A 110 kg scaffold is 7 m long. It is hanging with two wires, one from each end. A 520 kg box sits 1 m from the left end. What is the tension in the left hand side wire?
Mass of scaffold supported on each and each twine = (one hundred fifteen/2) = fifty seven.5kg. share of 520kg. supported on marvelous twine = 520/((7/a million.4) +a million) = 86.67kg. entire mass supported marvelous = (fifty seven.5 + 86.sixty seven) = one hundred forty four.17kg. (one hundred forty four.17 x 9.8) = a million,412.87N. tension.
Solve by balancing torques Choose the point of rotation to be the center of the scaffold. Let the tension in the left wire be T1 Let the tension in the right wire be T2 T2*3.5+ 520g*2.5 = T1*3.5 This is because the tension in T2 and the weight of the box are both in the counter clockwise direction with respect to the point of rotation while T1 is in the clockwise direction. T2*3.5 + 1300g = T1*3.5 We know T1 + T2 = 520g + 110g = 630g => T2 = 630g -T1 So now we have two equations and two unknowns (T1 and T2) Substitute T2 in terms of T1 into the equation above 3.5[630g-T1] + 1300g = 3.5T1 3505g = 7T1 T1 = 4907N 630g - 4907 = T2 = 1267N It might seem like we are neglecting the 110kg weight modeled as being hung from the center of the scaffold. When we use T1+T2=520g+110g, it is accounted for. You can prove this by choosing another point of rotation (0.5m to the right of the center for example). You get the same answer.
