A horizontal uniform board of weight 125

A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is hanging from the board. The tension in the right chain is 250 N.
Which of the following describes where the person is hanging?

Answer:

let Tension in the left chain is F

here, Fnet = 0

⇒F – 500N + 250 N – 125 N = 0

⇒F – 375 N = 0

⇒F = 375 N

let r from the left end of the board is the person sitting.

now torque at left end, τ = 0.

⇒375 × (0) – 500N × (r) + 250N × (4m) – 125 N × (2m) = 0

⇒0 – 500r + 1000 – 250 = 0

⇒750 = 500r

⇒r = 750/500 = 1.5 m

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