RESULTANT OF NON-CONCURRENT FORCE SYSTEMS:OPTFOPTFPPTTFF==OPTF+OPTFPTF=ORMO

RESULTANT OF NON-CONCURRENT FORCE SYSTEMS:OPTFOPTFPPTTFF==ORMO=ORdxy

PRACTICE PROBLEMS:The effect of a certain non-concurrent force system is defined by the following data: X = +90 lb; Y = 60 lb; and MO = 360 lb-ft counterclockwise. Determine the point at which the resultant intersects the Y axis.OxyR= (X)2+(Y)2Rx = X = +90Ry = Y = -60MO= 360 lb-ft MO= 360 lb-ft Rx = X = +90yMO= Rx(y)360 = 90(y)y = 4 ft below O

PRACTICE PROBLEMS:In a certain non-concurrent force system it is found that X = 80 lb, Y = + 160 lb, and MO = 480 lb-ft in a counterclockwise sense. Determine the point at which the resultant intersects the X axis.OxyR= (X)2+(Y)2Rx = X = -80Ry = Y = +160MO= 480 lb-ft xMO= Ry(x)480 = 160(x)x = 3 ft right of ORy = Y = +160

PRACTICE PROBLEMS:Determine completely the resultant of the forces acting on the step pulley shown in the figure.X = RXRX = 750cos30 +250Y = RYRY = 750sin30 - 1250RX = 899.52 lbRY = 875 lb1Moment of components at the center of pulleysMO = MO = 0R = (RX)2 + (RY)2R = (899.52)2 + (875)2R = 1254.89 lbtan = 875899.52= 0.9727410174 0.9727420417 = 44.2083777 = 441230.16R = 1254.89 lbRX = 899.52RY = 875 = 44.2083777MO = 750(MO = 750(1.25)MO = 750(1.25) 250(MO = 750(1.25) 250(1.25)MO = 750(1.25) 250(1.25) 1250(MO = 750(1.25) 250(1.25) 1250(0.5)

PRACTICE PROBLEMS:Determine the resultant of the force system shown in Fig. P-263 and its x and y intercepts.Fig. P-263RX = XRX = +300sin30RX = +149.90 lbRY = YRY = +300cos30RY = +59.61 lbR = (RX)2 + (RY)2R = (149.90)2 + (59.61)2R = 161.32 lbtan = 149.9059.61= 0.3976651101 = 21.6859897 = 214109.56MO = MO = 100.60 lb-ftMO = 100.60 lb-ft ixiyMO = RY(ix)100.60 = 59.61(ix)ix = 100.60/59.61ix = 1.69 ft.MO = RX(iy)100.60 = 149.90(iy)iy = 100.60/149.90iY = 0.67 ft.RX = +300sin30 224(2/5)RX = +300sin30 224(2/5) + 361(2/13)RY = +300cos30 + 224(1/5)RY = +300cos30 + 224(1/5) - 361(3/13)MO = 300sin30(2)MO = 300sin30(2) 224(1/5)(2) MO = 300sin30(2) 224(1/5)(2) 361(2/13)(1) RXRYixiy

PRACTICE PROBLEMS:Completely determine the resultant with respect to point O of the force system shown in Fig. P-264.Fig. P-264RX = XRX = RX = 548.50RY = YRY = RY = 208.98R = (548.50)2 + (208.98)2R = 586.96 lbMO = MO = 2277.11 ft-lbtan=RY/RX = 208.98/548.50 = 20.85697743 = 205125.12MO = RY(x)2277.11 = x = 10.90 ft. left of Origin2277.11 =MO = RX(y)y = 4.15 ft. above Origin RX = 141.4(RX = 141.4(1/2)RX = 141.4(1/2) + 390RX = 141.4(1/2) + 390sin60RX = 141.4(1/2) + 390sin60 + 250(RX = 141.4(1/2) + 390sin60 + 250(12/13)RX = 141.4(1/2) + 390sin60 + 250(12/13) 240RX = 141.4(1/2) + 390sin60 + 250(12/13) 240sin30RY = 141.4(RY = 141.4(1/2) RY = 141.4(1/2) 390RY = 141.4(1/2) 390cos60 RY = 141.4(1/2) 390cos60 + 250(RY = 141.4(1/2) 390cos60 + 250(5/13)RY = 141.4(1/2) 390cos60 + 250(5/13) + 240RY = 141.4(1/2) 390cos60 + 250(5/13) + 240cos30MO = 141.4(1/2)(MO = 141.4(1/2)(3)MO = 141.4(1/2)(3) + 390sin60(MO = 141.4(1/2)(3) + 390sin60(4)MO = 141.4(1/2)(3) + 390sin60(4) + 390cos60(MO = 141.4(1/2)(3) + 390sin60(4) + 390cos60(4) MO = 141.4(1/2)(3) + 390sin60(4) + 390cos60(4) + 250(12/13)(MO = 141.4(1/2)(3) + 390sin60(4) + 390cos60(4) + 250(12/13)(1) MO = 141.4(1/2)(3) + 390sin60(4) + 390cos60(4) + 250(12/13)(1) 250(5/13)(MO = 141.4(1/2)(3) + 390sin60(4) + 390cos60(4) + 250(12/13)(1) 250(5/13)(4) 2277.11 = 208.98(x)2277.11 = 548.50(y)

PRACTICE PROBLEMS:Compute the resultant of the three forces shown in Fig. P-265. Locate its intersection with the X and Y axes.RX = FXMR = Mcomponents = MORX =RX = - 300RX = - 300sin30RX = - 300sin30 + 390(RX = - 300sin30 + 390(12/13)RX = - 300sin30 + 390(12/13) + 722(RX = - 300sin30 + 390(12/13) + 722(3/13)RY = FYRY = RY = - 300RY = - 300cos30RY = - 300cos30 + 390(RY = - 300cos30 + 390(5/13)RY = - 300cos30 + 390(5/13) - 722(RY = - 300cos30 + 390(5/13) - 722(2/13)RX = 810.74 lbRY = - 510.30 lbR = (810.74)2 + (510.30)2R = 957.97 lbMcomponents = tan = RY / RXtan = 510.30 / 810.74 = 32.18733595 = 32 11 14.41

PRACTICE PROBLEMS:Determine the resultant of the three forces acting on the dam shown in Fig. P-266 and locate its intersection with the base AB. For good design, this intersection should occur within the middle third of the base. Does it?

PRACTICE PROBLEMS:The resultant of four forces, of which three are shown in Fig. P-268, is a couple of 480 lb-ft clockwise in sense. If each square is 1 ft. on a side, determine the fourth force completely.MO = 480 = MF = 480MF = 480 680MF = 200 F=200 lbMF = F(x)200 = 200(x)x = 1 ft. right of OF=200 lbMF = F(y)200 = 200(y)y = 1 ft. above O if F=200 lb if F=200 lb MF = F(x)200 = 200(x)x = 1 ft. left of OMF = F(y)200 = 200(y)y = 1 ft. below O if F=200 lb if F=200 lb F=200 lbF=200 lbMF = 200 ft-lbMO = 110(MO = 110(4)MO = 110(4) + 120(MO = 110(4) + 120(2)MO = 110(4) + 120(2) + MF 480 = 440 + 480 = 440 + 240 + 480 = 440 + 240 + MF MF = 480 440MF = 480 440 240x = 200/200x = 200/200x = 200/200x = 200/200

PRACTICE PROBLEMS:The Howe roof truss shown in Fig. P-267 carries the given loads. The wind loads are perpendicular to the inclined members. Determine the magnitude of the resultant, its inclination with the horizontal, and where it intersects AB.

PRACTICE PROBLEMS:Repeat Prob. 268 if the resultant is 390 lb directed down to the right at a slope of 5 to 12 passing through point A. Also determine the x and y intercepts of the missing force.

PRACTICE PROBLEMS:The three forces shown in Fig. P-270 are required to cause a horizontal resultant acting through point A. If F = 316 lb, determine the value of P and T. Hint: Apply MR = MB to determine R, then MR = MC to find P, and finally either MR = MD or Ry = Y to compute T.The three forces in Fig. P-270 create a vertical resultant acting through point A. If T is known to be 361 lb, compute the values of F and P.

END OF PRESENTATION