# Tension in two ropes holding a board

• Mark Taylor
In summary, the tension in the ropes is 9.8N and the board will be accelerated by 9.8N if the painter steps off of it.
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.

## 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 Ftension is equal to Fgravity at both ends of the board.

T1+T2 = Fgravity

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 = Fg1(the force of gravity on the left side of the board and Fg2 on the right.)

T1 = Fg1 = (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.

Last edited by a moderator:
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.

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

## What is tension?

Tension is the force that is transmitted through a rope, cable, or other similar object when it is pulled tight by forces acting from opposite ends.

## How is tension calculated?

Tension is calculated by dividing the force applied to an object by the area over which the force is distributed.

## What factors affect tension in two ropes holding a board?

The tension in two ropes holding a board is affected by the weight of the board, the angle of the ropes, and the strength and elasticity of the ropes.

## What happens if one of the ropes breaks?

If one of the ropes holding a board breaks, the tension in the other rope will increase and may cause the board to tilt or fall depending on the remaining tension and the weight of the board.

## How can tension in two ropes holding a board be adjusted?

Tension in two ropes holding a board can be adjusted by changing the angle of the ropes or by using ropes with different strengths or elasticity.

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