anonymous
  • anonymous
A block of mass 'm' is placed on a smooth wedge of inclination \(theta\). The whole system is accelerated horizontally so that the block doesn't slip on the wedge. What is the magnitude of force exerted by the wedge on the block?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Can you draw a picture for your question ?
anonymous
  • anonymous
|dw:1377703976149:dw|
anonymous
  • anonymous
you just have to resolve it

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More answers

anonymous
  • anonymous
I can't speak or read English very well.so I didn't understand the question! If you tell your question with mass and physics symbols , i will help you :)
anonymous
  • anonymous
so what u want me to write...? do N= mg cos theta and something like that
Fifciol
  • Fifciol
|dw:1377708059237:dw|when wedge is accelerated towards left block is exerted by fictitious force due to that accelarated frame of reference. Object is in equilibrium so Fcosθ=mgsinθ⇒F=mgtanθ
anonymous
  • anonymous
answer is mg/cos theta
anonymous
  • anonymous
I think this is good for you ;) http://en.wikipedia.org/wiki/Inclined_plane and this picture:
anonymous
  • anonymous
this is not helping me to get the answer..sorry. :|
anonymous
  • anonymous
pseudo force will also be applied here..so please take care of that
AkashdeepDeb
  • AkashdeepDeb
See as the force applied is horizontal! All we have to find is the component of the force ON THE BLOCK!! So we have this... |dw:1377713495492:dw|
anonymous
  • anonymous
so the answer?
AkashdeepDeb
  • AkashdeepDeb
Did you understand this?
anonymous
  • anonymous
yeah..but whats the answer u got after doing all this?
AkashdeepDeb
  • AkashdeepDeb
If you understood it, I want you to find out the answer! I'll help! :)
anonymous
  • anonymous
please tell me if you get mg/cos theta..?
anonymous
  • anonymous
did you?
AkashdeepDeb
  • AkashdeepDeb
I believe the actual answer is mg.sin theta.
anonymous
  • anonymous
nope...then you are wrong somewhere,
anonymous
  • anonymous
well, ask your teacher and then do tell me
AkashdeepDeb
  • AkashdeepDeb
Are you even TRYING to understand this?
AkashdeepDeb
  • AkashdeepDeb
If you understood my explanation above you would understand it and might as well bring up your own answer.!!! :)
anonymous
  • anonymous
first of all you are wrong..ok..i'll solve it by myself ..thanks for the "not so helpful" help
AkashdeepDeb
  • AkashdeepDeb
:)
anonymous
  • anonymous
I asked somebody yesterday and realized that the answer is: \[mg \sin \theta\]
anonymous
  • anonymous
Tell me whether there is "friction" or not.

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