• This topic has 62 replies, 35 voices, and was last updated 12 years ago by IA.
Viewing 40 posts - 1 through 40 (of 63 total)
  • One for all the armchair engineers, physicists, mathematicians, etc. etc.
  • RealMan
    Free Member

    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.

    nonk
    Free Member

    how depressing i have no idea how to do it. 🙁

    CHB
    Full Member

    About 15.7 kilos?

    nonk
    Free Member

    how?

    futonrivercrossing
    Free Member

    My guess is 0 😀

    porter_jamie
    Full Member

    5.5kg?

    deadlydarcy
    Free Member

    27.43 Newtons.

    sambob
    Free Member

    42. Same as the meaning of life, not 42kg.

    CHB
    Full Member

    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.

    redthunder
    Free Member

    92

    CHB
    Full Member

    Is slope any form of conveyor belt? This would naturally change the calculations.

    deadlydarcy
    Free Member

    Can we at least get the units of force correct.

    kg = mass.

    CHB
    Full Member

    Is my answer similar to any of your mates?

    amt27
    Free Member

    252.76 N

    nonk
    Free Member

    CHB is the only one to show workings so marks awarded even if it’s wrong.

    mogrim
    Full Member

    I’m with amt on this one.

    futonrivercrossing
    Free Member

    So if I lean a book against the left hand side of my iMac – there is a force on the right hand side??? Puzzled.

    portlyone
    Full Member

    I’d imagine you’re looking at a cosine of 20 degrees not 2/9ths…?

    float
    Free Member

    i think 340.5N

    ianpinder
    Free Member

    106.4N

    portlyone
    Full Member

    460.44N by B
    386.77N by A

    73.67N

    IanMunro
    Free Member

    😀

    redthunder
    Free Member

    I new there was a 2 in the answer.

    portlyone
    Full Member

    Ah, wrong quadrant

    Spin
    Free Member
    aracer
    Free Member

    <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 you your 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 = 70.71g

    Resolving 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>

    CHB
    Full Member

    <engineer> So about 15.7kg </engineer>

    aracer
    Free Member

    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>

    CountZero
    Full Member

    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’

    Rorschach
    Free Member

    <bike mechanic> my nuber 3 hammer,an M6 bolt and some blue threadlock <bike mechanic>

    andrewh
    Free Member

    10 years ago I would have understood that (and maybe even worked it out) Quite worrying how much I’ve forgotton…

    bullheart
    Free Member

    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…?

    CHB
    Full Member

    <field engineer> Hmmm 15.7kg eh? Well lets add 20% to be sure…erm thats…oh bugger, lets call it 19kg.</field engineer>

    anto164
    Free Member

    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.

    TheBrick
    Free Member

    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.

    TheFunkyMonkey
    Free Member

    Zero, second word of the question.

    RealMan
    Free Member

    Zero, second word of the question.

    What?

    Resolving in the vertical direction for the top block (which removes N2 from the equation):
    N1*cos45 = 50g
    N1 = 70.71g

    Resolving 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

    This looks pretty good, cheers.

    TheFunkyMonkey
    Free Member

    You’re confusing mass with weight. from the information given, the objects have no weight, only mass.

    RealMan
    Free Member

    Sigh. I think it’s fairly safe to assume you’re on the surface of the earth, and not in space. But yes I’m sure everyone thinks you’re very clever.

    TheFunkyMonkey
    Free Member

    Why? In that case the question is worded incorrectly and should say ‘weighs’ not ‘has a mass of’

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