I am just not sure why I am getting this wrong. Please show all steps to solutio
ID: 2889743 • Letter: I
Question
I am just not sure why I am getting this wrong. Please show all steps to solutions.
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. The hoist weighs 420 N. The ropes, fastened at different heights, make angles of 50° and 38° with the horizontal. Find the tension in each rope and the magnitude of each tension. (Let T2 and T3 represent the tension vectors corresponding to the ropes of length 2 m and 3 m respectively. Round all numerical values to two decimal places.)
T2 = T3 = | 270.10 ITal = | 270.10 X N 38° 50° 2 mExplanation / Answer
The horizontal and vertical components of T2 and T3 are
T2x = -T2cos50o, T3x = T3cos38o
T2y = -T2sin50o, T3y = T3sin38o
Now for equibrium, |T2x| = |T3x|
=> T2cos50o = T3cos38o
=> T2 = T3 (cos38o)/cos50o ...(i)
Also sum of vertical components of all forces is equal to the weight that is acting downwards
=> T2sin50o + T3sin38o = 420
Using above equation (i)
=> T3cos38o tan50o + T3 sin38o = 420
=> T3 = 420/(cos38o tan50o +sin38o ) = 420/ (0.7880 * 1.1918+0.6157) = 420/1.5548384
=> |T 3 |= 270.12 N
|T2| = |T3| (cos38o)/(cos50o) = 270.12(0.7880/0.6428) = 331.14
T2 = -T2 cos500 i+ T2 sin500 j= 212.85 i + 253.67 j
T3 = T3 cos380i + T3 sin380j = 212.86 i + 166.30 j
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