So, if in the lake's frame the CM of the man and the boat does not move even if they do. This may suggest the displacement of the man is somehow proportional to the boat's?
I mean, if the man walks to the right, then the boar moves to the left?
What if I tried a different approach, as CMo=CMf=1.43 m And calling the displacement "x";
1.43 ⋅ 120 m⋅kg = 55 ⋅ 3.25 m⋅kg + 65 ⋅ x kg ; solving for x i get x ≈-0.11 m
And this would be the displacement of the canoe, I think.
So, If I applied the reference frame, I guess I should undo it. Then by substracting it to each center of mass I should be given the result; so:
dcanoo= (xocanoo-ΔCM)= (2 - 1.14)m = 0.86 m
Would it be correct?
Sorry :frown: Here goes the complete reasoning;
CMo=(1/M)⋅∑xn ⋅ mn= (1/120 kg)⋅[(55⋅0.75 kg⋅m)+(65⋅2 kg⋅m)] = 1.43 m
CMf=(1/M)⋅∑xn ⋅ mn= (1/120 kg)⋅[(55⋅ 3.25 kg⋅m)+(65⋅2 kg⋅m)] = 2.57 m
And calculate ΔCM=CMf-CMo= (2.57-1.43)m = 1.14 m
Would be this the displacement of the boat?
So; I should compute CMo = CMf?
CMo=(1/M)⋅∑xn ⋅ mn
CMf=(1/M)⋅∑xn ⋅ mn
And calculate ΔCM=Cf - Co
With this I get ΔCM= -0.33 but I do not know how to proceed forward.
Homework Statement
A 55 kg man is standing on a 65 kg and 4.0 m length canoe that floats without friction on water.
The man walks from a 0.75m from one end of the canoe to another point 0.75 from the other end of the canoe.
What distance does the canoe move?
Homework Equations
Center of mass...
Homework Statement
Considering the vector Field F(x,y,z))(zx, zy, z2), and the domain whose boundary is provided by S=S1∪S2 with exterior orientation and
S1={(x,y,z)∈ℝ3 : z=6-2(x2+y2), 0≤z≤6},
S2={(x,y,z)∈ℝ3 : z=-6+2(x2+y2, -6≤z≤0}.
Compute the total flux of F across S.
Homework Equations...