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Chapter 2

Q. 2.5

The force F of magnitude 195\: kN acts along the line AB. (1) Determine the moments M_{x}, M_{y}, and M_{z} of F about the coordinate axes by the scalar method; and (2) find the moment of F about point O by the vector method and verify that M_{O} = M_{x}i + M_{y}j + M_{z}k.

The force F of magnitude 195 kN acts along the line AB. (1) Determine the moments Mx, My, and Mz of F about the coordinate axes by the scalar method; and (2) find the moment of F about point O by the vector method and verify that MO = Mxi + Myj + Mzk.

Step-by-Step

Verified Solution

We start by computing the rectangular components of F:
F=Fλ_{AB}=F\frac{\overrightarrow{AB}}{|\overrightarrow{AB}|}=195\left(\frac{3i+12j-4k}{\sqrt{3^2+12^2+(-4)^2}}\right)=45i+180j−60k\:kN

When calculating the moment of a force, the force may be placed at any point on its line of action. As shown in Fig. (a), we chose to have the force acting at point A.

Part 1
The moment of F about a coordinate axis can be computed by summing the moments of the components of F about that axis (the principle of moments).

Moment about the x-Axis Figure (b) represents a two-dimensional version of Fig. (a), showing the yz-plane. We see that the 45-kN and the 60-kN components of the force contribute nothing to the moment about the x-axis (the former is parallel to the axis, and the latter intersects the axis). The perpendicular distance (moment arm) between the 180-kN component and the x-axis is 4\:m. Therefore, the moment of this component about the x-axis (which is also the moment of F) is 180(4)=720\:kN\cdot m, clockwise. According to the right-hand rule, the positive sense of M_{x} is counterclockwise, which means that M_{x} is negative; that is,

M_{x}=−720\:kN\cdot m

Moment about the y-Axis To compute the moment about the y-axis, we refer to Fig. (c), which represents the xz-plane. We note that only the 45-kN force component has a moment about the y-axis, because the 180-kN component is parallel to the y-axis and the 60-kN component intersects the y-axis. Because the moment arm of the 45-kN component is 4\:m, the moment of F about the y-axis is 45(4)=180\:kN\cdot m, counterclockwise. Therefore, we have
M_{y}=45(4)=180\:kN\cdot m
The sign of the moment is positive, because the right-hand rule determines positive M_{y} to be counterclockwise.
Moment about the z-Axis The moment of F about the z-axis is zero, because F intersects that axis. Hence
M_{z}= 0

Part 2
Recognizing that the vector from O to A in Fig. (a) is r_{OA}=4k\:m, the moment of F about point O can be computed as follows.
M_{O}=r_{OA}×F=\left|\begin{matrix}i&j&k\\0&0&4\\45&180&-60\end{matrix} \right|=−i(4)(180)+j(4)(45)=−720i+180j\:kN\cdot m

Comparing with M_{O}=M_{x}i+M_{y}j+M_{z}k, we see that
M_{x}=-720\:kN\cdot m         M_{y}=180\:kN\cdot m         M_{z}=0
which agree with the results obtained in Part 1.

The force F of magnitude 195 kN acts along the line AB. (1) Determine the moments Mx, My, and Mz of F about the coordinate axes by the scalar method; and (2) find the moment of F about point O by the vector method and verify that MO = Mxi + Myj + Mzk.
The force F of magnitude 195 kN acts along the line AB. (1) Determine the moments Mx, My, and Mz of F about the coordinate axes by the scalar method; and (2) find the moment of F about point O by the vector method and verify that MO = Mxi + Myj + Mzk.