Tension in two ropes holding a board

[Mark Taylor]

Homework Statement

A 12m long board weighing 4kg is suspended by ropes on each end. 3m from the left side a painter is standing, closer to the first rope.

Find the tension in both ropes.

If the painter were to get off of the board, and the second rope snapped immediately after, find the instantaneous acceleration of the board.

Homework Equations

The Attempt at a Solution

I'm pretty lost as to where to even start, or if this is solvable without the mass of the painter, but here's my line of thinking:

If the system is in a state of equilibrium, then F_tension is equal to F_gravity at both ends of the board.

T₁+T₂ = F_gravity

If I want to solve for T₁ or T₂, I'm assuming that the center of mass is located where the painter is standing, which would mean that:

T₁ = F_g1(the force of gravity on the left side of the board and F_g2 on the right.)

T₁ = F_g1 = (3m/12m) * (4kg) * (9.8m/s) = 9.8N

If I carry this out for T₂:

T₂ = 29.4N

But this seems too easy/simple to be right, and I'm sure that the correct answer would lead to the answer of the third question where the rope snaps.

 

[andrewkirk]

It is not solvable without the mass of the painter. Without any calcs, we can see that if the the painter were a hippopotamus the tensions would be much greater than if the painter were a flea. The second part doesn't need the painter's mass though, because she has stepped off the board.

In the first part, the centre of mass is not where the painter is standing. It will be a little bit towards the centre of the board from where she is.

To solve the first part, assign some variables names, say P for the painter's mass, TL for the total tension in the left ropes and TR for the total tension in the right ropes. We assume that the ropes are vertical. If not, the tensions would be greater and we'd need to know the rope angles.

We have three unknowns. We can get two equations: one by equating the net force on the board, from the four forces on it (own weight, painter's weight, TL and TR) to zero, and the other by equating the net torque on the board around a suitable point - say the middle of the board - to zero. Those equations allow us to eliminate two unknowns, say TL and TR. P will remain unknown, but we can rearrange to express TL and TR in terms of P.

 

[Andrew Mason]

Hi Mark. Welcome to PF!

I think you found the correct answer to this problem: it is not solvable without knowing the mass of the painter (and assuming the board is horizontal).

AM