My mates and I keep arguing about this one. I'm not particularly sure to be honest. Got a few very different answers.
No conveyor belts involved.
Chat Forum
One for all the armchair engineers, physicists, mathematicians, etc. etc.
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Posted 5 months ago #
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how depressing i have no idea how to do it.
Posted 5 months ago # -
About 15.7 kilos?
Posted 5 months ago # -
how?
Posted 5 months ago # -
My guess is 0
Posted 5 months ago # -
5.5kg?
Posted 5 months ago # -
27.43 Newtons.
Posted 5 months ago # -
42. Same as the meaning of life, not 42kg.
Posted 5 months ago # -
Logic (please rip appart as its 20 years since I had to work owt like this out!)
Block B pushes in the direction opposite F by 25 kilos.
Block A has a force of 9.3 kilos in the direction of F.
Reason for this is that 90 degrees would be 100% of mass so 20 degrees is 2/9ths of mass (is this right).F = difference of 25kg less the 9.3kg from block A slope.
Probably wrong.
Posted 5 months ago # -
92
Posted 5 months ago # -
Is slope any form of conveyor belt? This would naturally change the calculations.
Posted 5 months ago # -
Can we at least get the units of force correct.
kg = mass.
Posted 5 months ago # -
Is my answer similar to any of your mates?
Posted 5 months ago # -
252.76 N
Posted 5 months ago # -
CHB is the only one to show workings so marks awarded even if it's wrong.
Posted 5 months ago # -
I'm with amt on this one.
Posted 5 months ago # -
So if I lean a book against the left hand side of my iMac - there is a force on the right hand side??? Puzzled.
Posted 5 months ago # -
I'd imagine you're looking at a cosine of 20 degrees not 2/9ths...?
Posted 5 months ago # -
i think 340.5N
Posted 5 months ago # -
106.4N
Posted 5 months ago # -
460.44N by B
386.77N by A73.67N
Posted 5 months ago # -
Posted 5 months ago # -
I new there was a 2 in the answer.
Posted 5 months ago # -
Ah, wrong quadrant
Posted 5 months ago # -
Posted 5 months ago #
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<engineer>Of course the question says the surfaces are smooth, it doesn't say they're zero friction (such surfaces don't exist). As it doesn't give the coefficient of friction the problem is insoluble</engineer>
<mathematician>We'll assume smooth means zero friction. Correct answer, IanMunro (surprised it took so long and so many attempts), though
youyour textbook overcomplicates things by solving in the normal axes and using simultaneous equations - if you turn your reference frames you can do without.Borrowing your diagrams:
Resolving in the vertical direction for the top block (which removes N2 from the equation):
N1*cos45 = 50g
N1 = 70.71gResolving parallel to the inclined plane for the bottom block (which removes N3 from the equation):
F*cos20 = N1*sin25 - 42g*sin20
F = (70.71g*0.4226 - 42g*0.3420) / 0.9397
F = 16.51g = 162N
</mathematician>Posted 5 months ago # -
<engineer> So about 15.7kg </engineer>
Posted 5 months ago # -
So about 15.7kg
<QC>I suppose at least you're in the right order of magnitude, though you're a good 5% out. That's not what we call engineering round here.</QC>Posted 5 months ago # -
My brains have just tried to leak out of my ears trying to make head or tail of that lot.
As Ian Dury once so eloquently put it 'there ain't 'alf some clever buggers'Posted 5 months ago # -
<bike mechanic> my nuber 3 hammer,an M6 bolt and some blue threadlock <bike mechanic>
Posted 5 months ago # -
10 years ago I would have understood that (and maybe even worked it out) Quite worrying how much I've forgotton...
Posted 5 months ago # -
If only we knew of someone qualified to answer such a puzzle? Maybe with membership of an organisation dedicated to dealing with the intricate nature of such mysteries...?
Posted 5 months ago # -
<field engineer> Hmmm 15.7kg eh? Well lets add 20% to be sure...erm thats...oh bugger, lets call it 19kg.</field engineer>
Posted 5 months ago # -
If you're a proper physicists, mathematician or Engineer, you'd have let a computer do all that maths for you..
Sitting there doing derivations is less cost effective than letting a computer do it.
Posted 5 months ago # -
anto164 - Member
If you're a proper physicists, mathematician or Engineer, you'd have let a computer do all that maths for you..Why bother for something like this?
Sitting there doing derivations is less cost effective than letting a computer do it.
There is no need to use any calculus to solve this problem.
Posted 5 months ago #
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