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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.