Can someone balance these weights?

[chemistrymole]

Can someone please help balance these weights?

I would say the bottom and second to bottom connect as I circled them so they are balanced, but it gets really confusing after this.

 

[CWatters]

Try simplifying the drawing at each step. For example replace the bottom arm with a 2kg weight hanging from E on the next one up.

Remember that the suspension points need not be mid way between two weights. eg 4*1=2*2

 

[chemistrymole]

Try simplifying the drawing at each step. For example replace the bottom arm with a 2kg weight hanging from E on the next one up.

Remember that the suspension points need not be mid way between two weights. eg 4*1=2*2

I have tried this no luck. The bottom connects the way in the pick I did which balances out the 2 kg with 2 kg. Then from there I just seem to not figure it out.

 

[Nugatory]

I have tried this no luck. The bottom connects the way in the pick I did which balances out the 2 kg with 2 kg. Then from there I just seem to not figure it out.

The string that supports the second one up is carrying the entire weight of the first and the second one. How much is that? Where on the third does that need to connect to keep the third balanced? Why isn't this in the homework forum?

 

[chemistrymole]

The string that supports the second one up is carrying the entire weight of the first and the second one. How much is that? Where on the third does that need to connect to keep the third balanced? Why isn't this in the homework forum?

Thank you for trying to explain. I have tried that method.

I am a new user. I saw physics and posted in it. I did not realize there was a homework section. Can you please move my thread there if possible?

 

[HallsofIvy]

Okay, the first has two equal masses of "1" so they balance at the center point, B. Now that is a total mass of "2" and the second also has "2" so, yes, in order that both arms balance at F the two weight must be equally spaced from F so the first arm must be attached at E and the two arm system has a weight 4.

In order that these two arms balance the "2" on the third arm, that "2" must be twice as far from the fulcrum as two arms being attached. That means that the fulcrum (where the third arm will be attached to the next above) cannot be at the center point, J. Perhaps you were thinking that each arm must be attached to the next higher at its center. If so that was your error. In order that the third arm balance the first two arms must be attached at H and the three arms must be attached to the next higher at I and has total mass "6".

The next higher arm has a mass of "2" and we must attach the first three arms and hook to the next higher arm so that the "moment" on each side of the fulcrum is the same. That is, the mass times the distance from the fulcrum must be the same on each side of the fulcrum. The given "2" is at the far left end and we want it to be three times as far from the fulcrum as the three attached arms (because 2(3)= 1(6)). We an do that by making N the fulcrum and attaching the lower arms at O.

Do you get the idea?

 

[chemistrymole]

Okay, the first has two equal masses of "1" so they balance at the center point, B. Now that is a total mass of "2" and the second also has "2" so, yes, in order that both arms balance at F the two weight must be equally spaced from F so the first arm must be attached at E and the two arm system has a weight 4.

In order that these two arms balance the "2" on the third arm, that "2" must be twice as far from the fulcrum as two arms being attached. That means that the fulcrum (where the third arm will be attached to the next above) cannot be at the center point, J. Perhaps you were thinking that each arm must be attached to the next higher at its center. If so that was your error. In order that the third arm balance the first two arms must be attached at H and the three arms must be attached to the next higher at I and has total mass "6".

The next higher arm has a mass of "2" and we must attach the first three arms and hook to the next higher arm so that the "moment" on each side of the fulcrum is the same. That is, the mass times the distance from the fulcrum must be the same on each side of the fulcrum. The given "2" is at the far left end and we want it to be three times as far from the fulcrum as the three attached arms (because 2(3)= 1(6)). We an do that by making N the fulcrum and attaching the lower arms at O.

Do you get the idea?

No I don't sorry. You say to connect first two to third at H. Which letter to which letter would be more helpful.

This is what I have so far

 

[CWatters]

Looks right to me.