# A weight of 20 KN supported by two cords, one 3 m long and the other 4m long with points of support 5 m apart find the tensions T1 and T2 in the cords

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can u plz give the diagram

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can u plz give the diagram

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See diagram.

Perhaps the trianlge ABC a right angle triangle .

As the forces are in equilibrium, we can use triangle law of forces:

T1/sin (β+90) = T2/sin (α+90) = W/sin(α+β)

So T1/cos β = T2 / cos α = W / sin (α+β)

From the triangle ABC, sin α / sin β = 4/3

3 cos α + 4 cos β = 5

So T2/T1 = 4/3

solving the above trigonometric equations, we get β = 36.87 deg α = 53.13 deg

sinα = 4/5 = 0.8, cos α = 3/5 =0.6 , sin β = 3/5 = 0.6 cos β = 4/5 = 0.8

T1/0.6 = T2/0.8 = W/1 = 20 kN

T1 = 12 kN

T2 = 16 kN

Perhaps the trianlge ABC a right angle triangle .

As the forces are in equilibrium, we can use triangle law of forces:

T1/sin (β+90) = T2/sin (α+90) = W/sin(α+β)

So T1/cos β = T2 / cos α = W / sin (α+β)

From the triangle ABC, sin α / sin β = 4/3

3 cos α + 4 cos β = 5

So T2/T1 = 4/3

solving the above trigonometric equations, we get β = 36.87 deg α = 53.13 deg

sinα = 4/5 = 0.8, cos α = 3/5 =0.6 , sin β = 3/5 = 0.6 cos β = 4/5 = 0.8

T1/0.6 = T2/0.8 = W/1 = 20 kN

T1 = 12 kN

T2 = 16 kN

So T1/cos β = T2 / cos α = W / sin (α+β)

From the triangle ABC, sin α / sin β = 4/3

3 cos α + 4 cos β = 5

So T2/T1 = 4/3

β = 36.87 deg α = 53.13 deg

sinα = 4/5 = 0.8, cos α = 3/5 =0.6 , sin β = 3/5 = 0.6 cos β = 4/5 = 0.

T1/0.6 = T2/0.8 = W/1 = 20 kN

T1 = 12 kN

T2 = 16 kN