Lecture Notes For Section 4.5

MOMENT ABOUT AN AXIS Today’s Objectives : 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 : Check Homework Reading Quiz Applications Scalar Analysis Vector Analysis Concept Quiz Group Problem Solving Attention Quiz

READING QUIZ 1. When determining the moment of a force about a specified axis, the axis must be along _____________. A) the x axis B) the y axis C) the z axis D) any line in 3-D space E) any line in the x-y plane 2. The triple scalar product u • ( r  F ) results in A) a scalar quantity ( + or - ). B) a vector quantity. C) zero. D) a unit vector. E) an imaginary number.

APPLICATIONS With the force F , a person is creating the moment M A . What portion of M A is used in turning the socket? The force F is creating the moment M O . How much of M O acts to unscrew the pipe?

SCALAR ANALYSIS Recall that the moment of a force about any point A is M A = F d A where d A 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. In the figure above, the moment about the y-axis would be M y = 20 (0 . 3) = 6 N ·m. However, this calculation is not always trivial and vector analysis may be preferable.

VECTOR ANALYSIS Our goal is to find the moment of F (the tendency to rotate the body) about the axis a’-a . 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 axis a’-a using the dot product. M a = u a • M O

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 axis a’-a axis, r is the position vector from any point on the a’-a axis to any point A on the line of action of the force , and F is the force vector.

EXAMPLE Given: A force is applied to the tool to open a gas valve. Find: The magnitude of the moment of this force about the z axis of the value. Plan : 1) We need to use M z = u • (r  F). 2) Note that u = 1 k. 3) The vector r is the position vector from A to B. 4) Force F is already given in Cartesian vector form. A B

EXAMPLE (continued) u = 1 k r AB = {0 . 25 sin 30 ° i + 0 . 25 cos30 ° j } m = {0.125 i + 0 . 2165 j } m F = {-60 i + 20 j + 15 k } N M z = u • (r AB  F) A B 0 0 1 0.125 0.2165 0 -60 20 15 M z = = 1{0.125(20) – 0.2165(-60)} N · m = 15.5 N · m

CONCEPT QUIZ 1. The vector operation ( P  Q ) • R equals A) P  ( Q • R ). B) R • ( P  Q ). C) ( P • R )  ( Q • R ). D) ( P  R ) • ( Q  R ).

CONCEPT QUIZ 2. The force F is acting along DC. Using the triple product to determine the moment of F about the bar BA, you could use any of the following position vectors except ______. A) r BC B) r AD C) r AC D) r DB E) r BD

GROUP PROBLEM SOLVING Given : A force of 80 lb acts along the edge DB. Find : The magnitude of the moment of this force about the axis AC. Plan : 1) We need to use M AC = u AC • ( r AB  F DB ) 2) Find u AC = r AC / r AC 3) Find F DB = 80 lb u DB = 80 lb ( r DB / r DB ) 4) Complete the triple scalar product.

SOLUTION r AB = { 20 j } ft r AC = { 13 i + 16 j } ft r DB = { -5 i + 10 j – 15 k } ft u AC = ( 13 i + 16 j ) ft / (13 2 + 16 2 ) ½ ft = 0 . 6306 i + 0 . 7761 j F DB = 80 { r DB / (5 2 + 10 2 + 15 2 ) ½ } lb = {-21 . 38 i + 42 . 76 j – 64 . 14 k } lb

M AC = M AC = 0 . 6306 {20 (-64 . 14) – 0 – 0 . 7706 (0 – 0)} lb ·ft = -809 lb ·ft The negative sign indicates that the sense of M AC is opposite to that of u AC Solution (continued) Now find the triple product, M AC = u AC • ( r AB  F DB ) ft lb

ATTENTION QUIZ 1. For finding the moment of the force F about the x-axis, the position vector in the triple scalar product should be ___ . A) r AC B) r BA C) r AB D) r BC 2. If r = {1 i + 2 j } m and F = {10 i + 20 j + 30 k } N, then the moment of F about the y-axis is ____ N ·m. A) 10 B) -30 C) -40 D) None of the above.

End of the Lecture Let Learning Continue

Lecture Notes For Section 4.5

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