Answer:
T1 = 499.9N, T2 = 865.8N, T3 = 1000N
Step-by-step explanation:
To find the tensions we need to find the vertical and horizontal components of T1 and T2
T1x = T1 cos60⁰, T1y = T1 sin60⁰
Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰
For the forces to be in equilibrium,
the sum of vertical forces must be zero and the sum of horizontal forces must also be zero
Sum of Fx = 0
That is, T1x - T2x=0
NB: T2x is being subtracted because T1x and T2x are in opposite directions
T1 cos60⁰ - T2 cos30⁰ = 0
0.866T1 - 0.5T2 = 0 ............ (1)
Sum of Fy = 0
T1y + T2y - 1000 = 0
T1 sin60⁰ + T2 sin30⁰ - 1000 = 0
NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.
0.5T1 - 0.866T2 - 1000 = 0 ........(2)
From (1)
make T1 the subject
T1 = 0.5T2/0.866
Substitute T1 into (2)
0.5 (0.5T2/0.866) - 0.866T2 = 1000
(0.25/0.866)T2 - 0.866T2 = 1000
0.289T2 - 0.866T2 = 1000
1.155T2 = 1000
T2 = 865.8N
Then T1 = 0.5 x 865.8 / 0.866
T1 = 499.9N
T3 = 1000N
NB: The weight of the bag is the Tension above the rope, which is T3