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