[RD]

Gratuitous photo to demonstrate my lightning reflexes (and stupidity)...

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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.

[syhprum]

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.

[Chemistry4me (OP)]

It's looks like it's about to knock him out.


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[Chemistry4me (OP)]

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.

[Chemistry4me (OP)]

No one wants a go?

[syhprum]

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 ?.

[Chemistry4me (OP)]

The answer is:

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

[erickejah]

cool, another one please

[RD]

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)...

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[turnipsock]

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.

[Chemistry4me (OP)]

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.

[erickejah]

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
 
 []

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  []  [] []

[Chemistry4me (OP)]

I did not get 3.28 N.

[erickejah]

Can you show me the path, dont I need the AoA, or the time in which it changes?

[Chemistry4me (OP)]

Have you done your research?

[erickejah]

kind of,,

[Chemistry4me (OP)]

It look's something like this:

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

[erickejah]

nice

[erickejah]

another one?

[Chemistry4me (OP)]

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 45^o to the ground. There is a 35 N friction force on the trolley. Calculate the size of the trolley's acceleration.