What Are the Tensions in the Two Cables at Point C?

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The discussion focuses on determining the tensions in two cables, AC and BC, tied together at point C. The calculated angles for the cables are approximately 38.66 degrees and 53.13 degrees. The x-components of the tensions cancel each other, leading to the equation T1*cos(38.66) = T2*cos(53.13). Additionally, the y-components of the tensions must sum to 200 N. The solution involves solving these equations to find the tensions, which are given as 120.1 N for AC and 156.3 N for BC.
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Homework Statement



2 cables are tied together at C as shown. Determine the tension in AC and BC.
(ans: 120.1N, 156.3N)

tutorial2_7.jpg


Homework Equations





The Attempt at a Solution



tan-1 4/5 = 38.66
tan-1 4/3 = 53.13

100/tan 38.66 = ACx
ACx = 124.99

AC = sqrt(124.99^2 + 100^2)
AC = 160 (wrong)
 
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If T1 is the tension in AC and T2 that of BC, then x-components of T1 and T2 cancel each other. Hence

T1*cos(38.66) = T2*cos(53.13) ...(1)

Similarly y-components of tensions add up to 200 N.

Solve the two equations and find T1.
 
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