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I think it is mg + Fsin \[\theta\]
no, i'm very sure it is not sinθ...
F= frictional force
smooth plane.. = no friction ...
not that im an expert at this, but from looking at the web it looks like Fn = mg cos(t)
not correct neither :(
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
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..
The answer to this question is Mg+mg(cosθ)^2... can anyone explain it???
hey callissto can u tell me how u got Mg + Mgcostheta
resolve the component mg
show me ur work plz
|dw:1332686941820:dw| Ff is frictional force so wat now is normal force?
even i also agree with @Callisto i m also getting the same ans
That's what I thought |dw:1332687170257:dw|
its a smooth surface and here u can put pseudo force ok instead of ff
N=? WAT AS the block isnt moving up or dwn(in the vertical plane, of slope wat is N?
never bring in pseudo forces when it is not present for solving sums actual forces are more than enough @calli mgcos@ thats it
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]\
the vertical component of N ... :(
N = mgcosθ Vertical component of N =(mgcosθ)cosθ = mg(cosθ)^2...
r u sure coz i also dunno anyway ask ur teacher den
normal force is mgcos@ thats all wat else do u have to add?
actually.. my friend just sent me her workings ... the answer must be correct, because it's the answer from the exam. authority... :(
so R = Mg + mg(cosθ)^2...
oh den find james he is the only hope ...
Thank you very much...
@heena who's james?
wat is the answer given?
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?
Mg + mg(cosθ)^2
This is actually a MC question, so no working is needed to be shown
which is ground/
? lower thing is ground?
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
The option is correct, Mg + mgcos^2theta... I am not sure what's tripping you guys?
Why are you considering the small block, just consider the wedge, forces on the wedge.
\[N_g = Mg + mg\cos^2 \theta\]
Did I do something wrong?
where does the ^2 come into play?
exactly the same as my friend's working :P
So, was your friend right?
@amistre64 that's the problem, it's not the the block is not at static..
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.
(delete the words 'it's not the')
im sure im wrong about it:) but i still cant make sense of it with whats been presented :/
In case of static small block, you will need to consider the static friction as well.
smooth plane, no friction..
then a glue or nail lol and then yes the whole body would behave as (M+m)
nice...thanks to callisto first time there wer more than seven ppl lookin into the same question in p6 group!
always it is math group that is popular!
Yuup, i was surprised to see too!!
Thank you for all your efforts :)