a. Cos A = 0.3/0.5 = 0.6.
A = 53.13o = B.
T1*sin53.13 + T2*sin(180-53.13) = +15.
T2 = T1:
T1*sin53.13 + T1*sin126.9 = 15
0.8T1 + 0.8T1 = 15
T1 = 9.375 N. = Tens1on in the string.
b. Cos A = 0.3/0.33 = 0.90909,
A = 24.62o.
T1*sin24.62 + T2*sin(180-24.62) = +15.
T2 = T1
T1 = ?.
Use the same procedure as part "a" to finish the problem.
a framed picture of weight 15N is to be hung on a wall using a price of string . the end of the string are tied to two points ,0.60m apart on the same horizontal level ,on the back of the picture .Find the tension in the string if the string is (a) 1.0 m long
(b)0.66m long
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