A wooden frame weighing 20N is hung on a nail using a two ropes as shown below find the tension in the rope.

The left side has 50¡ã under it is T and The right side hast 50¡ã under it is T

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/ \
|50¡ã/ \50¡ã|
_|__/_T_____T_\_|_
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|_________|_________|
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20N

T1 = T2, T1*Cos45 - T2*Cos45 = 0.

So, the hor. components cancel.

T1*sin45 + T2*sin45 = 20, Replace T2 with T1: T1*sin45 + T1*sin45 = 20, 2T1*sin45 = 20,
1.414T1 = 20, T1 = 14.14 N. = T2.

To find the tension in the ropes, we can set up equations based on the equilibrium of forces.

Let's assume that the left rope has a tension T1 and the right rope has a tension T2.

For the vertical equilibrium:
Sum of vertical forces = 0
T1 × cos(50°) + T2 × cos(50°) - 20N = 0

For the horizontal equilibrium:
Sum of horizontal forces = 0
T1 × sin(50°) - T2 × sin(50°) = 0

Solving these two equations will give us the values of T1 and T2.

Step 1: Calculate the value of cos(50°) and sin(50°)
cos(50°) = 0.6428 (rounded to four decimal places)
sin(50°) = 0.7660 (rounded to four decimal places)

Step 2: Set up the equations:
T1 × 0.6428 + T2 × 0.6428 = 20N (equation 1)
T1 × 0.7660 - T2 × 0.7660 = 0 (equation 2)

Step 3: Solve the equations simultaneously:
Using equation 2, rearrange it to get T1 in terms of T2:
T1 = (0.7660 × T2) / 0.6428

Substitute T1 in equation 1:
(0.7660 × T2) / 0.6428 × 0.6428 + T2 × 0.6428 = 20N
0.7660 × T2 + 0.6428 × T2 = 20N
1.4088 × T2 = 20N
T2 = 20N / 1.4088
T2 ≈ 14.212N (rounded to three decimal places)

Step 4: Substitute T2 back into equation 2 to find T1:
T1 = (0.7660 × 14.212N) / 0.6428
T1 ≈ 16.968N (rounded to three decimal places)

Therefore, the tension in the left rope (T1) is approximately 16.968N, and the tension in the right rope (T2) is approximately 14.212N.