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  1. Naked Science Forum
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  3. Physics, Astronomy & Cosmology
  4. Who is up for a challenge?
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Who is up for a challenge?

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Offline RD

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

 [ Invalid Attachment ]


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.

* porridge.jpg (36.68 kB, 251x397 - viewed 679 times.)
« Last Edit: 02/06/2009 01:54:13 by RD »
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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.
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Offline Chemistry4me (OP)

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


 [ Invalid Attachment ]

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Offline Chemistry4me (OP)

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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.

[diagram=465_0]

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.
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Offline Chemistry4me (OP)

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« Reply #44 on: 06/06/2009 07:50:51 »
No one wants a go?
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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 »
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Offline Chemistry4me (OP)

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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
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Offline erickejah

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

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« Reply #48 on: 10/06/2009 21:52:00 »
Quote from: Chemistry4me on 02/06/2009 06:15:28
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)...

 [ Invalid Attachment ]

* bar.jpg (22.28 kB, 186x400 - viewed 615 times.)
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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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Beeswax: Natures petrol tank sealant.

When things are in 3D, is it always the same three dimensions?
 

Offline Chemistry4me (OP)

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« Reply #50 on: 12/06/2009 09:14:31 »
Quote from: erickejah on 10/06/2009 01:30:06
cool, another one please
Alright, see if you can work this one out then.

Phugoid oscillations.
A boy is flying his radio-controlled aeroplane (mass = 3.67 kg) at a speed of 36.0 ms–1 in a straight and level flight at an altitude of 50.0 m. The plane suddenly experiences a region of turbulence, causing it to lift several metres above its original altitude. The amplitude of the vertical oscillations is 4.56 m. Calculate the maximum upward force acting on the plane during the time that the plane is oscillating.
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Offline erickejah

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« Reply #51 on: 13/06/2009 03:45:19 »
F=ma
F=(3.67Kg)(9.8m/s^2)
F=35.966N

PE1=mgh
PE1=(3.67Kg)(9.8m/s^2)(50m)     PE2=(3.67Kg)(9.8m/s^2)(54.67m)   
PE1=1798.3J                              PE2=1962.3J
 
 [:P]

F  = Fx
PE1   PE2

35.966N =  Fx
-------   ----
1798.3J   1962.3J

Fx=39.246

ΔF= Fx-F → 39.246-35.966 = 3.28N

This is my assumption without researching about Phugoid oscillations. I will try to look that theory ASAP to do it right. I wonder how different that answer will be  [???]  [:D] [:D]
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Offline Chemistry4me (OP)

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« Reply #52 on: 13/06/2009 03:52:12 »
I did not get 3.28 N.
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Offline erickejah

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« Reply #53 on: 13/06/2009 23:27:10 »
Can you show me the path, dont I need the AoA, or the time in which it changes?
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« Reply #54 on: 13/06/2009 23:36:20 »
Have you done your research?
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Offline erickejah

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« Reply #55 on: 13/06/2009 23:36:55 »
kind of,,
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« Reply #56 on: 13/06/2009 23:48:02 »
It look's something like this:

 [ Invalid Attachment ]

But the plane experiences an upward force of 3.67 x 9.8 N just to keep it from falling, therefore Fmax(total) = 38.45 N

* Capture.JPG (20.85 kB, 380x233 - viewed 694 times.)
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Offline erickejah

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« Reply #57 on: 13/06/2009 23:59:04 »
nice :)
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« Reply #58 on: 14/06/2009 18:55:07 »
another one?
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« Reply #59 on: 15/06/2009 03:42:42 »
John is sitting in a trolley and Sam pulls him along with a rope. John's mass is 65 kg, the trolley's mass is 11 kg. The tension force in the rope attached to the trolley is 95 N, and the rope is at an angle of 45o to the ground. There is a 35 N friction force on the trolley. Calculate the size of the trolley's acceleration.
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