LAB Statics &Torque; -Equilibrium LAB Name The Painter\'s Problem Two painters a
ID: 2304862 • Letter: L
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LAB Statics &Torque; -Equilibrium LAB Name The Painter's Problem Two painters are working on a platform hanging off the side of a building. The platform is 10 meters long and supported by two cables. The painters are below. The weights of the objects are: cable when the painters are in the position To start, locate the pivot point where Cable 1 attached to the platform. Next add the force about the pivot point together and set equal to zero represented by vector forces in the free-body drawing shown below Case 1). t-0 or better & Fr Remember that are negative old your finger on the pivot point so that your paper can rotate but not -5 slide. Then use you other hand to move the force in the direction of its arrow. If your paper rotates CW the torque should have a negative sign Ifit rotates CCW use a plus sign. Note that in calculating torques both the 30 force and distance are always positive. The +1- in front of each torque comes from the direction of impending rotation 207x5 downwe ack OYLRS Cable1 220*2 S 184N 30 N 22ON 2.5m5.0m2.5m 20N suL 4.om 53 toro u weight 84N 5 20 eath As Case 2, find out how far Sue can walk towards Jack before there is trouble (platform starts to tip) In solving Case 2, make a free body diagrams (FBD) showing the external forces and the direction of a Also calculate the force in the cables at this critical point. Hint-When Cable 2 is loose, the force in it is zero. positive torque and the pivot point you selected use same pivot point as in Case 1. Show you equation work in an orderly fashion and enter your results on the bottom of this sheet. Your lab grade will depend on the accuracy of your work and the clarity of your solution. After your group has completed the calculations you will test a model to determine if your numbers are correct. You will turn this lab in before you check your solution on the scale model. Case 1 CALCULATED MEASURED Cable 1 Cable 2 Case 2 Cable 1 Cable 2 Sue's Position (Measure from left end of platform)Explanation / Answer
Given:
Platform weight P = 184 N
Sue weight S = 130 N
jack weight J = 220 N
To solve such type of problem take a moment of any one point (where the cable is attached) , i just select the left side of the cable (1), assuming CW direction is positive
- 220 x 2.5 + 184 x 2.5 + 130 x 4 = T2 x 5
T2 = 86 N
we also know that sum all the vertical forcs should be zero.
T1 + T2 = 184 +220 + 130
T1 = 448 N
Case 2:
Basically, trouble will start when T2 = 0, so we have to find the value of distance x.
- (2.5)(220) + (184)(2.5) + (130)(4.0 - x) - 550 + 460 + 520 - 130x = 0
x = 3.31m
At this point, T1 = 448 + 86 = 534N and T2 should be zero.
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