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## Q. 2.4

Determine the moment of the force F in Fig. (a) about point A.

## Verified Solution

The force F and point A lie in the xy-plane. Problems of this type may be solved using either the vector method (r × F) or the scalar method (Fd). For illustrative purposes, we use both methods.

Vector Solution

Recall that the three steps in the vector method are to write F in vector form, choose r and write it in vector form, and then evaluate the cross product r × F.

Writing F in vector form, we get

$F = −(\frac{4}{ 5}) 200i + (\frac{3}{ 5})200j \\ =−160i+120j\:lb$

There are several good choices for $r$ in this problem, three of which are $r_{AB}$, $r_{AC}$, and $r_{AD}$. Choosing

$r=r_{AB}=−4i+6j\: in$.

the moment about point $A$ is

$M_{A}=r×F=r_{AB}×F=\left|\begin{matrix}i&j&k\\-4&6&0\\-160 &120&0\end{matrix} \right|$

Expanding this determinant, we obtain

$M_{A}=k[(120)(−4)+(160)(6)]=480k\:lb\cdot in$.

The magnitude of $M_{A}$ is $480\:lb\cdot in$. Note that the direction of $M_{A}$ is the positive $z$ direction, which by the right-hand rule means that the moment about point $A$ is counterclockwise.

Scalar Solution

In Fig. (b), we have resolved the force into the rectangular components $F_{1}$ and $F_{2}$ at point $B$. The moment arm of each component about point $A$ (the perpendicular distance between $A$ and the line of action of the force) can be determined by inspection. The moment arms are $d_{1}=6\:in$. for $F_{1}$ and $d_{2}=4\:in$. for $F_{2}$, as shown in Fig. (b).

The moment of $F$ about $A$ now can be obtained by the principle of moments. Nothing that the moment of $F_{1}$ is counterclockwise, whereas the moment of $F_{2}$ is clockwise, we obtain

$\underset{+}{\large\curvearrowleft}\:\:M_{A} = F_{1}d_{1} − F_{2}d_{2}\\=160(6)−120(4)=480\:lb\cdot in$.

Note that the sense of $M_{A}$ is counterclockwise. Applying the right-hand rule, the vector representation of the moment is

$M_{A}=480k\:lb\cdot in$.

Recall that a force, being a sliding vector, can be moved to any point on its line of action without changing its moment. In Fig. (c) we have moved $F$ to point $C$.

Now the moment arm of $F_{1}$ about $A$ is $d_{1}=3\:in$., and the moment arm of $F_{2}$ is zero. Hence, the moment of $F$ about $A$ is

$\underset{+}{\large\curvearrowleft}\:\: M_{A}=F_{1}d_{1}=160(3)=480\:lb\cdot in$.

counterclockwise, as before.

Another convenient location for $F$ would be point $D$ in Fig. (c). Here the moment arm of $F_{1}$ about $A$ is zero, whereas the moment arm of $F_{2}$ is $4$ in., which again yields $M_{A}=480\:lb\cdot in$. counterclockwise.