Answer:
Explanation:
Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.
Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components
U = Tcos(41°)/cos(63°) . . . . write an expression for U
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The vertical components must total 200 N, so we have ....
Tsin(41°) +Usin(63°) = 200
Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200
T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200
T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons
U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons
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The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.
Then the tensions in the 3 cables are ...
41°: 93.6 N
63°: 155.6 N
90°: 200 N
The tension in each of the three cables are 94.29, 155.56 and 200 Newton respectively.
Given the following data:
First of all, we would determine the third tension force based on the vertical component as follows:
[tex]\sum F_y = 0\\\\T_3 - F_g =0\\\\T_3 - F_g=200\;N[/tex]
Next, we would apply Lami's theorem to resolve the forces acting on the traffic light at equilibrium:
For the horizontal component:
[tex]\sum F_x = -T_1cos41+T_2cos 63=0\\\\0.7547T_1=0.4540T_2\\\\T_1=\frac{0.4540T_2}{0.7547}\\\\T_1 = 0.6016T_2[/tex] ....equation 1.
For the vertical component:
[tex]\sum F_y = T_1sin41+T_2sin 63-T_3=0\\\\\sum F_y = T_1sin41+T_2sin 63-200=0\\\\0.6561T_1+0.8910T_2 =200[/tex] ...equation 2.
Substituting eqn. 1 into eqn. 2, we have:
[tex]0.6561 \times (0.6016T_2)+0.8910T_2 =200\\\\0.3947T_2+0.8910T_2 =200\\\\1.2857T_2 =200\\\\T_2 = \frac{200}{1.2857} \\\\T_2 = 155.56\;Newton[/tex]
For the first tension:
[tex]T_1 = 0.6061T_2\\\\T_1 = 0.6061 \times 155.56\\\\T_1 = 94.29\;Newton[/tex]
Read more on tension here: brainly.com/question/4080400
