Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The declaration has a mass of 5kg.

York

York

Answered question

2020-10-28

Ropes 3m and 5m in length are fastened to a holiday decoration that is suspended over a town square. The declaration has a mass of 5kg. The ropes, fastened at different heights, make angles of 52 degrees and 40 degrees with the horizontal. Find the tension in each wire and the magnitude of each tension.
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Answer & Explanation

krolaniaN

krolaniaN

Skilled2020-10-29Added 86 answers

This is how I solved it, which has similar solutions to above:
T1=|T1|cos52i+|T1|sin52j
T2=|T2|cos40i+|T2|sin40j
T1+T2=mg
(|T1|cos52i+|T1|sin52j)+(|T2|cos40i+|T2|sin40j)=mg
Horizontal comp:
|T1|cos52=|T2|cos40
|T1|=|T2|cos40cos52
Sub into the vertical comp equation
|T1|sin52+|T2|sin40=mg
(|T2|cos40cos52)sin52+|T2|sin40=(5)(9.8)
Magnitudes:
If I solve for |T2|,|T2|=30.1 N
solving for |T1|=37.55
Tension vector form:
T1=|T1|cos52i+|T1|sin52j=23.1i+29.60j
T2=|T2|cos40i+|T2|sin40j=23.06i+19.347j
Jeffrey Jordon

Jeffrey Jordon

Expert2021-09-28Added 2605 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 (in Newtons) is 5 x 9.8 = 49N in the negative j direction (straight down). If we set the problem up in the same way as in the example, then T1 is the tension in the left rope and T2 is the tension in the right 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, we get that

|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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