Engineering Mechanics : STATICS BDA1023 Lecture #10 University Tun Hussein Onn Malaysia (UTHM), Faculty of Mechanical and Manufacturing, Department of Mechanical Engineering Expected Lecture Date : 1 February 2007

MOMENT ABOUT AN AXIS (Section 4.5) 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. Learning Topic: Applications Scalar analysis Vector analysis

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

SCALAR ANALYSIS Recall that the moment of a force F about point O has a magnitude Mo= rA x F where rA is a distance from the point to the force’s line of action. Its axis are always perpendicular to the plane containing the force and the moment arm, Mo

It may be necessary to determine the component of moment, M2 which rotate about Y-axis since this component tends to unscrew the bolt MO = (20N)(0.5m) = 10N.m My = 3/5(10N.m) = 6N.m Rather than performing two steps, it is necessary to determine perpendicular distance from F to Y-axis. My = Fda My = (20) (0.3) = 6N.m

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. MO = r  F Now, find the component of MO along the axis a’-a using the dot product. Ma = ua • MO

VECTOR ANALYSIS (continued) Ma can also be obtained as The above equation is also called the triple scalar product. In the this equation, ua 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.

Vector Analysis Example MO = rA X F = (0.3i +0.4j) X (-20k) = {-8i + 6j}N.m Since unit vector for this axis is ua = j, My = MO.ua = (-8i + 6j)·j = 6N.m

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 Mz = 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.

EXAMPLE (continued) u = 1 k rAB = {0.25sin30°i + 0.25cos30°j + 0.4k}m = {0.125 i + 0.2165 j + 0.4k} m F = {-60 i + 20 j + 15 k} N Mz = u • (rAB  F) Mz = = 1{0.125(20) – 0.2165(-60)} N·m = 15.5 N·m

4.5 Moment of a Force about a Specified Axis Example 4.8 The force F = {-40i + 20j + 10k} N acts on the point A. Determine the moments of this force about the x and a axes.

4.5 Moment of a Force about a Specified Axis Solution Method 1 Negative sign indicates that sense of Mx is opposite to i

We can also compute Ma using rA as rA extends from a point on the a axis to the force

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 ). Answers: 1.B

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) rBC B) rAD C) rAC D) rDB E) rBD ANSWERS: 2. D

IN CLASS TUTORIAL (GROUP PROBLEM SOLVING) Given: The hood of the automobile is supported by the strut AB, which exerts a force of F = 100N on the hoof Find: Determine the moment of this force about the hinged axis y. Plan: 1) We need to use MY = u • (r  F). 2) Note that u = 1 j. 3) The vector r is the position vector from O to A. 4) Force F can be find in Cartesian vector form. O

GROUP PROBLEM SOLVING (Continue) MY = uY • (rA  F) uY = 1 j rA = {1.2i + 0j + 0k}m, FAB = FABuAB = FAB ( rAB / rAB )

GROUP PROBLEM SOLVING (Continue) rAB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m = {(0.6 – 1.2) i + ( 0.6 – 0) j + ( 1.2 – 0 ) k }m = {-0.6i + 0.6j + 1.2k }m

GROUP PROBLEM SOLVING (Continue) uAB = rAB / rAB = -0.41i + 0.4j + 0.82k FAB = FABuAB = 100{-0.4i + 0.4j +0.82k} = -40i + 40j + 82k MY = uY • (rA  FAB) = 98.4 Nm

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.

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) rAC B) rBA C) rAB D) rBC 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. ANSWERS: 1. C 2. B

HOMEWORK TUTORIAL Q1 (4-56): The RollerBall skate is an in-line tandem skate that uses two large spherical wheels on each skate, rather than traditional wafer-shape wheels. During skating the two forces acting on the wheel of one skate consist of a normal force F2 and a friction force F1. Determine the moment of both of these forces about the axle AB of the wheel.

HOMEWORK TUTORIAL (continued) Q2 (4-57) : The cutting tool on the lathe exerts a force F on the shaft in the direction shown. Determine the moment of this force about the y axis of the shaft.

HOMEWORK TUTORIAL (continued) Q4 (4-64): The flex-headed ratchet wrench is subjected to force P, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical axis of the bolt at A.

HOMEWORK TUTORIAL (continued) Q5 (4-65) : If a torque or moment M is required to loosen the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench.