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PLZ HELP Will mark Brainliest!!
In the picture below, the block is being hung in equilibrium. What is the tension needed in each of the ropes to keep the box in equilibrium? Ignore all friction and show all work

PLZ HELP Will mark BrainliestIn the picture below the block is being hung in equilibrium What is the tension needed in each of the ropes to keep the box in equi class=

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MathPhys

Answer:

311 N

Explanation:

Apply Newton's second law to the box in the y direction.

∑F = ma

2 T sin 60° − mg = 0

2 T sin 60° = mg

T = mg / (2 sin 60°)

T = (55 kg) (9.8 m/s²) / (√3)

T ≈ 311 N

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