Zero, second word of the question.
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One for all the armchair engineers, physicists, mathematicians, etc. etc.
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Posted 5 months ago #
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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 = 162NThis looks pretty good, cheers.
Posted 5 months ago # -
You're confusing mass with weight. from the information given, the objects have no weight, only mass.
Posted 5 months ago # -
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.
Posted 5 months ago # -
Why? In that case the question is worded incorrectly and should say 'weighs' not 'has a mass of'
Posted 5 months ago # -
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.
Posted 5 months ago # -
Posted 5 months ago # -
I cba to do this. It's not hard, there's just a lot of it. Not the same thing!
Posted 5 months ago # -
overcomplication factor
It says the block is smooth
are the surfaces frictionless
Posted 5 months ago # -
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?
Posted 5 months ago # -
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?
Posted 5 months ago # -
Woh woh woh woh, how far from the centre of the earth is it?
About 6371km
Posted 5 months ago # -
Its like an episode of The Big Bang Theory on here
Posted 5 months ago # -
2 things -
There are some clever people on here. I wouldn't have known where to even start.
Why?Posted 5 months ago # -
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.
Posted 5 months ago # -
aracer is correct, I'm doing a first year engineering degree module, it's one of the questions from that.
Posted 5 months ago # -
The correct answer is - just build one and measure it.
Posted 5 months ago # -
Oliver...is that how they make space shuttles n formula 1 cars? Build one and see what happens?
Posted 5 months ago # -
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...
Posted 5 months ago # -
I guess the space programme built lots and lots of prototypes and tested them to destruction
Yeah, this. There's a massive scrapyard with about 1000 slightly strange looking, bent space shuttles just outside NASA's back gate.Now, what if two baby Robins, with a rope made out of Coconut fibres ......
Posted 5 months ago # -
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.
Posted 5 months ago # -
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.
Posted 5 months ago # -
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?
Posted 5 months ago # -
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
Posted 5 months ago # -
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.
Posted 5 months ago # -
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!
Posted 5 months ago # -
^ He means K.A Stroud, and consider this a +1
Posted 5 months ago # -
Ah, the big boys book of maths - another +1 here.
Posted 5 months ago #
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