MegaSack DRAW - This year's winner is user - rgwb
We will be in touch
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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.
how depressing i have no idea how to do it. 🙁
About 15.7 kilos?
how?
My guess is 0 😀
5.5kg?
27.43 Newtons.
42. Same as the meaning of life, not 42kg.
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.
92
Is slope any form of conveyor belt? This would naturally change the calculations.
Can we at least get the units of force correct.
kg = mass.
Is my answer similar to any of your mates?
252.76 N
CHB is the only one to show workings so marks awarded even if it's wrong.
I'm with amt on this one.
So if I lean a book against the left hand side of my iMac - there is a force on the right hand side??? Puzzled.
I'd imagine you're looking at a cosine of 20 degrees not 2/9ths...?
i think 340.5N
106.4N
460.44N by B
386.77N by A
73.67N
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😀
I new there was a 2 in the answer.
Ah, wrong quadrant
[url= 
]No. There is an Elephant in the way.[/url]
<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 [s]you[/s] 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>
<engineer> [b]So about 15.7kg [/b]</engineer>
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>
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'
<bike mechanic> my nuber 3 hammer,an M6 bolt and some blue threadlock <bike mechanic>
10 years ago I would have understood that (and maybe even worked it out) Quite worrying how much I've forgotton...
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...?
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<field engineer>[b] Hmmm 15.7kg eh? Well lets add 20% to be sure...erm thats...oh bugger, lets call it 19kg.[/b]</field engineer>
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.
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.
Zero, second word of the question.
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 = 70.71gResolving 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
This looks pretty good, cheers.
You're confusing mass with weight. from the information given, the objects have no weight, only mass.
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.
Why? In that case the question is worded incorrectly and should say 'weighs' not 'has a mass of'
Why? In that case the question is worded incorrectly and should say 'weighs' not 'has a mass of'
But yes I'm sure everyone thinks you're very clever.
I'm taking this back lol.
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I cba to do this. It's not hard, there's just a lot of it. Not the same thing!
overcomplication factor
It says the block is smooth
are the surfaces frictionless
You're confusing mass with weight. from the information given, the objects have no weight, only mass.
Who, me? You'll note my solution uses 'g' until the last line, so substitute whatever value for 'g' you like there (it can even be in zero gravity if you like).
In that case the question is worded incorrectly and should say 'weighs' not 'has a mass of'
Not really. The intrinsic property is mass. You only have to consider weight if you want to know the force (as in this case you do). From an engineering perspective it's perfectly normal to define stuff in terms of mass in the earth's gravitational field and expect to know force.
How about an easier one for you:
A block of mass 52.3kg sits on a flat surface perpendicular to the earth's gravitational field. What force does the block exert on the surface?
A block of mass 52.3kg sits on a flat surface perpendicular to the earth's gravitational field. What force does the block exert on the surface?
Woh woh woh woh, how far from the centre of the earth is it?
Woh woh woh woh, how far from the centre of the earth is it?
About 6371km 😉
Its like an episode of The Big Bang Theory on here 😉
2 things -
There are some clever people on here. I wouldn't have known where to even start.
Why?
There are some clever people on here. I wouldn't have known where to even start.
I was going to suggest it's applied maths A level (back when such exams were hard 😛 ), but realistically the method I used is probably actually 1st year degree engineering maths - I do remember being taught new methods to solve the same problems and wondering why we'd not been taught them before.
aracer is correct, I'm doing a first year engineering degree module, it's one of the questions from that.
The correct answer is - just build one and measure it.
Oliver...is that how they make space shuttles n formula 1 cars? Build one and see what happens?
A space shutlle doesn't need 92kg of scrap material and some bricks under a sheet of (very polished) wood though. I guess the space programme built lots and lots of prototypes and tested them to destruction - if you took the time to just work everything out from first principles... hmm probably wouldn't have made it out of the stone age.
I love these problems though it's like - yeah we've weighed everything precisely, measured the slope - but using a load cell against the last force? - no, that's simply not cricket...
Yeah, this. There's a massive scrapyard with about 1000 slightly strange looking, bent space shuttles just outside NASA's back gate.I guess the space programme built lots and lots of prototypes and tested them to destruction
Now, what if two baby Robins, with a rope made out of Coconut fibres ......
I guess the space programme built lots and lots of prototypes and tested them to destruction
It's how they make motorway bridges as well. They just throw them together and then some old gaffer in a cap comes along and squints at it and says "yeah, that'll be fine". Sometimes he pushes on the big concretey bits to see if they wobble or not.
Engineering really is that easy. It's just people like aracer who try to make it complicated with "maths" and "equations" and other such nonsense.
Engineering really is that easy. It's just people like aracer who try to make it complicated with "maths" and "equations" and other such nonsense.
Someone somewhere built a motorway bridge from substandard materials. just to see if it would stay up, or blew one up with a petrol tanker to see what would happen. They probably enjoy their job.
Yeah, this. There's a massive scrapyard with about 1000 slightly strange looking, bent space shuttles just outside NASA's back gate.
I've seen Neil Armstrong crashing that practice lander - NASA have a hoot. I'm sure they let 3M blow up lots of materials on their behalf.
I got 340N. I could be wrong, it's been a while and I may have missed something but here's how I went about it. Someone else came up with the same number which makes me a little more confident.
Imagine free body diagram. Where block A meets slope there is a reaction Ra, perpendicular. That can be split into hori and vert components (vectors) Rah and Rav. The assumption I've made I'm not totally sure about is that Rav is equal to force on block due to gravity, Ma.g.
Likewise for where block A meets block B there is a reaction perpendicular, Rb. Split this into hori and vert components and again assuming vert component is equal to force due to gravity on B, Mb.g.
Now, sum of forces in horizontal = 0
0 = F + Rah - Rbh (1)
Rah; tan 20 = Rah/Rav
Assuming Rav = Ma.g = 42 x 9.8
Rah = Ma.g.tan20 = 149.81
Rbh; Assuming Rbv = Mb.g and knowing at 45 degrees Rbv = Rbh
Rbh = 50 x 9.8 = 490
So from (1)
F = Rbh - Rah
= 490 - 149
= 340.19 N
Anyone got any thought on assuming vertical component of reaction force is equal to the force of gravity acting on object?
Anyone got any thought on assuming vertical component of reaction force is equal to the force of gravity acting on object?
That's an incorrect assumption for the bottom block, as it also has the force from the top block acting in a vertical direction on it. In fact it's actually fairly simple - the vertical component of Ra is the weight* of the top block plus the weight of the bottom block (the only forces acting on the top block with a vertical component are gravity and Rb, so trivially the vertical component of Rb is the weight of the top block).
Your method is actually good apart from this issue - all you need to do is:
Rav = (Ma + Mb).g = (42 + 50) x 9.8
FTFY as they say
...and just substitute the change in from there down.
Arguably your method is neater than mine, as you can produce a much simpler answer before doing numbers:
F = Mb.g - (Ma + Mb)*g*tan20
F = (50 - 92*tan20)*9.81
*weight being shorthand for force due to gravity
Cheers Aracer, I was treating the blocks as isolated objects for working out the forces acting on them, forgeting the effect of block b resting on a would impact the forces where block a met the plane.
FWIW This reminds me more of A level applied maths (i did 3 modules pure and 3 of mechanics) than first year uni.
aracer is correct, I'm doing a first year engineering degree module, it's one of the questions from that.
Buy K.A.Stround, worth it's weight in pass marks.
You would probably lose marks for quoting the answer in newtons though, we were always told to convert any answer back tothe same units (or system of units) used in the question. So kgf would be acceptable.
Ohh and it kg not KG or Kg, they really hate that!
^ He means K.A Stroud, and consider this a +1
Ah, the big boys book of maths - another +1 here.