- #1

Mark Taylor

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## 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.

T

_{1}+T

_{2}= F

_{gravity}

If I want to solve for T

_{1}or T

_{2}, I'm assuming that the center of mass is located where the painter is standing, which would mean that:

T

_{1}= F

_{g1}(the force of gravity on the left side of the board and F

_{g2}on the right.)

T

_{1}= F

_{g1}= (3m/12m) * (4kg) * (9.8m/s) = 9.8N

If I carry this out for T

_{2}:

T

_{2}= 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.

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