A 10 kg sign is being held up by two ropes. One of the ropes is attached to the top right corner of the sign (rope B) and the other is attached to the top left corner of the sign (rope A). If we use the top of the sign as our x axis, rope A is at a 45 degree angle in relation to the x axis and rope B is at a 30 degree angle. What is the tension on each rope?
4 Answers

Solution : (Draw the free body diagram)
Let :
Ta = Tension in Rope A
Tb = Tension in Rope B
Summation of Forces Horizontal = 0
Ta (cos 45) = Tb (cos 30)
Ta = (Tb cos 30) / (cos 45) – – (Equation # 1)
Summation of Forces Vertical = 0
Ta (sin 45) + Tb (sin 30) = 10
Ta = (10 – Tb sin 30) / (sin 45) – – (Equation # 2)
Equating the 2 equations :
Ta = Ta
(Tb cos 30) / (cos 45) = (10 – Tb sin 30) / (sin 45)
1.224745Tb = (10 – 0.50Tb) / (0.707107)
0.866Tb = 10 – 0.50Tb
0.866Tb + 0.50Tb = 10
1.366Tb = 10
Tb = 10/1.366
Tb = 7.321 Kg (Answer) or
Tb = 7.321 x 9.81 = 71.82 N (Answer)
Substitute Value of Tb to Any Equation to Solve Ta :
(Using Equation # 1)
Ta = (Tb cos 30) / (cos 45)
Ta = (7.321 x cos 30) / (cos 45)
Ta = 8.966 Kg (Answer) or
Ta = 8.966 x 9.81 = 87.96 N (Answer)

the basic principal is that the tension in the x direction in each wire must be equal and opposite because other wise it would move to the left or to the right. Also that the sum of the y tensions must equal the weight of the sign.
Wt=T(Ay)+T(By)
Wt=sin(45)T(A)+sin(30)T(B)
98.7071T(A)=.5T(B)
1961.4142T(A)=T(B)
T(Ax)=T(Bx)
cos(45)T(A)=cos(30)T(B)
.7071T(A)=.8660T(B)
.816512T(A)=T(B)
1961.4142T(A)=.816512T(A)
196=2.23071T(A)
T(A)=87.86 N
t(B)=71.742 N

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