a pivot point 𝐵 at three, four and a force 𝐅 equal to negative three 𝐢 plus 𝑚𝐣 acting on the point 𝐴 five, one. We know that the moment 𝐌 of 𝐅 about 𝐵 is equal to 𝐤. And we are asked to find the perpendicular distance 𝐿 between the line of action of the force and the point 𝐵, which is this distance here.
To do this, we will need to rearrange this equation for 𝐿. But we don’t currently know the value of the magnitude of 𝐅. To find this, we can use the first equation 𝐌 equals 𝐫 cross 𝐅 because we already know 𝐌 and 𝐫 is fairly easy to determine. 𝐫 is equal to the vector 𝐵 to 𝐴. And 𝐵 to 𝐴 is equal to the position vector of 𝐴 five, one minus the position vector of 𝐵 three, four. This gives us 𝐫 equals two, negative three. Now we can take the cross product 𝐫 cross 𝐅, which is equal to the determinant of the three-by-three matrix 𝐢, 𝐣, 𝐤, two, negative three, zero, negative three, 𝑚, zero. Both vectors are in the 𝑥𝑦-plane and have a 𝐤-component of zero. Therefore, only the 𝐤-component of the cross product will be nonzero.
Taking this determinant by expanding along the top row gives us two 𝑚 minus nine 𝐤. The question tells us that the moment of the force about the point 𝐵 is 𝐤. Therefore, two 𝑚 minus nine is equal to one. Solving for 𝑚 gives us 𝑚 equals five. We now have the values of the components for 𝐅, negative three 𝐢 and five 𝐣. From this, we can find the magnitude of 𝐅 and then the perpendicular distance 𝐿. The magnitude of 𝐅 is given by the Pythagorean theorem as the square root of the sum of the squares of its components. So we have the square root of negative three squared plus five squared. This comes to the square root of 34, which cannot be simplified any further.
Since the moment of the force about the point 𝐵 is equal to 𝐤, the magnitude of the moment is just one. We can now rearrange the second equation, giving 𝐿 equals the magnitude of 𝐌 over the magnitude of 𝐅. So 𝐿 is equal to one over the square root of 34, which we can rewrite by multiplying the numerator and denominator by the square root of 34, giving root 34 over 34 length units.