Moment of force acting at a point about other point

Question

$F=4i-2j$ acts at a point $A=(2,5)$ then the perpendicular distance between $B=(10,3)$ and line of action if the force $F=..$? I found this kind of questions disturbing I don't know if I should start with $F x(A-B)$ or $F x (B-A)$ then finding the $P$. Distance. Can anyone explain the answer? With telling me a good method to solve the questions of this lesson. :) my exams are after 8 months and I still have other loads. Answer is: $\frac {15}{5}\sqrt{5}$

Answer 1

By line of action, they probably mean line passing through $A(2,5)$ in direction of $\vec{F} = <4,-2>$.

So line of action can be written both in vector form or cartesian form. Going the cartesian way, the line is:

$$y-5 = -\frac{1}{2}(x-2)\\ x+2y-12 = 0$$

Perpendicular distance from $B(10,3)$ can be found using this formula

$$\dfrac{10+6-12}{\sqrt{5}} \\= \frac{4}{\sqrt{5}}$$

Answer 2

Cross product: $\vec M= \vec r × \vec F$.

This problem:

Let $\vec a$ be the vector pointing from $ (0,0) $ to $ A$:

$\vec a = (2,5)$.

Let $ \vec b$ be the vector pointing from $ (0,0)$ to $B.$

$\vec b = (10,3)$

1) Force is acting at $A(2,5)$

2) $\vec F = (4,-2)$.

2) Take moment $\vec M$ about $B(10,3)$.

Vector pointing from $B$ to $A$,

foot point $B$, head $A:$

$\vec a - \vec b = (-8,2)$.

$\vec M = (-8,2,0)×(4,-2,0) = 8\hat{k}.$

Length of $ \vec M$ is $8$ pointing

in the $z$- direction.

Check: $ |\vec F|$ • perpendicular distance from $B$ to line of action of force:

$|\vec M |$ = $|\vec F|$ • perpendicular distance $=$

$\sqrt{20} \dfrac{4}{√5} = 8.$

Used: Samjoe's distance.

https://en.m.wikipedia.org/wiki/Cross_product