Monday, 17 December 2007 - 4:21 PM EST

I dont have my packet home with me but i kind of remember the question from class. The two parts of the broom do not have teh same weight, but it does balance according to the equation m1d1=m2d2. One side has greater mass but has a much shorter distance, and the other side has less mass but a greater distance. THerefore making them balance

Monday, 17 December 2007 - 4:30 PM EST

The end of the broom would weigh more than the handle part of the broom. The two ends do not necessarily have the same weight because the center of gravity doesn't mean that both ends are equal in weight. They still balance since the heavier end is shorter and the lighter end is longer, following the equation m1d1=m2d2..

Monday, 17 December 2007 - 4:43 PM EST

I agree with Dan, each end of the broom is a different weight. The handle part of the broom weighs less than the actual head of the broom. The broom can be balanced at the center of gravity because, in this case, m1d1=m2d2. Since the distance from the fulcrum is different for both sides, the masses must be different also. The side that is closer to the fulcrum has more mass and vice versa.

Monday, 17 December 2007 - 4:47 PM EST

If you cut the broom at the center of gravity, the right end (the head of the broom) would weigh more than the left end (the handle). They do not have the same weight because the head of the broom is closer to the center of gravity. Following the equation m1d1=m2d2, the mass of the broom head must be greater than that of the handle because its distance from the CG is smaller.

I agree with both Kevin and Dan, and I'm impressed that Kevin remembered the question so well without his packet.

Monday, 17 December 2007 - 5:08 PM EST

If the broom was cut at the center of gravity the head of the broom would weigh more than the handle. The reason why it balances is because the handle has more distance on its side than the head does so even though the mass is smaller it evens out in the m1d1 = m2d2 equation.

Monday, 17 December 2007 - 5:09 PM EST

I agree with Allison, I was thinking pretty much the same thing. 

Monday, 17 December 2007 - 5:20 PM EST

The part of the broom with the handle would weigh less then the other end of the broom.  They would still balance as long as the heavier end is shorter then the lighter end.  You use the equation m1d1=m2d2

Monday, 17 December 2007 - 5:21 PM EST

I agree with dan

Monday, 17 December 2007 - 5:36 PM EST

The system is balanced based on the equation m1d2 = m2d2.  The mass of the head of the broom is more than that of its handle.  This is because the distance span of the head of the broom from the fulcrum is smaller than the distance from the fulcrum to the end of the handle.  In order to satifsy the equation, the larger mass is paired with the shorter distance and the smaller mass is paired with the longer distance.

I agree with what everyone has said so far.

Monday, 17 December 2007 - 5:38 PM EST

That's supposed to say d1 not d2.... my bad

Monday, 17 December 2007 - 6:25 PM EST

The broom is currently in equillibreum (Yeah I can spell..) because the mass of the stick side and it's distance is equal to that of the distance from the CG to the head of the broom and its mass. So when you cut the stick side in half, the sides are no longer equal, giving the head of the broom a greater mass and making it fall. I basically agree with everything Val said before me.

Monday, 17 December 2007 - 6:33 PM EST

I agree with Dan Vu and Joe Duffy.

Monday, 17 December 2007 - 6:52 PM EST

I agree with Meredith and mostly everybody who commented so far. The broom is able to be balanced because of the equation m1d1=m2d2. Since the broom head has more mass it is paired with the shorter distance from the fulcrum and the lighter handle has a longer distance, the broom in general is able to balanced despite the fact that one end is heavier than the other.

Monday, 17 December 2007 - 6:55 PM EST

if you cut the broom in half the would tip towords the side with the head of the broom because it weighs more. this is because the broom handle had more weight because it was longer hand had more mass. I agree with cory

Monday, 17 December 2007 - 7:01 PM EST

Because this is in equilibrium when balanced on that point, you can tell that both sides would weigh exactly the same. I agree with what Meredith said. 

Monday, 17 December 2007 - 7:09 PM EST

according to the quation m1d1=m2d2 the side of the broom with the handle will be alot less then the side of the broom with the straw used for sweeping, so the distance will be shorter on the heavier end, even though the center of gravity isnt where the pivet is.

I agree with dan Vu and Joe duffy

Monday, 17 December 2007 - 8:03 PM EST

If you cut the broom down the middle the side with the brushes will way more.  the center of gravity is more towards the brushes side so if u cut it there the weights would be the same just not down the middle.

Monday, 17 December 2007 - 8:05 PM EST

i completly agree with kevins statement because he uses the equation that we were using and it makes complete sense

Monday, 17 December 2007 - 8:10 PM EST

If you cut the broom in half, the right side would weigh more.  The broom was in equilibrium before you cut it, implying that the right side was heavier because it was closer to the fulcrum.  Even after you cut it in half, the handle's distance from the fulcrum will still be greater.  If you refer to the equation m1d1=m2d2, the side with the larger mass must have a smaller distance to the fulcrum, and vice versa.

Monday, 17 December 2007 - 8:16 PM EST

I disagree with what Matt Harper said.  Since your cutting half of the broom, only the side to the left of the CG will be effected.  Only the handle end's mass will be changed, the broom head's mass will remain constant.

Monday, 17 December 2007 - 8:30 PM EST

The right side would weigh more because even though that is where the broom balances, it just means there is an equal amount of torque is equal so Tnet=0. The equation would be F1r1sin0=F2r2sin0 with F as the force and r as the radius, or the distance from the center of gravity. To have the broom reach equilibrium, and since the right side of the broom would be heavier since it has a greater mass, the radius/distance would have to be greater than the torque on the other side. r1=F2r2sin0/F1sin0. Therefore, the weight wouldn't have changed but the distance did, keeping the broom balanced. So, the right side still has a greater weight.

I agree with what Meredith said.

Monday, 17 December 2007 - 8:30 PM EST

I agree with what Dan said.

Since the center of gravity is not in the middle of the broomstick it evens it out. m1d1=m2d2 makes this true. Because the center of gravity is more towards the side with the brush on it, it would weigh the same as the other side.  This is because the broomstick also has mass that will even out when the center of gravity is in the right place.

Monday, 17 December 2007 - 8:35 PM EST

If the broom was cut at the center of gravity the right side would weigh more. The right side has less distance than the left there for it must have a larger mass because of the equation m1d1=m2d2

Monday, 17 December 2007 - 8:39 PM EST

I agree with allison who in turn agrees with kevin and dan so i agree with them too.

Monday, 17 December 2007 - 8:40 PM EST

That Rob Vucelich kid is so profound. He speaks with the wisdom of the ancients.

Monday, 17 December 2007 - 8:55 PM EST

The side with the broom head would weigh more, because it is closer to the center of gravity, meaning the corresponding distance in the equation m1d1=m2d2 would be smaller than d1, so m2 would have to be greater than m1. 

Monday, 17 December 2007 - 9:05 PM EST

I'd say that the heavier en piece of the broken broom would be the end with the broom head. i say this because when they were connected, that end directed the center of gravity towards that end of the broom.

Monday, 17 December 2007 - 9:05 PM EST

i agree with everyone except matt harper.... 

Monday, 17 December 2007 - 9:06 PM EST

ha... pat... lol i did the same... i agree completely with your answer....

Monday, 17 December 2007 - 9:07 PM EST

I dont have my packet, but from looking at the responses I agree with Kevin and Corey's.

Monday, 17 December 2007 - 9:43 PM EST

The broomstick balances or evens out because the center of gravity is not in the middle. The center of gravity is more towards the side with the brush on it, it would weigh the same as the other side.  The broomstick also has a even mass that will balance the center of hte gravity. 

i agree with brittany burns

Monday, 17 December 2007 - 10:06 PM EST

I agree with Rob.

Monday, 17 December 2007 - 10:46 PM EST

The handle of the broom would weigh less than the end of the broom because even if they're balanced it doesn't mean they have the same weight. They follow the equation m1d1=m2d2

Monday, 17 December 2007 - 10:51 PM EST

The part of the broom that would weigh the most would be the part that has the actual broom and its brushes.  The reason for this is due to the equation m1d1=m2d2.  the actual broom is closer to the fulcrum than the handle which means in order to blance the equation the mass of the actualy broom must be greater. 

Monday, 17 December 2007 - 10:54 PM EST

I agree with Kevin Dougherty.  I feel that he addressed a good point of how the object is in equilibrium and that once you cut it, if you use the equation m1d1=m2d2, the mass must be greater if the distance is smaller.

Monday, 17 December 2007 - 11:11 PM EST

The head of the broom would weigh more than the handle of the broom because the distance from the head of the broom to the fulcrum is shorter than the handle of the broom to the fulcrum. Because the center of gravity is not on either side, the equation for this situation is m1d1=m2d2, which justifies why the head of the broom would weigh more. I agree with Joe Duffy's statement.

Monday, 17 December 2007 - 11:26 PM EST

I agree with what joe duffy said, that the bottom part of the broom would weigh more than the handle. The reason why the broom balances at the center of gravity is because of the distances from either end of the broom the CG compared with each mass. This is shown in the equation, m1d1=m2d2. If the center of gravity were in the middle of the broom the parts would weigh the same because that would mean that the masses and distances of both sides of the broom were the same. Since the center of gravity is closer to the bottom part of the broom, that part would have a larger mass and weight.

Tuesday, 18 December 2007 - 1:15 AM EST

The center of gravity would not be half the length of the broom due to the head of the broom which weights much more than the stick. While using the equation m1d1=m2d2 it's clear that the mass of the broom head and the stick are different due to the difference between the distances to the center of gravity. Because the head is closer to the center of gravity it must have more mass, which in turn means more weight, which also means that the stick must have less mass because it is farther away from the center of gravity meaning less mass in turn meaning less weight.

I agree with Allison completely. (block 4)

Tuesday, 18 December 2007 - 7:25 AM EST

The broom with the sweeping part would weigh more than the handle part.

Because of the center of gravity, where it is located, does not mean both sides weigh the same either.

According to m1d1=m2d2, it balances out though

Tuesday, 18 December 2007 - 7:26 AM EST

I think each end has the same weight because you are balancing the broom with more distance on the one side, so the weight will equal the side of the broom with the other side.

Tuesday, 18 December 2007 - 7:26 AM EST

i agree with dan, i think he thought the same thing that i thought, in that the broom ends are not actually equal in weight, but balance out due to where the center of gravity it.  Also he mentioned m1d1=m2d2.  this is the formula i mentioned that made sense for the problem.

Tuesday, 18 December 2007 - 7:27 AM EST

i agree with allison

Tuesday, 18 December 2007 - 11:10 AM EST

If the broom was cut at the center of gravity, then the part at the right (the head of the broom) has greater weight than the part at the left (the handle of the broom) because the formula for torque is force times distance times the sin of the angle formed by the force and displacement vectors. Because the handle spans a greater distance than the broom head, the force must be greater on the head than on the handle. Therefore, the head of the broom would weigh more.

Tuesday, 18 December 2007 - 11:12 AM EST

I agree with Gina Ranalli (Honors Block 4) and I think that she answered the question well.