A scaffold of mass 51 kg and length 6.0 m is supported in a horizontal position by a vertical cable at each end. A window washer of mass 87 kg stands at a point 1.8 m from one end.(a) What is the tension in the cable closer to the painter? (b) What is the tension in the cable further from the painter?
m? = mass of the scaffold = 51 kg m? = mass of the painter = 87 kg L = length of the scaffold = 6 m d = distance between the painter and end of the scaffold = 1.8 m g = acceleration by gravity = 9.8 m/s? Using the end of the scaffold that is the furthest away from the painter as pivot point, you can find the tension in the cable that is closest to the painter. m?×g×L/2 + m?×g×(L - d) = Tclose×L Tclose = 846.72 N < - - - - - - - - - - - - - - - - - answer a Using the end of the scaffold that is closest to the painter as pivot point, you can find the tension in the cable that is the furthest away from the painter. m?×g×L/2 + m?×g×d = Tfar×L Tfar = 505.68 N < - - - - - - - - - - - - - - - - - - - answer b Check: Tclose + Tfar = (m? + m?)×g (846.72 N) + (505.68) = ((51 kg) + (87 kg))×(9.8 m/s?) . . . . ? 1352.4 N = 1352.4 N . . . . . . . . . . . . . . . . . . . . . . . . . . . .! Seems OK
