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Answer:
The mass of the stick is:
The center of mass is at the 48,41cm mark
Explanation:
To be balanced in any case the resultant torque of the system must be zero.
For the first situation:
We have a 0,15kg mass al -0,4m from the pivot point, and a 0,3kg mass at +0,25m of the pivot point. On the other hand we know that the center of gravity of the stick is not at the 50cm mark point, so it will be at a X distance with a mass M, the torque sum:
So:
(1)
For the second situation:
Now the weights are interchanged, and as the pivot point has change, the distances from the pivot point are different:
We have a 0,3kg mass al -0,33m from the pivot point, and a 0,15kg mass at +0,32m of the pivot point. As the pivot has moved we have to move our reference for X: The center of gravity of the stick is at a X+0,07m distance with a mass M, the torque sum:
replacing X*M from (1):
The mass of the stick is:
and replacing this in (1):
As we took the 50cm mark as the reference, the center of mass is at the 48,41cm mark