- #1

ago01

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- 8

- Homework Statement
- A heavy, 6m long uniform plank has a mass of 30 kg. It is positioned so that 4m is supported on the deck of a ship and 2m sticks out over the water. It is held in place only by its own weight. You have a mass of 70kg and walk the plank past the edge of the ship.

- Relevant Equations
- Torque, Forces in Y, Forces in X. Static equilibrium.

I first created the following diagram (after watching some videos, I had initially combined masses and this is wrong).

##F_p## Is the gravitational force of the board on the boat.

##N_0## Is the corresponding normal force the board experiences from the boat.

##F_m## Is the force of me at the point I step over the pivot.

##N_1## Is the normal force I experience from the plank after the pivot.

##m_p = 30 kg##

##m_m = 70 kg##

The pivot point is the red outlined black dot. The point exactly where the plank begins to overhang.

If this is in static equilibrium then at any point after the pivot that doesn't cause the plank to rotate I must be experiencing some normal force from the plank as well.

So, using the static equilibrium equations:

##\sum{F_x} = 0##

##\sum{F_y} = N_0 + N_1 - m_pg - m_mg = 0##

##\sum{T} = 0##

Call the left end of the plank ##A## and where I am standing ##B##:

##T_{A} = -4m_pg##

##T_{B} = -rm_mg##

##T_{N_0} = 0## (since it is at the lever arm)

##T_{N_1} = rN_1## (since this torque counters my torque in equilibrium).

So ##\sum{T} = -4m_pg - rm_mg + rN_1 = 0##

I understand that in order for the board to rotate it must be the case that ##N_1## = 0. But if that's the case, solving for ##N_0## using the sum of the y-forces does not help me solve for ##r## since in this case ##T_{N_1} = 0## and so:

##-4m_pg = rm_mg##

## r = -1.71m##

Which is obviously wrong. I'm sure I'm staring right at the answer and missing a critical piece of insight to solve this.

##F_p## Is the gravitational force of the board on the boat.

##N_0## Is the corresponding normal force the board experiences from the boat.

##F_m## Is the force of me at the point I step over the pivot.

##N_1## Is the normal force I experience from the plank after the pivot.

##m_p = 30 kg##

##m_m = 70 kg##

The pivot point is the red outlined black dot. The point exactly where the plank begins to overhang.

If this is in static equilibrium then at any point after the pivot that doesn't cause the plank to rotate I must be experiencing some normal force from the plank as well.

So, using the static equilibrium equations:

##\sum{F_x} = 0##

##\sum{F_y} = N_0 + N_1 - m_pg - m_mg = 0##

##\sum{T} = 0##

Call the left end of the plank ##A## and where I am standing ##B##:

##T_{A} = -4m_pg##

##T_{B} = -rm_mg##

##T_{N_0} = 0## (since it is at the lever arm)

##T_{N_1} = rN_1## (since this torque counters my torque in equilibrium).

So ##\sum{T} = -4m_pg - rm_mg + rN_1 = 0##

I understand that in order for the board to rotate it must be the case that ##N_1## = 0. But if that's the case, solving for ##N_0## using the sum of the y-forces does not help me solve for ##r## since in this case ##T_{N_1} = 0## and so:

##-4m_pg = rm_mg##

## r = -1.71m##

Which is obviously wrong. I'm sure I'm staring right at the answer and missing a critical piece of insight to solve this.