Cartesian Vector Addition Learning Goal: To find the resultantof multiple forces

Question

Cartesian Vector Addition Learning Goal: To find the resultantof multiple forces using Cartesian components and to determine the direction ofthis resultant from its components. As shown, three forces acton the tip ofa pole F1 -50 i 105 j 35 k lb. F2 150 lb and forms the following angles with the and zaxes, respectively: a* 54.7 66.1 and Y 135. i F's 110 lb. forms the angle 28 with the zaxis, and forms the angle 6 27 between the xaxis and the projection of F in the zy plane. Part A Resultant of adding F1 and F3 Find the resultant ofadding Fi and F3.Express your answer in component form. Express your answers, separated by commas, to three significant figures. ilb, jib, k lb Hints My Answers Give Up Review Part Incorrect, Try Again; 4 attempts remaining Part B Resultant of adding F and F2 and F3 Find the resultant ofadding F1, 2,and Fs.Express your answer in component form. Express your answers, separated by commas, to three significant figures. ilb, jib, k lb R12s Hints My Answera Glve Up Review Part Part C Direction of a resultant For the given forces F1.F and Fs.find the three direction anglesa and TR between the resultant force R12s found in Part B and the x,y.andzaxes, respectively. Express your answers, separated by commas, in rees to three significant figures. Hints My Answera Glve Up Review Part Part D Find the direction angle a when given direction angles and Y Asshown, a force of 225 lb acts at the tip of a tower. Two of the force's direction angles are known, B 39 and 64.5 that define the angles between the force and the yandzaxes, respectively. Find aRI, the angle between thexaxis and the force Express your answer in degrees to three significant figures.

Explanation / Answer

F1 = -50i + 105j + 35k lb

magnitude lF2l = 150 lb

F2x = cosalfa*F2 = cos54.7*150 = 86.7 lb

F2y = cosbeat*lF2l = cos66.1*150 = 60.1 lb

F2z = cosgamma*lF2l = cos135*150 = -106.1 lb

F2 = 86.7i + 60.1 j - 106.1 k

lF3l = 110 lb

F3z = cosphi*lF3l = cos28*110 = 97.1 lb

projection of F3 = F*sin28 = 110*sin28 = 51.6 lb

F3x = 51.6*cos27 = 45.6 lb

F3y = -51.6*sin27 = -23.4 lb

F3 = 97.1 i + 45.6 j - 23.4 k

partA)

F1 + F3 = -50i + 105 j + 35k + 97.1i + 45.6j - 23.4k

R13 = + 47.1 i + 150.6j + 11.6 k   <---------answer

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part B

R123 = R13 + F2 = + 47.1 i + 150.6j + 11.6 k + 86.7i + 60.1 j - 106.1 k

R123 = 133.8i + 210.7k - 94.5k     <---------answer

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lR123l = sqrt(133.8^2+210.7^2+94.5^2) = 267 lb

cosalfaR = R123x/lR123 = 133.8/267

alfaR = 59.9 = 60   <---------answer

cosbetaR = R123y/lR123l = 210.7/267

betaR = 37.8 <---------answer

cosgammaR = R123z/lR123l = 94.5/267

gammaR = 69.3 <---------answer

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part D)

cosbeta = Ry/lRl

225 *cos39 = Rx

Ry = 174.8 lb

Rz = lRl*cosgamma = 225*cos64.5 = 96.8

lRl = sqrt(Rx^2 + Ry^2 + Rz^2)

Rx = sqrt(R^2 - Ry^2 - Rz^2 )

Rx = sqrt(225^2 - 174.8^2-96.8^2)

Rx = 103.4

cos alfa = Rx/lRl = 103.4/225

alfa = 62.6 <---------answer

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