Question

A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chain at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N. How far from the left end of the board is the person sitting?

Answer 1

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

hence, 1.5 m far from the left end of the board is the person sitting.

Answer 2

$\huge\bold\purple{Answer:-}$

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