Statics FE review 032712

1 FE Sta'cs review h0 Scroll down to: Sta'cs review - Slides Torch Ellio0 (801) 587- 9016 MCE room 2016 (through 2000B door) Posi'on and Unit Vectors Y (- 2,6,3). A rAB O x B .(5,- 1,2) z If you wanted to express a 100 lb force that was in the direction from A to B in vector form, then you would find eAB and multiply it by 100 lb. rAB = 7i - 7j k (units of length) eAB = (7i-7j k)/sqrt[99] (unitless) Then: F = F eAB F = 100 lb {(7i-7j k)/sqrt[99]} Another Way to Define a Unit Vectors Y F y O x x z z Direction Cosines The values are the angles from the coordinate axes to the vector F.

2 The cosines of these values are the coefficients of the unit vector in the direction of F. If cos( x) = , cos( y) = , and cos( z) = - , then: F = F ( + ) Trigonometry hypotenuse opposite right triangle adjacent Sin = opposite/hypotenuse Cos = adjacent/hypotenuse Tan = opposite/adjacent A c B any triangle b C a angles = a + b + c = 180o Law of sines: (sin a)/A = (sin b)/B = (sin c)/C Law of cosines.

3 C2 = A2 + B2 2AB(cos c) Trigonometry Con'nued Products of Vectors Dot Product U . V = UVcos( ) for 0 <= <= 180 ( This is a scalar.) U In Cartesian coordinates: U . V = UxVx + UyVy + UzVz V In Cartesian coordinates to find ? UVcos( ) = UxVx + UyVy + UzVz = cos- 1 [(UxVx + UyVy + UzVz)/UV] Products of Vectors Dot Product UVcos( ) = UxVx + UyVy + UzVz = cos- 1 [(UxVx + UyVy + UzVz)/UV] However, usually we use vectors so that we do not have to deal with the angle between the vectors.

4 Products of Vectors Dot Product U . V = UVcos( ) for 0 <= <= 180 How could you find the U Projec'on of vector U, in The direc'on of vector V? V eV = V/V eV = 1 U . eV = (U)(1)cos( ) = Ucos( )= the answer The projec'on of U in the direc'on of V is U do0ed with the unit vector in the direc'on of V. 90o Ucos( ) Products of Vectors Cross Product Also called the vector product V V U x V = UVsin( )e e for 0 <= <= 180o U U Note: U x V = - V x U + If = 0 U x V = 0 i j If = 90o U x V = UVe In Cartesian coordinates: k i x i = 0 i x j = k j x i = - k etc.

5 Products of Vectors Cross Product If U = 2i + 3j k and V = - i + 2j k, then find U x V. i j k i j U x V = 2 3 - 1 2 3 - 1 2 - 1 - 1 2 Mul'ply and use opposite same sign. 3k 2i 2j - 3i j 4k Then add all six values together, but note that i can add to i, j to j, and k to k, but i cannot add to j, etc. U x V = - i + 3j + 7k units Products of Vectors Mixed Triple Product U x V.

6 W W V U x V = UVsin( )e U Projec'on of UVsin UVsin( )e W V on the direc'on U of W Products of Vectors Mixed Triple Product Note: No i j k Otherwise, same procedure Ux Uy Uz U x V . W = Vx Vy Vz Wx Wy Wz m T m mA Pulley Problem (frictionless pins) T T Fy = 0 = 2T T1 mg T1 = 2T mg T1 T1 Fy = 0 = 2T1 mg mg mAg = 2(2T mg) T1 mg mAg mg mAg T = (3m + mA)g/4 Moment of a Force 2- dimensional, in the plane of the screen y.

7 Z x O. F The moment of F D 90o about point O is the perpendicular distance from point O to the line of ac'on of F 'mes the magnitude |F| of F, where posi've is counter clockwise (CCW). Moment of a Force z 3- dimensional y x eO. r F Then: D 90o r x F = rFsin( )e = rsin( )Fe = (D)(F)e = M Note that since M = DF D = M/F The magnitude of the force F is 100N.

8 Determine the moment of F about point O and about the x- axis. y MO = rOB x F B C rOB 360mm O x F 500mm D A 600mm z You should determine: MO = (- + ) ?? j term?? Then find Mo x axis Mo x axis = - ?? vector form?? Couples F causes translation and, in general, rotation. Let F be: Equal in magnitude to F Opposite direction of F Not collinear with F.

9 Then F and F form a plane, and cause rotation, but no translation. _____ Pick an arbitrary point in 3-D space and draw position vectors from that point to a point on the line of action of F (point A), and a point on the line of action of -F (point B). Draw position vector rBA. Then: r0B + rBA = r0A Or: rBA = r0A - r0B _____ M0 = (r0A x F) + (r0B x (-F)) = (r0A - r0B) x F M0 = rBA x F The forces F and -F form a couple. The moment of a couple is the same about every point in space. Therefore, the moment of a couple is a free vector. Also, two couples that have the same moment are equivalent. A A B arbitrary point 0 r0B r0A rBA F -F A . A . A . A . Equivalent Systems 1 2 F F F - F 3 4 F F Mc F - F equivalent to 1 review For sta'c equilibrium: F = 0 and M = 0 In 3- D you have 6 independent equa'ons: 3 force and 3 moment in 2- D you have 3 independent equa'ons.

10 2 force and 1 moment or 1 force and 2 moment (omen) or 0 force and 3 moment (omen) Reactions (2-D) Rough surface or F F pinned support A B Ax B Ay Smooth surface or roller support F F Fixed, built-in, or MA Ax cantilever support Ay Rope or cable (tension) Spring (tension or compression) 2-D Beam Solution 10 N 800 4 m 2 m 10 N MA Ax 800 4 m Ay Fx = 0 = Ax 10cos(80) Ax = N Fy = 0 = Ay 10sin(80) Ay = N A = 0 = MA 10sin(80)(4) MA = 3-D Beam Solution Given T find MA y Is this bar properly constrained and statically determinate?