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Statics (MET 2214) Prof. S. Nasseri Forces and Moments MET 2214.

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Presentation on theme: "Statics (MET 2214) Prof. S. Nasseri Forces and Moments MET 2214."— Presentation transcript:

1 Statics (MET 2214) Prof. S. Nasseri Forces and Moments MET 2214

2 Statics (MET 2214) Prof. S. Nasseri Moments and Forces Part 2

3 Statics (MET 2214) Prof. S. Nasseri Calculating the moment using rectangular components The moment of a force F about the axis passing through point O and perpendicular to the plane containing O and F can be expressed using the cross product: Mo = r x F The magnitude of the moment is the area shown below:

4 Statics (MET 2214) Prof. S. Nasseri Mo = | r x F | = r F sinθ r = x i + y j + z k F = Fx i + Fy j + Fz k Calculating the moment using rectangular components

5 Statics (MET 2214) Prof. S. Nasseri Resultant Moment

6 Statics (MET 2214) Prof. S. Nasseri Moments and Forces Part 3

7 Statics (MET 2214) Prof. S. Nasseri Moment about an axis Sometimes the moment about a point is known and you are supposed to calculate its component about an axis. To find the moment, consider the dot product of Mo and unit vector along axis a: O: any point on a-a

8 Statics (MET 2214) Prof. S. Nasseri Moment about an axis You can also find the tangent force F θ and then r x F θ is the moment about aa:

9 Statics (MET 2214) Prof. S. Nasseri Example 1 Force F causes a moment M O about point O. What is the component of M O along axis oy (M y )?

10 Statics (MET 2214) Prof. S. Nasseri Scalar Analysis There are 2 methods to find M y Scalar Analysis Vector Analysis (1) Scalar Analysis (first way): M O = (20)(0.5) = 10 N.m Imagine that we have found the direction of this moment (shown in figure) M O tends to turn the pipe around axis ob. The component of M O along the y-axis, M y, tends to unscrew the pipe from the flange at O. Thus it is important to know its value. M y = (3/5)(10) = 6 N.m

11 Statics (MET 2214) Prof. S. Nasseri Scalar Analysis Scalar Analysis (second way): To find M y directly (not form M O ) it is necessary to determine the moment-arm, knowing that the distance from F to the y-axis is 0.3m: M y = (20)(0.3) = 6 N.m In general, If the line of action of a force F is perpendicular to any specific axis aa thus: Ma = F.da

12 Statics (MET 2214) Prof. S. Nasseri Vector Analysis First, use the cross product formula to calculate the moment about O: M O = r A × F M O = (0.3i + 0.4j) × (-20k) M O = {-8i + 6j} N.m Then use the dot product of MO and the unit vector along y-axis to get My: M y = M O. u a M y = (-8i + 6j). (j) M y = 6 N.m

13 Statics (MET 2214) Prof. S. Nasseri We can always combine the two previous equations that you saw into one: M O = r A × F yielding a scalar M y = M O. U a Vector Analysis M y = (r A × F). u a Or My = u a. (r A × F) Which is called triple scalar product

14 Statics (MET 2214) Prof. S. Nasseri Remember the triple scalar product from Math section?!! Magnitude of moment M about axis aa’ Vector of moment M about axis aa’

15 Statics (MET 2214) Prof. S. Nasseri Example 2

16 Statics (MET 2214) Prof. S. Nasseri Example 2

17 Statics (MET 2214) Prof. S. Nasseri Example 2


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