A worker stands a distance d = 0.325 m from the left end of a plank...

Tension force T₁ is resolved into two components, vertical component (T₁ sin40) and

horizontal component ( T₁ cos40 )

At equilibrium, Sum of forces is zero.

Sum of vertical forces is zero, i.e., sum of upward forces equals sum of downward forces.

T₂ + T₁ sin40 = W = 780 N ..................................(1)

Sum of horizontal forces is zero, i.e, sum of forces acting in left equals the sum of forces acting on right.

T₁ cos40 = T₃..........................(2)

At equilibrium, sum of Torques in counterclockwise direction equals sum of Torques in clockwise direction.

Let us consider the torque equations, by considering the rotation axis at left end of plank.

T₁ sin40 2 = w d ..................(3)

T₁ sin40 2 = 780 0.325 = 253.5

Hence T₁ = 253.5 / (2 sin40 ) = 197.19 N ................(4)

Tension force T₂ is obtained from eqn.(1) using known T₁

T₂ = 780 - T₁ sin40 = 780 - (197.19 sin40 ) = 653.25 N

Tension force T₃ is obtained from eqn.(2), T₃ = T₁ cos40 = 197.19 cos40 = 151.06 N

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As the worker walks towards right , distance d increases.

It can be seen from eqn.(3), Tension force T₁ is proportional to d .

Hence as d increases, T₁ increases

It can be seen from eqn. (1), when distance d increases, T₁ increases and that makes T₂ to decrease.

From eqn.(2), we have T₁ > T₃ always.

Due to all these reasons, T₁ is the tension force that is going to exceed the limit 780 N

The distance d for which the right side cord breaks is obtained from eqn.(3), by substituting T₁ = 780

d = 2 T₁ sin40 / w = (2 780 sin40)/780 = 1.286 m