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University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS Instructor: Eng. Eman Al.Swaity 1 Lecture 5 Chapter 4: Force resultant Chapter(4) Force System Resultants Chapter Objectives • To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions. • To provide a method for finding the moment of a force about a specified axis. • To define the moment of a couple. • To present methods for determining the resultants of nonconcurrent force systems. • To indicate how to reduce a simple distributed loading to a resultant force having a specified location. 2 Lecture 5 Chapter 4: Force resultant Today’s Objectives : Students will be able to: a) understand and define moment, and, b) determine moments of a force in 2-D and 3-D cases. The moment (Torque) of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis. For example, consider the horizontal force Fx, which acts perpendicular to the handle of the wrench and is located a distance dy from point 0, is seen that this force tends to cause the pipe to turn about the Z axis. 3 Lecture 5 Chapter 4: Force resultant Magnitude: The magnitude of Mo is where d is referred to as the moment arm or perpendicular distance from the axis at point 0 to the line of action of the force. Units of moment magnitude consist of force times distance, e.g., N. m or lb. ft. 4 Lecture 5 Chapter 4: Force resultant Direction: The direction of Mo will be specified by using the "right-hand rule." To do this, the fingers of the right hand are curled such that they follow the sense of rotation, which would occur if the force could rotate about point 0, The thumb then points along the moment axis so that it gives the direction and sense of the moment vector, which is upward and perpendicular to the shaded plane containing F and d. 5 Lecture 5 Chapter 4: Force resultant READING F = 10 N 1. What is the moment of the 10 N force about point A (MA)? A) 10 N·m B) 30 N·m C) 13 N·m d=3m D) (10/3) N·m E) 7 N·m • A 6 Lecture 5 Chapter 4: Force resultant APPLICATIONS (continued) What is the effect of the 30 N force on the lug nut? 7 Lecture 5 Chapter 4: Force resultant MOMENT IN 2-D The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque). 8 Lecture 5 Chapter 4: Force resultant MOMENT IN 2-D (continued) In the 2-D case, the magnitude of the moment is Mo = F d As shown, d is the perpendicular distance from point O to the line of action of the force. In 2-D, the direction of MO is either clockwise or counter-clockwise depending on the tendency for rotation. 9 Lecture 5 Chapter 4: Force resultant MOMENT IN 2-D (continued) F a b For example, MO = F d and the O direction is counter-clockwise. d Often it is easier to determine MO by using the components of F as shown. Fy F Fx b a O Using this approach, MO = (FY a) – (FX b). Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the force. 10 Lecture 5 Chapter 4: Force resultant EXAMPLE 1 Given: A 400 N force is applied to the frame and = 20°. Find: The moment of the force at A. Plan: 1) Resolve the force along x and y axes. 2) Determine MA using scalar analysis. 11 Lecture 5 Chapter 4: Force resultant EXAMPLE 1 (continued) Solution + Fy = -400 cos 20° N + Fx = -400 sin 20° N + MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m = 1160 N·m 12 Lecture 5 Chapter 4: Force resultant EXAMPLE 2 Find the moment of F1 at point o F1=100kN F2=200kN + O 4.5m M = F.d M = 100 4.5 = 450 kN.m Find the moment of F2 at point o M = F.d M = 200 0 = 0 kN.m 13 Lecture 5 Chapter 4: Force resultant Example 3 Fine the moment of all forces at point A 50N 1m 100N 100N 50N 1m A 50N - Don’t forget the signs 3m 3m 2m 2m Moment at A = F1d1 + F2d2 + F3d3 +…. = Fi .d i Five forces means five moments MA= (-)500 + (-)1002 + 1003 + (-)506 + 501 MA= -150 N.m 14 Lecture 5 Chapter 4: Force resultant Moment theorem (Varignon’s theorem) The moment of force F at point O is equal to the sum of moments of the force components at the same point O Example 4 Fined the magnitude of the moment about point O in four deferent ways Technique I the moment arm d d = 4cos40o + 2sin40o d = 4.35m the moment = Fd - M = -6004.35 = -2610 N.m 15 Lecture 5 Chapter 4: Force resultant Example 4 Technique II Replace the force by its components F1 = 600cos40o = 460N - F2 = 600sin40o = 386N Moment theorem M0= (-)4604 + (-)3862 M0 = -2610N.m Technique III the principle of transmissibility d1 = 4 + 2tan40o = 5.68m M0= (-)4605.68 + 3860 - M0 = -2610N.m 16 Lecture 5 Chapter 4: Force resultant Example 4 Technique IV the principle of transmissibility d1 = 2 + 4tan50o = 6.77m M0= 4600 + (-)3866.77 = -2610N.m - 17 Lecture 5 Chapter 4: Force resultant GROUP PROBLEM SOLVING Given: A 40 N force is applied to the wrench. Find: The moment of the force at O. Plan: 1) Resolve the force along x and y axes. 2) Determine MO using scalar analysis. Solution: + Fy = - 40 cos 20° N + Fx = - 40 sin 20° N + MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm = -7107 N·mm = - 7.11 N·m 18 Lecture 5 Chapter 4: Force resultant For each case illustrated in Fig. 4-4, determine the moment of the force about point O. 19 Lecture 5 Chapter 4: Force resultant 20 Lecture 5 Chapter 4: Force resultant In general, the cross product of two vectors A and B results in another vector C , i.e., C = A B. The magnitude and direction of the resulting vector can be written as C = A B = A B sin UC Here UC is the unit vector perpendicular to both A and B vectors as shown (or to the plane containing the A and B vectors). 21 Lecture 5 Chapter 4: Force resultant CROSS PRODUCT The right hand rule is a useful tool for determining the direction of the vector resulting from a cross product. For example: i j = k Note that a vector crossed into itself is zero, e.g., i i = 0 22 Lecture 5 Chapter 4: Force resultant CROSS PRODUCT (continued) Of even more utility, the cross product can be written as Each component can be determined using 2 2 determinants. 23 Lecture 5 Chapter 4: Force resultant The moment of a force F about point 0, or actually about the moment axis passing through 0 and perpendicular to the plane containing 0 and F, Fig. 4-12a, can be expressed using the vector cross product, namely, 24 Lecture 5 Chapter 4: Force resultant 25 Lecture 5 Chapter 4: Force resultant 26 Lecture 5 Chapter 4: Force resultant Moment M=rF r = xi + yj + zk F = FXi + FYj + FZk i j k M rx ry rz FX FY FZ M = (ryFz – rzFy)i + (rzFx – rxFz)j + (rxFy – ryFx)k Mx = ryFz – rzFy My = rzFx – rxFz Mz = rxFy – ryFx 27 Lecture 5 Chapter 4: Force resultant EXAMPLE 2 Given: a = 3 in, b = 6 in and c = 2 in. Find: Moment of F about point O. o Plan: 1) Find rOA. 2) Determine MO = rOA F . Solution rOA = {3 i + 6 j – 0 k} in MO = i j k 3 6 0 = [{6(-1) – 0(2)} i – {3(-1) – 0(3)} j + 3 2 -1 {3(2) – 6(3)} k] lb·in Lecture 5 = {-6 i + 3 j – 12 k} lb·in 4: Force resultant Chapter 28 29 Lecture 5 Chapter 4: Force resultant 30 Lecture 5 Chapter 4: Force resultant 31 Lecture 5 Chapter 4: Force resultant 32 Lecture 5 Chapter 4: Force resultant 33 Lecture 5 Chapter 4: Force resultant Recall that when the moment of a force is computed about a point, the moment and its axis are always perpendicular to the plane containing the force and the moment arm. In some problems it is important to find the component of this moment along a specified axis that passes through the point. To solve this problem either a scalar or vector analysis can be used. 34 Lecture 5 Chapter 4: Force resultant APPLICATIONS With the force F, a person is creating the moment MA. What portion of MA is used in turning the socket? 35 Lecture 5 Chapter 4: Force resultant SCALAR ANALYSIS Recall that the moment of a force about any point A is MA= F dA where dA 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 My= 20 (0.3) = 6 N·m. However this calculation is not always trivial and vector analysis may be preferable. 36 Lecture 5 Chapter 4: Force resultant VECTOR ANALYSIS 37 Lecture 5 Chapter 4: Force resultant 38 Lecture 5 Chapter 4: Force resultant 39 Lecture 5 Chapter 4: Force resultant 40 Lecture 5 Chapter 4: Force resultant Example Determine the moment MZ of T about z-axis passing O Technique I AB (12 0)i (0 15) j (9 0)k T T .n F T 10 AB (12 0) 2 (0 15) 2 (9 0) 2 T 10 0.566i 0.707 j 0.424k T 5.66i 7.07 j 4.24k MO r F M O 15 j (5.66i 7.07 j 4.24k ) M O 84.9k 63.6i M Z 84.9kN .m Minus sign indicate that the Mz is in the negative Z direction 41 Lecture 5 Chapter 4: Force resultant Technique II The force T resolved into two components Tz & Txy 9 X Y tan 1 25.1O 122 152 T XY T cos xy 9.06kN T Z T sin xy 4.24kN TZ is parallel to the Z-axis. That’s means no moment comes from TZ about Z-axis. MZ is due only to TXY M Z T XY .d 12 d .15 9.37 12 15 2 2 M Z 9.06 9.37 84.9kN .m 42 Lecture 5 Chapter 4: Force resultant EXAMPLE Given: A force is applied to A the tool to open a gas valve. Find: The magnitude of the B 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. 43 EXAMPLE (continued) A rAB = {0.25 sin 30° i + 0.25 cos30° j} m B = {0.125 i + 0.2165 j} m F = {-60 i + 20 j + 15 k} N Mz = (rAB F) 0 0 K1 = with the 0.2165 SmartDraw 5. MzCreated0.125 Trial Edition of0 -60 20 15 = 1{0.125(20) – 0.2165(-60)} N·m = 15.5 N·m 44 Lecture 5 Chapter 4: Force resultant University of Palestine College of Engineering & Urban Planning ENGINEERING MECHANICS STATICS & DYNAMICS 45 Lecture 5 Chapter 4: Force resultant