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.

Question

krolaniaN · Accepted Answer

This is how I solved it, which has similar solutions to above:  T1=−|T1|cos⁡52i+|T1|sin⁡52j  T2=|T2|cos⁡40i+|T2|sin⁡40j  T1+T2=mg  (−|T1|cos⁡52i+|T1|sin⁡52j)+(|T2|cos⁡40i+|T2|sin⁡40j)=mg  Horizontal comp:  |T1|cos⁡52=|T2|cos⁡40  |T1|=|T2|cos⁡40cos⁡52  Sub into the vertical comp equation  |T1|sin⁡52+|T2|sin⁡40=mg  (|T2|cos⁡40cos⁡52)⋅sin⁡52+|T2|sin⁡40=(5)(9.8)  Magnitudes:  If I solve for |T2|,|T2|=30.1 N  solving for |T1|=37.55  Tension vector form:  T1=−|T1|cos⁡52i+|T1|sin⁡52j=23.1i+29.60j  T2=|T2|cos⁡40i+|T2|sin⁡40j=23.06i+19.347j

Jeffrey Jordon · Accepted Answer

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|cos⁡52i+|T1|sin⁡52j  T2=|T2|cos⁡40i+|T2|sin⁡40j  w=−49j  Since the decoration is not moving, we have  0=T1+T2+w  =−|T1|cos⁡52i+|T1|sin⁡52j+|T2|cos⁡40i+|T2|sin⁡40j−49j  =(−|T1|cos⁡52+|T2|cos⁡40)i+(|T1|sin⁡52+|T2|sin⁡40−49)j  This gives us a system of equations:  −|T1|cos⁡52+|T2|cos⁡40=0  |T1|sin⁡52+|T2|sin⁡40−49=0  Solving for |T2| in the first equation, we get that  |T2|=|T1|cos⁡52cos⁡40  Plugging this into the second equation and solving for |T1|, we get  |T1|=49sin⁡52+cos⁡52tan⁡40≈37.6  We then get that  |T2|=|T1|cos⁡52cos⁡40≈30.2  Therefore we get  T1=−23.1i+29.6j  T2=23.1i+19.4j

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.

Answered question

Answer & Explanation

New Questions in Force, Motion and Energy