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Chapter 2

Q. 2.4

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

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.

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