Question
A bag of cement of weight 475 N hangs from three wires as suggested in the figure below. Two of the wires make angles ?1 = 50.0° and ?2 = 28.0° with the horizontal. Assuming the system is in equilibrium, find the tensions in the wires.
T1 =______ N
T2 =______ N
T3 =______ N
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Answer #1
Given that
weight of cement bag W = 475 N
--------------------------------------------------------
The tenion in the third string
        T3= 475 N
The net horizontal force along x- direction is zero.
T2 cos28 = T1 cos50
T2 = (cos 50/cos28) T1
The net vertical force along y- direction is zero.
         T1 sin50 + T2 sin28 = T3
          T1 sin 50 + ((cos 50/cos28) T1)sin28 = 475 N
               T1 [sin50 + ((cos 50/cos28) )sin28] = 475 N
            T1= 428.769754 N
Now T2 = (cos 50/cos28) (428.769754 N) = 312.1452 N
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