A hypothetical plank of 3 m that has no mass balances at the extremities will have a pivot point located exactly halfway along the length of the plank. Then, at one end of the plank a mass of 1 kg is added and the other end a mass of 2 kg is suspended. Which direction and how far does the pivot point of the plank have to move to allow it to maintain its balance?

Refer to figure.

Consider a weightless beam resting on a central fulcrum point, the beam has a length of 3 m and no mass positioned at each end.

The Centre of Gravity (CG) in the centre of the beam and moments at either extremity are equal. Therefore, the beam’s pivot point must be in the middle of the beam.

If the position of the CG is unknown (like in our case) we can find it by taking moments from Mass A and find the total mass involved:

*Mass (B) 2 kg x lever 3 m = 6 kg m**Mass (A) 1 kg + Mass (B) 2 kg = 3 kg**Arm of CG = Total Moment / Total Mass**Arm of CG = 6 kg m / 3 kg = 2 m*

This tells us that the CG position is 2 m from Mass (A) 1 kg. Therefore, the beam’s __ pivot point has to move 0.5 m towards Mass (B) 2 kg__.

__NOTE__:

*When finding out the moment of Mass (A) and then calculating the Arm of CG, you will find out the CG positions distance from Mass (B).**If we now utilise the principle of our simple beam and superimpose an aircraft over the top of it, we can use the same method to calculate the aircraft’s CG. The nose wheel is used as a datum for the calculation, just as the end of the simple beam was used.*

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