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## Q. 2.3

Determine (1) the moment of the force $F$ about point $C$; and (2) the perpendicular distance between $C$ and the line of action of $F$.

## Verified Solution

Part 1

The moment of a force about point $C$ can be computed by either the scalar method $(M_{C} = Fd)$, or the vector method $(M_{C} = r×F)$. In this problem the scalar method would be inconvenient, because we have no easy means of determining $d$ (the perpendicular distance between $C$ and the line $AB$). Therefore, we use the vector method, which consists of the following three steps: (1) write $F$ in vector form; (2) choose an $r$, and write it in vector form; and (3) compute $M_{C} =r×F$.

Step 1: Write $F$ in vector form.

Referring to the figure, we obtain

$F = 500λ_{AB} = 500\frac{\overrightarrow{AB} }{|\overrightarrow{AB} |}=500 \left(\frac{2i−4j+3k}{5.385}\right)$

which yields

$F=185.7i−371.4j+278.6k\:N$

Step 2: Choose an $r$, and write it in vector form.

The vector $r$ is a vector from point $C$ to any point on the line of action of $F$. From the figure we see that there are two convenient choices for $r$—the vector from point $C$ to either point $A$ or point $B$. As shown in the figure, let us choose $r$ to be $r_{CA}$. (As an exercise, you may wish to solve this problem by choosing $r$ to be the vector from point $C$ to point $B$.) Now we have

$r=r_{CA}=−2i\:m$

Step 3: Calculate $M_{C}=r×F$.

The easiest method for evaluating the cross product is to use the determinant expansion:

$M_{C} = r×F= r_{CA}×F=\left|\begin{matrix}i&j&k\\-2&0&0\\185.7&−371.4& 278.6\end{matrix} \right|$

Expanding this determinant gives

$M_{C}=557.2j+742.8k\:N\cdot m$

Part 2

The magnitude of $M_{C}$ is

$M_{C} = \sqrt{(557.2)^2 + (742.8)^2 }= 928.6\:N\cdot m$

The perpendicular distance $d$ from point $C$ to the line of action of $F$ may be determined by

$d=\frac{M_{C}}{F}=\frac{928.6}{500}= 1.857\:$m

Observe that, instead of using the perpendicular distance to determine the moment, we have used the moment to determine the perpendicular distance.
Caution $A$ common mistake is choosing the wrong sense for $r$ in Eq. (2.4).

$M_O =r×F$                 (2.4)

Note that $r$ is directed from the moment center to the line of action of $F$. If the sense of $r$ is reversed, $r×F$ will yield the correct magnitude of the moment, but the wrong sense. To avoid this pitfall, it is strongly recommended that you draw $r$ on your sketch before attempting to write it in vector form.