Calculating Center of Mass for a System of Objects with Unequal Masses

AI Thread Summary
The discussion focuses on calculating the center of mass for a system consisting of a 3.0m plank and two blocks of unequal masses. Participants emphasize the importance of understanding the placement of the blocks and their distances from the center of the plank to accurately determine the center of mass. It is noted that the third block, positioned at the center of gravity, does not affect the overall center of mass calculation. The solution involves considering torques and the distances of the blocks from a chosen reference point. Ultimately, the consensus is to simplify the problem by ignoring the third block and focusing on the plank and the two blocks.
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Homework Statement



A 3.0m long 10 kg plank has a 10 kg block mounted flush with the left end and a 20 kg block flush with the other end i.e. The outside edges of the blocks are 3.0m apart. A third block is set with its center of gravity right a the center of gravity of the system. If these blocks are all .1m wide, what is the center of mass of the system.

The Attempt at a Solution



First of all, the question say 1 block is at the center of gravity of the system, but i though finding the center of gravity was the whole point to the problem, as such, i can't even begin since i don't even have a proper diagram. additionally, i though if you place the center of gravity of one object at the center of gravity of a second object, the center of gravity won't change. Also it asks for center of gravity/mass, but it looks like torques could be used to solve the problem. Please help is needed, thank
 
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Placing the 3rd block on the center of mass of the first two blocks shouldn't change the center of mass, like you said.

So what's the center of mass of a 3 meter plank with weight evenly distributed on both sides?
 
oops, i wrote the question wrong, they are not of equal weight.
 
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A good diagram would help, but doable without if one is careful to account for the blocks size and distance to their own cg, for instance, the leftblock should be considered as a mass 1.45m from center of plank. Its torque acting thru the center would be force (mg) times 1.45.

The third block can be ignored since its distance from the cg of the system is zero. The plank cannot be ignored. So pick a point somewhere to the right of the physical center, and call the distance to the left end D. The distance to the right edge will be 3-D. Then you have two torques which would act to rotate conterclockwise (assume these are negative) and one acting to rotate clockwise, this positive. By definition of center of gravity of the entire system will be when sum of torques acting at that point produces no rotation.
 
i just combined the center of mass of the plank and the center of mass of the right block into one mass, and then treated the question as a torque question. thanks.
 
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thats all there ever are--torque problems. anyway you can simplify is good!
 
Ignore the third block. What is the center of gravity of the system with just the plank and first two blocks?
 
HallsofIvy said:
Ignore the third block. What is the center of gravity of the system with just the plank and first two blocks?

thats what i did, answer seems good! :cool:
 
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