Question

A uniform plank of length 2.00 m and mass 29.5 kg Is supported by three ropes as Indicated by the blue vectors In the figure below. Find the tension In each rope when a 685 N person is d = 0.525 m from the left end. |T vector_1| = N |T vector_2| = N |T vector_3| = N

Answer 1

We need three equations for the three unknown tensions.

Taking torques about the left end and equating it to zero we get,

685*0.525 + 295*1 - T₁*2*sin(40) = 0 { i took clockwise torques to be positive)

The first term is due to weight oof person, second term from the weight of plank and third due tension in the rope

This gives, T₁=509.21N

The net force in horizontal direction will be zero:

T₃=T₁cos40 = 390.08N

The force in the vertical direction is zero:

T₂=685+295-T₁*sin(40) = 652.69N