A7.1 kg watermelon is placed at one end of a 4.8 m, 260 n scaffolding supported by two cables. one supporting cable is at the opposite end of the scaffolding, and the other is 0.53 m from the watermelon. how much tension is in the cable at the end of the scaffolding? the acceleration of gravity is 9.8 m/s 2 . answer in units of n

Answer

given,

mass of watermelon = 7.1 Kg

weight of melon = 7.1 x 9.8 = 69.58 N

Assume T₁ and T₂ are the tension on the cable

Weight of  scaffolding = 260 N

now,

assuming the system is in equilibrium

sum of vertical force = 0

T₁ + T₂ = 69.58 + 260

T₁ + T₂ = 329.58 N.......(1)

taking moment about T₂ cable

As moment about the second wire will be zero because system is in equilibrium.

69.58 x 4.8 - T₁ x 4.27 + 260 x 2.4 = 0

T₁ x 4.27 = 957.98

T₁  = 224.35 N

from equation (1)

T₂ = 329.58 - 224.35

T₂ = 105.23 N

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