Ropes 3 m and 5 m in length are fastened to a holiday decoration that

Dowqueuestbew1j 2021-12-17 Answered
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension.
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Expert Answer

Toni Scott
Answered 2021-12-18 Author has 32 answers
There are two things to notice. The first is that the length of the rope is not important in terms of finding the tension vectors. The second is that the mass of the decoration is 5 kg. That means that the force is 5×9.8=49 N in the negative j direction. If we set the problem up in the same way as in the example, then T1 is then tension in the left rope and T2 is the tension in the tight rope. We see that
T1=|T1|cos52i+|T1|sin52j
T2=|T2|cos40i+|T2|sin40j
w=49j
Since the decoration is not moving, we have
0=T1+T2+w
=|T1|cos52i+|T1|sin52j+|T2|cos40i+|T2|sin40j49j
=(|T1|cos52+|T2|cos40)i+(|T1|sin52+|T2|sin4049)j
This gives us a system of equations:
|T1|cos52+|T2|cos40=0
|T1|sin52+|T2|sin4049=0
Solving for |T2| in the first equation and solving for |T1|, we get
|T2|=|T1|cos52cos40
Plugging this into the second equation and solving for |T1|, we get
|T1|=49sin52+cos52tan4037.6
We then get that
|T2|=|T1|cos52cos4030.2
Therefore we get
T1=23.1i+29.6j
T2=23.1i+19.4j

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