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Unformatted text preview: FORCE VECTORS LEARNING OBJECTIVES Be able to resolve each force into its rectangular components Fx and Fy. Be able to represent the components of force in terms of the Cartesian unit vectors i and j. Be able to determine coplanar force resultants using the Cartesian unit vectors. PREREQUISITE KNOWLEDGE Units of measurements. Trigonometry concepts. Vector concepts. REVIEW (SI UNITS)
Basic Quantities Length is measured in meter (m) Time is measured in second (s) Mass is measured in kilogram (kg) Force is measured in Newton (N) REVIEW (FPS)
Basic Quantities Length is measured in foot (ft) Time is measured in second (s) Force is measured in pound (lb) Mass is measured in slug REVIEW
Vector Addition
Vectors A and B can be added using the parallelogram law or triangle construction (head to tail)
A A B B R=A+B R=A+B Triangle construction Parallelogram Law REVIEW
What is the magnitude (FC) of the force FC? Fc
FA= 35 N
135 FB=60 N
A) 43.1 N B) 60.7 N C) 88.3 N D) 95.0 N REVIEW
Determine the magnitude of the resultant force
4 kN 30o
A)6.84 kN 10 kN
B) 8.72 kN D) 12.5 kN C) 10.5 kN RECTANGULAR COMPONENTS OF A FORCE VECTOR F =F +F
X Y Y F
F Fx X = F cos X F = F sin Y
Fy X CARTESIAN UNIT VECTORS
In the diagram below, i denotes the Cartesian unit vector in the xdirection whereas j denotes the Cartesian unit vector in the ydirection
Y
1 unit j i 1 unit X USES OF CARTESIAN UNIT VECTORS
Y F =F i+F j
X Y F = 5i + 2j
1
X F F
X F = 5i  4j
2 3 F = 6i F COPLANAR FORCE RESULTANTS
FR = F FR = FRX i + FRY j FR = FRX + FRy FRY = tan F RX
1 2 2 Y FRX = F X ; FRY = FY FRy FR FRx X COPLANAR FORCE RESULTANTS
Y Example F1 = 5 i + 2 j F2 = 5 i  4 j F3 = 6 i
X FR = F1 + F2 + F3 FRx = i(5+56)= 4i FRy = j(2 4) = 2j F F F
F = 4i  2j
R 2 2 1  2 FR F = 4 + (  2 ) = 4.47; = tan = 26.2 R 4 o REVIEW QUESTION
Resolve F along the x and y axes and write it in vector form. F = { ___________ } N
y x 30 F = 80 N A) 80 cos (30) i  80 sin (30) j B) 80 sin (30) i + 80 cos (30) j C) 80 sin (30) i  80 cos (30) j D) 80 cos (30) i + 80 sin (30) j REVIEW QUESTION
Determine the magnitude of the resultant vector FR = (F1 + F2) in N when F1 = {10i + 20j } N and F2 = { 20i + 20j } N A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N EXAMPLE 1 Determine the magnitude and the direction of F1 to lift the 800 N box vertically upward SOLUTION 1
Y 800 N 600 N 3 FW2 5 Draw a free body diagram (FBD) F 30 FW1 400 N X 4 FW First, because of symmetry FW1 = FW2; FW = FW1 + FW2 and FW = 800 N SOLUTION 1  continued
Y 800 N 600 N 3 FW2 5 Draw a free body diagram (FBD) F 30 FW1 400 N X 4 FW Since the box will be lifted vertically, the components of the resultant force must satisfy F RX = 0.0 N and F RY = 800 N SOLUTION 1  continued
Y 800 N 600 N 3 FW2 5 Draw a free body diagram (FBD) F 30 FW1 400 N X 4 = tan (3/4) = 36.9
1 o FW FRX = 400cos(30) 600cos(36.9) + F1sin() = 0.0 FRY = 400sin(30) + 600sin(36.9) + F1cos() = 800 SOLUTION 1  continued
Y 800 N 600 N 3 5 Draw a free body diagram (FBD) F 30 FW1 400 N X 4 F1X = F1sin() = 133.4 N F1Y = F1cos() = 239.7 N FW2 FW F12(sin2 + cos2 ) = 133.42 + 239.72 F1 = 274.3 N; sin() = 133.4/274.3 = 0.486; = 29o EXAMPLE 2 Determine the magnitude F of the force F so that the resultant FR of the three forces is as small as possible SOLUTION 2 Solution steps Calculate FRx and FRy Calculate FR Differentiate FR with respect to F, set dFR/dF equal to zero and solve for FR. Check that the second derivative (d2FR/dF2) > 0.0 SOLUTION 2
FRx = 20(4/5)  F cos(45) = 16.0  0.707F FRy = 20(3/5)  12 + Fsin(45) = 0.707F FR2 = (FRx )2 + (Fry)2 FR2 = F2  22.63F + 256; differentiating 2FR(dFR/dF) = 2F 22.63 = 0.0; (1) F = 11.31 kN FR2 = (11.31)2  22.63 (11.31) + 256 FR = 11.31 kN Substituting in equation 1 yields: 22.62(dFR/dF) = 2F 22.63; Differentiating 22.62(d2FR/dF2) = 2 (d2F the minimum 0.0 Hence, F = 11.31 kN producesR/dF2) = 0.0884 >resultant force. ...
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This note was uploaded on 07/26/2009 for the course CE 221 taught by Professor Buch during the Spring '08 term at Michigan State University.
 Spring '08
 Buch

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