Callisto
  • Callisto
A wedge of mass M is fixed on the horizontal ground. A block of mass m slides down the wedge's smooth surface as shown. What is the normal reaction exerted on the wedge by the ground?
Physics
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SOLVED
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katieb
  • katieb
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Callisto
  • Callisto
I thought it was Mg + mgcosθ but it is was wrong.. Can anyone explain the correct one?
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anonymous
  • anonymous
I think it is mg + Fsin \[\theta\]
Callisto
  • Callisto
no, i'm very sure it is not sinθ...

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anonymous
  • anonymous
F= frictional force
Callisto
  • Callisto
smooth plane.. = no friction ...
amistre64
  • amistre64
not that im an expert at this, but from looking at the web it looks like Fn = mg cos(t)
Callisto
  • Callisto
not correct neither :(
amistre64
  • amistre64
can you give a reason as to why it aint correct? i dont even know how to think thru the problem to determine a solution
Callisto
  • Callisto
Nah.. the reason is the answer to the question is not like this, and i don't know how the answer comes.. If i knew it , i wouldn't posted it here..
Callisto
  • Callisto
The answer to this question is Mg+mg(cosθ)^2... can anyone explain it???
anonymous
  • anonymous
hey callissto can u tell me how u got Mg + Mgcostheta
Callisto
  • Callisto
resolve the component mg
anonymous
  • anonymous
show me ur work plz
Callisto
  • Callisto
|dw:1332686392395:dw|
anonymous
  • anonymous
|dw:1332686941820:dw| Ff is frictional force so wat now is normal force?
anonymous
  • anonymous
even i also agree with @Callisto i m also getting the same ans
Callisto
  • Callisto
That's what I thought |dw:1332687170257:dw|
anonymous
  • anonymous
its a smooth surface and here u can put pseudo force ok instead of ff
anonymous
  • anonymous
N=? WAT AS the block isnt moving up or dwn(in the vertical plane, of slope wat is N?
Callisto
  • Callisto
mgcosθcosθ....
anonymous
  • anonymous
never bring in pseudo forces when it is not present for solving sums actual forces are more than enough @calli mgcos@ thats it
anonymous
  • anonymous
i think cos\[\theta\] cames from ome where like i had hint but not sure how to get dat 1-sin^2\[\theta\]=cos^2\[theta]\
Callisto
  • Callisto
the vertical component of N ... :(
Callisto
  • Callisto
N = mgcosθ Vertical component of N =(mgcosθ)cosθ = mg(cosθ)^2...
anonymous
  • anonymous
r u sure coz i also dunno anyway ask ur teacher den
anonymous
  • anonymous
normal force is mgcos@ thats all wat else do u have to add?
Callisto
  • Callisto
actually.. my friend just sent me her workings ... the answer must be correct, because it's the answer from the exam. authority... :(
Callisto
  • Callisto
Mg
Callisto
  • Callisto
so R = Mg + mg(cosθ)^2...
anonymous
  • anonymous
oh den find james he is the only hope ...
Callisto
  • Callisto
Thank you very much...
Callisto
  • Callisto
@heena who's james?
anonymous
  • anonymous
wat is the answer given?
anonymous
  • anonymous
uhmm he really good in physc i think he is not online right now but wenever he ll come i ll ask him (: wen u have to show the work i mean ur xam?
Callisto
  • Callisto
Mg + mg(cosθ)^2
Callisto
  • Callisto
This is actually a MC question, so no working is needed to be shown
anonymous
  • anonymous
|dw:1332688273405:dw|
anonymous
  • anonymous
which is ground/
Callisto
  • Callisto
|dw:1332688328574:dw|
anonymous
  • anonymous
? lower thing is ground?
Callisto
  • Callisto
|dw:1332688527311:dw|
anonymous
  • anonymous
there is no other possibillity in that case than normal forces exerted by ground on system=(m+M)g let the m move we dont care abt it
anonymous
  • anonymous
The option is correct, Mg + mgcos^2theta... I am not sure what's tripping you guys?
anonymous
  • anonymous
Why are you considering the small block, just consider the wedge, forces on the wedge.
anonymous
  • anonymous
|dw:1332689347529:dw|
anonymous
  • anonymous
\[N_g = Mg + mg\cos^2 \theta\]
anonymous
  • anonymous
Did I do something wrong?
amistre64
  • amistre64
where does the ^2 come into play?
anonymous
  • anonymous
|dw:1332689488796:dw|
Callisto
  • Callisto
exactly the same as my friend's working :P
amistre64
  • amistre64
|dw:1332689530792:dw|
anonymous
  • anonymous
So, was your friend right?
Callisto
  • Callisto
sure :)
Callisto
  • Callisto
@amistre64 that's the problem, it's not the the block is not at static..
anonymous
  • anonymous
No amistre, the force between the small block and the wedge is a contact force, contact forces act perpendicular to the surface of the contact.
Callisto
  • Callisto
(delete the words 'it's not the')
amistre64
  • amistre64
im sure im wrong about it:) but i still cant make sense of it with whats been presented :/
anonymous
  • anonymous
In case of static small block, you will need to consider the static friction as well.
Callisto
  • Callisto
smooth plane, no friction..
anonymous
  • anonymous
then a glue or nail lol and then yes the whole body would behave as (M+m)
anonymous
  • anonymous
nice...thanks to callisto first time there wer more than seven ppl lookin into the same question in p6 group!
anonymous
  • anonymous
always it is math group that is popular!
Callisto
  • Callisto
Yuup, i was surprised to see too!!
Callisto
  • Callisto
Thank you for all your efforts :)

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