How can i set up this a 50kg person is standing at the back of a 2.5m long rowboat if the person walk to the opposite end of the rowboat by how many m will the center of the mass of the boat change?

Answer 1

There are 6 variables to be considered in solving the problem:

1. The mass of the boat
2. The mass of the man
3. The location of the centre of mass of the boat
4. The location of the centre of mass of the man
5. The location of the centre of mass of the man and boat together
6. The length of the boat.

The change to variable 5 is what is wanted to be calculated.

The centre of mass of an object is such that there is no net turning moment about it.

A turning moment about a point of an object is calculated by multiplying the mass of the object by its distance from the point.

The distance is measured in a perpendicular direction to which gravity acts; normally gravity is considered to act down (the page), so the distance is measured across (the page). If the distance is to the right of the point, then there will be a clockwise turning moment; if to the left it will be an anticlockwise turning moment. To allow for this, distances in one direction are measured as positive and in the other negative so that clockwise and anticlockwise turning moments have opposite signs.

The mass and centre of mass of the boat do not move.

The mass of the man can be considered not to change, but the centre of mass of the man does move.

To find the centre of mass of the combined man-boat system you need to find the point around which there is no net turning moment by considering the distances of the centre of mass of the boat and the centre of mass of the man from this point - they will be unknown but a relationship can be made from the length of the boat.

Have a go before reading the solution below.

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Let the mass of the boat be B Kg.

Let the centre of mass of the boat be b metres from the stern of the boat.

Let the length of the boat be L metres.

Let the mass of the man be M kg.

The mass of the man shifts from the stern of the boat to the bow of the boat (a total distance of L metres).

By adding a mass to one side of another mass, the centre of mass of the two masses together will shift some distance towards the added mass.

Let the centre of mass of the boat and the man when the man is standing at the stern of the boat be x metres from the centre of mass of the boat. As the centre of mass of the boat is b metres from the stern, the man is standing b - x metres from this centre. There is no net turning moment about this centre; thus:

Bx = M(b - x)

→ Bx = Mb - Mx

→ Mb = Bx + Mx

→ Mb = (B + M)x

The centre of mass of the boat is b metres from the stern of the boat; thus it is L - b metres from the bow of the boat. Let the centre of mass of the boat and the man when the man is standing at the bow of the boat be y metres from the centre of mass of the boar. As the centre of mass of the boar is L - b metres from the bow of the boat, the man is standing at (L - b) - y metres from this centre. There is no net turning moment about this centre:

By = M(L - b - y)

→ By = ML - Mb - My

→ Mb = ML - By - My

→ Mb = ML - (B + M)y

From these two equations we can eliminate the distance b metres from the stern of the boat to the centre of mass of the boat, and then rearrange the result to find the shift in the centre of mass of the two together which is x + y metres:

(B + M)x = ML - (B + M)y

→ (B + M)x + (B + M)y = ML

→ (B + M)(x + y) = ML

→ x + y = ML/(B + M)

→ shift = ML/(B + M)

In this case:

shift is in metres

M = mass of man = 50 Kg

L = length of boat = 2.5 m

B = unknown mass of boat in Kg

Thus the centre of mass of both the boat and man will shift:

shift = (50 × 2.5)/(mass of boat + 50) = 125/(mass of boat + 50) metres

when the man walks form the stern to bow of the boat.

Answer 2

The answer depends on the mass of the rowboat itself. Since you have chosen not to provide that information, the question will remain unanswerable.