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Author Topic: Who is up for a challenge?  (Read 37663 times)

Offline erickejah

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« Reply #25 on: 01/06/2009 04:57:33 »
+2
 

Offline Chemistry4me

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« Reply #26 on: 01/06/2009 05:03:13 »
No. Pythagoras. 
 

Offline erickejah

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« Reply #27 on: 01/06/2009 05:06:28 »
I dont get it. I think that I need a lecture about rotating objects,, no two just in case.
 

Offline Chemistry4me

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« Reply #28 on: 01/06/2009 05:08:53 »
√(7.072 + 22) = 7.347 m
 

Offline erickejah

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« Reply #29 on: 01/06/2009 05:14:16 »
:) sweet. thanks now I feel less stupid

 :D now I better go to sleep some :P
 

Offline Chemistry4me

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« Reply #30 on: 01/06/2009 05:17:39 »
Buenas noches
 

Offline syhprum

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« Reply #31 on: 01/06/2009 06:31:42 »
D=GT^2/2........
Time for stone to fall 2.45 meters = ((2.45*2)/9.8)^0.5 = 0.7071 secs
F=MV^2........
Velocity of stone when string breaks = ((2*25)/.5)^0.5 = 10 M/sec
Therefore distance travelled before grounding = 7.07 meters.

calculation completed 17.55 31/05/09

My first attempt at a calculation not being able to remember the basic physics I used Kepler instead of Newton and somehow a factor Pi crept in. 
 

Offline syhprum

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« Reply #32 on: 01/06/2009 06:36:32 »
Then again I failed to read the question and omitted the Pythagoras bit!
 

Offline Chemistry4me

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« Reply #33 on: 01/06/2009 06:52:05 »
So you think this thread is worth continuing, or should it die? :D
If in the affirmative, I shall post another one next weekend.
 

Offline syhprum

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« Reply #34 on: 01/06/2009 07:39:22 »
It is surprising that out of the four answers submitted by well educated corresponds none were correct despite having access to Google and Wiki.
Most did not read the question and some like me did not remember the basic physics or made arithmetical errors.
I wonder how a group of Chinease or Japanese students would have done.
Please find some more problems like these to keep us on our toes and stop us worrying about black holes and dark matter etc.
 

Offline Chemistry4me

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« Reply #35 on: 01/06/2009 07:54:50 »
stop us worrying about black holes and dark matter etc.
There's nothing better than getting back to the basics!  :)
 

Offline erickejah

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« Reply #36 on: 01/06/2009 13:21:51 »
 

Offline LeeE

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« Reply #37 on: 01/06/2009 15:47:30 »
Lol - I get 4.994 metres
 

Offline syhprum

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« Reply #38 on: 01/06/2009 19:20:08 »
If we want greater precision allowance can be made for the reduced radius of the path that the stone takes, this is reduced to 1.961 meters by sagging and and feeds into the last part of the calculation as to how far from the operators feet the stone grounds.
This reduces the distance to 7.338 meters
« Last Edit: 01/06/2009 19:26:20 by syhprum »
 

Offline erickejah

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« Reply #39 on: 02/06/2009 00:56:47 »
:) sweet. thanks now I feel less stupid

 :D now I better go to sleep some :P

Hold on, Pythagoras theorem can be only use in right triangles how do we now that the paths of rock are perpendicular?
 

Offline RD

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« Reply #40 on: 02/06/2009 01:19:41 »
Gratuitous photo to demonstrate my lightning reflexes (and stupidity)...




I took this photo by sticking the camera lens through the safety cage,
(if the guy in the skirt had mistimed his throw I would have had to have the camera surgically removed from my skull).

BTW the weight here is 28lbs, they also fling a 56lb version.
« Last Edit: 02/06/2009 01:54:13 by RD »
 

Offline syhprum

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« Reply #41 on: 02/06/2009 05:42:46 »
erickejah

It is assumed that the circular path traced by the stone before its release is horizontal, if it was not that would be a whole new ball game.
In the case of the 'Highlands games competitor' the same circle would not be horizontal and the 'hammer' would be released when it was moving up.
 

Offline Chemistry4me

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« Reply #42 on: 02/06/2009 06:15:28 »
It's looks like it's about to knock him out.




 

Offline Chemistry4me

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« Reply #43 on: 02/06/2009 11:04:20 »
I might be away for the next few days so I'll post another one now:

Some painters are working in a warehouse. They have a uniform plank which is resting on two supports. The plank is 4 m long. It has a mass of 22 kg. The two legs that support the plank are 0.5 m from either end.



Calculate the support force on the plank at A if a painter of mass 60 kg sits 0.75 m from A and another painter of mass 75 kg sits at a distance of 0.8 m from B.
 

Offline Chemistry4me

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« Reply #44 on: 06/06/2009 07:50:51 »
No one wants a go?
 

Offline syhprum

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« Reply #45 on: 06/06/2009 08:52:26 »
I am very busy at the moment with holiday preparations but the problem seems pretty simple, just calculate the proportion of painters weight that falls on each support which will be partitioned in accordance to the ratio of their distances from the supports and the contribution of the plank itself and add them together

PS

Why did you not have some heavy equipment sitting right at the end of the plank, that would have made it more interesting ?.
« Last Edit: 06/06/2009 09:26:58 by syhprum »
 

Offline Chemistry4me

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« Reply #46 on: 08/06/2009 08:17:50 »
The answer is:

Taking moments about B.
FA x 3 = 588 x 2.25 + 735 x 0.8 + 215.6 x 1.5 = 2234.4 N
FA = 744.8 N
 

Offline erickejah

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« Reply #47 on: 10/06/2009 01:30:06 »
cool, another one please
 

Offline RD

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« Reply #48 on: 10/06/2009 21:52:00 »
It's looks like it's about to knock him out.

That is more likely in the "weight for height" contest: (throw 56lbs weight over polevault-type bar)...


 

Offline turnipsock

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« Reply #49 on: 11/06/2009 00:23:28 »
I thought if you threw anything horizontaly then gravity would be 9.81 m/s/s. The problem seems to boil down to the earth being bannana shaped. I don't think the mass has much to do with things.

I've been in a few highland games and the people don't come across as people that do these kinda sums. They are more worried about avoiding the highland dancers and getting the noise of the bagpipes out of their ears.
 

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« Reply #49 on: 11/06/2009 00:23:28 »

 

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