# Homework Help: Center of Mass problem.

1. Apr 23, 2007

### linuxux

1. The problem statement, all variables and given/known data

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.

3. 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 cant 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 wont 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

Last edited: Apr 23, 2007
2. Apr 23, 2007

### tony873004

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?

3. Apr 23, 2007

### linuxux

oops, i wrote the question wrong, they are not of equal weight.

Last edited: Apr 23, 2007
4. Apr 23, 2007

### denverdoc

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.

5. Apr 23, 2007

### linuxux

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.

Last edited: Apr 23, 2007
6. Apr 23, 2007

### denverdoc

thats all there ever are--torque problems. anyway you can simplify is good!!

7. Apr 23, 2007

### HallsofIvy

Ignore the third block. What is the center of gravity of the system with just the plank and first two blocks?

8. Apr 24, 2007

### linuxux

thats what i did, answer seems good!