Two steel wires support a moveable overhead camera weighing W = Ib (see figure p
ID: 1939147 • Letter: T
Question
Two steel wires support a moveable overhead camera weighing W = Ib (see figure part a) used for close up viewing of field action sporting events. At some instant, wire 1 is at angle sigma = 22 degree to the horizontal and wire 2 is at a n angle beta = 40 degree. Wires 1 and 2 line diameters of and 35 mils, respectively, (Wire 1 and 2 wire diameters are often expressed in mils; one mil equals 0.001 in) Determine the tensile stress sigma1 and sigma2 in the two wires. If the stresses in wires 1 and 2 must be the same, what is the required diameter of wire 1? Now, to stabilize the camera for windy outdoor conditions a third wire is added (see figure part b). Assume the three wires meet at a common point (coordinates (75 ft, 48 ft, 70 ft). Wire 2 is supported at (-70 ft, 55ft, 80 ft). Wire 3 is supported at (30 ft, -85 ft, 75 ft). Assume that all three wires have a diameter of 30 mils. Find the tensile stresses in wires 1 to 3.Explanation / Answer
Bold face = vector.
T1=(75i +48j +70k)/(752+482+702)T1 = (0.6622i +0.4238j +0.6180k)T1
T2=(-70i +55j +80k)/(702+552+802)T2 = (-0.5849i +0.40.4595j +0.6684k)T2
T3=(-10i -85j +75k)/(102+852+752)T3 = (-0.0879i -0.7469j +0.6590k)T3
From F =0, or, T1 +T2 + T3 + W=0, we have
(0.6622i +0.4238j +0.6180k)T1 +
(-0.5849i +0.40.4595j +0.6684k)T2+(-0.0879i -0.7469j +0.6590k)T3 - 28k =0
=> (gt-grad's work, don't copy)
0.6622T1 -0.5849T2-0.0879T3 =0
0.4238T1 +0.40.4595T2 -0.7469T3 =0
0.6180T1 +0.6684T2+0.6590T3 - 28 =0
Solving, we have
T1 =13.8539 lb T2=13.2768lb T3 =16.0284 lb
Now you can calculate the stresses.
Please note that the coordinates in the figure are different from those in the text, for wire 3.
In the text, it says (10, -85, 75). If this is right, then
T1 =11.5109 lb T2=15.441lb T3 =16.0305 lb
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