MOMENT ABOUT AN AXIS
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1 Today s Objectives: MOMENT ABOUT AN AXIS Students will be able to determine the moment of a force about an axis using a) scalar analysis, and b) vector analysis. In-Class Activities: Applications Scalar Analysis Vector Analysis Group Problem Solving
2 APPLICATIONS With the force P, a person is creating a moment M A using this flex-handle socket wrench. Does all of M A act to turn the socket? How would you calculate an answer to this question?
3 APPLICATIONS Sleeve A of this bracket can provide a maximum resisting moment of 125 N m about the x-axis. How would you determine the maximum magnitude of F before turning about the x-axis occurs?
4 SCALAR ANALYSIS Recall that the moment of a scalar force about any point O is M O = F d O where d O is the perpendicular (or shortest) distance from the point to the force s line of action. This concept can be extended to find the moment of a force about an axis. Finding the moment of a force about an axis can help answer the types of questions we just considered.
5 SCALAR ANALYSIS In the figure above, the moment about the y-axis would be M y = F z (d x ) = F (r cos θ). However, unless the force can easily be broken into components and the d x found quickly, such calculations are not always trivial and vector analysis may be much easier (and less likely to produce errors).
6 VECTOR ANALYSIS Our goal is to find the moment of F (the tendency to rotate the body) about the a-axis. First compute the moment of F about any arbitrary point O that lies on the a - a axis using the cross product. M O = r F Now, find the component of M O along the a-axis using the dot product. M a -a = u a M O
7 VECTOR ANALYSIS (continued) M a can also be obtained as The above equation is also called the triple scalar product. In the this equation, u a represents the unit vector along the a-axis, r is the position vector from any point on the a-axis to any point A on the line of action of the force, and F is the force vector.
8 A B EXAMPLE Given: A force is applied to the tool as shown. Find: Plan: The magnitude of the moment of this force about the x axis of the value. 1) Use M x = u (r F). 2) First, find F in Cartesian vector form. 3) Note that u = 1 i in this case. 4) The vector r is the position vector from O to A.
9 Solution: EXAMPLE (continued) u = 1 i r OA = {0 i j k} m F = 200 (cos 120 i + cos 60 j + cos 45 k) N = {-100 i j k} N Now find M x = u (r OA F) M x = = 1{0.3 (141.4) 0.25 (100) } N m M x = 17.4 N m CCW
10 GROUP PROBLEM SOLVING Given: The hood of the automobile is supported by the strut AB, which exerts a force F = 120 N. Find: The moment of F about the hinged axis y. Plan: 1) Use M y = u (r OA F) 2) Find u AB from r AB 3) Find F in Cartesian vector form using u AB
11 GROUP PROBLEM SOLVING (continued) Solution: u = j r AB = { 0.6 i j k} m r AB = (-0.6) 2 + (0.6) 2 + (1.2) 2 = m u AB = r AB / r AB F = F u AB = (120 / 1.470) { 0.6 i j k} N = { 49.0 i j k} N
12 GROUP PROBLEM SOLVING (continued) Now find the triple product, M y = u (r OA F) M y = N m M y = -1 {1.2 (98.0) 0 (-49.0)} = N m CCW = N m CW
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