- #1
ShaunPereira
- 40
- 4
- Homework Statement
- A uniform wooden plank of mass 150 kg and length 8 m is floating on still water with a man of 50 kg at one end of it . The man walks to the other end of the plank and stops. Than the distance covered by the plank is
- Relevant Equations
- X(com)=(m1x1 + m2x2)/m1+m2
Lets take the original position of the man to be our origin
The plank is uniform so we can assume its mass to be concentrated at its center i.e. 4m from the origin
Xcom= m1x1+m2x2/m1+m2
=50(0) +150(4) /50+150
=3m
There is no external force on the system so the centre of mass does not move
The man moves to the other side of the plank so the distance form origin becomes 8m
To compensate for this the plank moves a distance d. The original position of the centre of plank was 4m from the origin so its new distance from the origin is (4-d)
Xcom= 3 = 50(8)+ 150(4-d) / 150+50
this gives us d=8/3 m
The answer is incorrect according to the solution provided to me
It should be 2m according to it
The plank is uniform so we can assume its mass to be concentrated at its center i.e. 4m from the origin
Xcom= m1x1+m2x2/m1+m2
=50(0) +150(4) /50+150
=3m
There is no external force on the system so the centre of mass does not move
The man moves to the other side of the plank so the distance form origin becomes 8m
To compensate for this the plank moves a distance d. The original position of the centre of plank was 4m from the origin so its new distance from the origin is (4-d)
Xcom= 3 = 50(8)+ 150(4-d) / 150+50
this gives us d=8/3 m
The answer is incorrect according to the solution provided to me
It should be 2m according to it