Asked by Danny
21. A traffic signal of mass 25.0 kg is suspended by two cables as shown in the diagram below. Solve for the tension in each cable each of the following scenarios: (a) θ1 = θ2 = 20.0°, and (b) θ1 = 20.0°, θ2 = 30.0°.
Answers
Answered by
Henry
M*g = 25 * 9.8 = 235 N. = Wt. of signal.
a. Given: A1 = A2=20o. Therefore,T1=T2.
-235i + T1*sin(180-20) + T2*sin20 = 0.
Replace T2 with T1:
-235i + T1*sin(180-20) + T1*sin20 = 0.
0.342T1 + 0.342T1 = 235i.
0.684i*T1 = 235i.
T1 = 344N. = T2.
b. T1*Cos(180-20) + T2*Cos30 = 0.
-0.940T1 + 0.866T2 = 0.
T2 = 1.09T1.
T1*sin(180-20) + T2*sin30 = -(-235i).
Replace T2 with 1.09T1:
0.342i*T1 + 1.09T1*sin30 = 235i.
0.342i*T1 * 0.545i*T1 = 235i.
0.887i*T1 = 235i.
T1 = 265 N.
T2 = 1.09*265 = 289 N.
a. Given: A1 = A2=20o. Therefore,T1=T2.
-235i + T1*sin(180-20) + T2*sin20 = 0.
Replace T2 with T1:
-235i + T1*sin(180-20) + T1*sin20 = 0.
0.342T1 + 0.342T1 = 235i.
0.684i*T1 = 235i.
T1 = 344N. = T2.
b. T1*Cos(180-20) + T2*Cos30 = 0.
-0.940T1 + 0.866T2 = 0.
T2 = 1.09T1.
T1*sin(180-20) + T2*sin30 = -(-235i).
Replace T2 with 1.09T1:
0.342i*T1 + 1.09T1*sin30 = 235i.
0.342i*T1 * 0.545i*T1 = 235i.
0.887i*T1 = 235i.
T1 = 265 N.
T2 = 1.09*265 = 289 N.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.