Find'n Ten ion in Mu ti 6 S ' 8 When we have a system of strings tied to each other supporting objects, we still follow the rules of Static problems. ZFtotal : 0 zazo 25:0 However. often we can not take all ropes into consideration at once. We must solve a part of the problem, which has only two unknowns. :1, Find the tension in. each rope given that the weight of the bucket a; it is t: ibi; Eveniththg In the diagram [5 in static Equilibrium. Show ALL work for full credit. Hint: draw a free body ding ram of A, solve for everything; you can. Then draw a free body diagram of B and solve {or the rest\" Lets assume the hanging bucket has a mass of 60 kg (not 60 lb as mentioned in the question. What we are asked to find is the tension in each of the ropes. As you can see there are 5 ropes in total, AH, AE, AB, CB and BD. Where to start? First you know that gravity is the force pulling down on any hanging mass, so we can find the Tension in AH quite easily. Fg = mg 2 60(9.8) = 588N [ll] Therefore the tension in AH must also be 588N[TT] Note: the direction of a tension in a rope is always along the length of the rope, and can be considered pulling with that force in either direction (or both directions) at the same time. In order to solve this problem, we must start by drawing a free body diagram for each knot, or intersection. You can see there are knots at Point \"A" and at Point \"B\" We will start with Knot A since we know the tension of AH, and that leaves only 2 ropes of unknown tension. The labels on this diagram is backwards. I found a suitable image to represent the forces on this knot, but had to flip it horizontally to reflect the tensions in this question. The labels were part of the image, so they flipped as well. The blue arrow represents TAH The green arrow represents TAE The purple arrow represents TAB ZFK = O ZFy: 0 ZFLEFT : ZFRiGHT ZFoowu = ZFUP TABCOSSOO : TAE TAH = TABSln60 0.5 TAB = TAE 588 = 0.8660TAB TAB : SSS/0.8660 TAB I 678.96N[Left 60 up] Now we can substitute this value into the other equation to get TAE. 0.5 (678.96): TAE TAS : 339.48 N[Left] Now look at the second Knot, since we now know TAB, there are only two tensions we do not know for the second knot (TBC, TBD), but first we need to find the angle of rope BD. (always find angle to the horizontal so sine gives us the vertical component and cosine gives us the horizontal) tand = 3/ tan '(0.75) = 0 0 = 36.87 For knot B: EF = 0 EFy = 0 EFLEFT = EFRIGHT EFDOWN = EFUP TCB = TBDCOS36.87 + TABCOS60 TABSin60 = TBDSin36.870 678.96(0.8660) = 0.6000T BD TBD =587.98/0.6000 TBD =587.98/0.6000 TBD =979.97N[right36.87up] TCB = TBDCOS36.87 + TABCOS60 TCB = 979.97cos36.87 + 678.96cos60 TCB = 979.97(0.8000) + 678.96(0.5) TCB = 783.97 + 339.48 TCB = 783.97 + 339.48 TCB = 1123.45N[Left] Notice that had you chosen to solve knot "B" first, there would have been an equation with three unknowns, which cannot be solved.Try these: 1. Batman, the caped crusader in grey, is hanging from the ceiling on a massless cable. Captain Red Boots-n-Gitch is hanging on the same cable below Batman. If Batman has a mass of 95kg and Captain Red Boots-nGitch has a mass of 120kg, what is the tension in the cable below Batman, and what is the tension in the cable above Batman? 2. What is the tension in the cables in the diagram shown