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Hey goys, working on some takehome physics shit and wondering

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Thread replies: 26
Thread images: 5

File: takehome.png (56KB, 1001x621px) Image search: [Google]
takehome.png
56KB, 1001x621px
Hey goys, working on some takehome physics shit and wondering if anyone would mind working this one out and explain their stuff to me? I'm trying to do it, but this one is more complicated than any I've done previously. Thanks in advance
>>
Keeping this bumped. Will post if I get anywhere significant on this.
>>
maybe study more?
>>
>>269063
I've been studying this stuff a lot, it just isn't coming easily to me.
>>
>>269040
Where specifically are you struggling, the hint given is pretty helpful imo.
>>
>>269075
I'm just unfamiliar with these types of problems and am having trouble wrapping my head around it. I was hoping someone could walk me through it and show me the logic behind each of their steps, I'll show what I have so far in my next post.
>>
First, we figure out what the maximum
acceleration that m2 can take before it slides
out from under m3. This means that a must be
less than or equal to the force of static
friction. The force of static friction
between m3 and m2 is mu_s * m3 * g * cos23, right?
Which means a is <= 12.18m/s^2
>>
I just realized that is wrong, the number I gave isn't a, it's just Fs, so 12.18n, which means the max acceleration going to be that divided by the mass... In this case is that going to be 12.18/(m2 + m3)?
>>
>>269081
12.18 is the force of friction, not the acceleration, also do you know how to make a free-body diagram?
>>
>>269083
no just m3, you are only applying force to m3, make a free-body diagram
>>
>>269083
And again, I realize that both masses would only apply if I'm talking about the friction between m2 and the ground. So it's just 12.18/1.5 = 8.12m/s^2
>>
>>269089
That is correct
>>
>>269088
>>269084
>>269090
I'm working on the free-body right now, I'll post it in a second.
>>
File: freebadbody1.png (49KB, 1859x766px) Image search: [Google]
freebadbody1.png
49KB, 1859x766px
I think that's right..
>>
>>269093
I meant to cross out m2g on the last one as I divided that one into components as well.
>>
Anyways... once I've got my a_max (8.12m/s^2) isn't this problem as simple as setting sigmaF/Mtot = to it and solving for m1?
>>
File: physics.png (18KB, 777x427px) Image search: [Google]
physics.png
18KB, 777x427px
>>269093
This is the diagram I made
>>
File: mayberightkek.png (17KB, 1288x202px) Image search: [Google]
mayberightkek.png
17KB, 1288x202px
>>269098
Beyond that, if it is accelerating slowly enough that m3 and m2 stay together, can't I consider them a single object and only consider the friction between the ground and m2? what I mean is pictured:
>>
>>269099
Shit, I forgot to consider friction for m1... Besides that, does what I have look right?
>>
>>269103
>>269104
This appear to be correct, I haven't spotted anything wrong with it
>>
File: thesolutionmaybekek.png (22KB, 1359x291px) Image search: [Google]
thesolutionmaybekek.png
22KB, 1359x291px
>>269106
Pictured is what I arranged accounting for all the friction, but then my m1 would end up being negative.. I'm not sure what I've done wrong here.
>>
>>269108
I'm not really good at physics either, but are you sure m2 and m3 can be treated as the same object?
>>
>>269110
I'm not 100%, but it makes sense. If the acceleration is such that they stick together, then the only thing to consider is the gravity of both ((m2+m3)gsintheta) and the kinetic friction from m2 ((m2+m3)gcostheta*mu_k).
>>
>>269114
In that case, do
8.12 x (m2+m3)
and you get the force of the two blocks, then
divide by gsin(65) to find the mass
>>
>>269119
I'm not sure exactly what you're doing there, sorry..
>>
Okay holy carp, after working on this some more I got that m1 is 100kg. If anyone is still up that actually worked this out (or wouldn't mind doing it quickly) would you tell me if that's right? The math works out, it just seems so strange...
Thread posts: 26
Thread images: 5


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