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anonymous
 3 years ago
In the given figure all the surfaces are smooth. The ratio of forces exerted by the wedge on the mass M WHEN F is not applied and when the force F is applied such that M is at rest with respect to the wedge is 
Answers options1,1:2,sec^2 theta,cos^2 theta.
anonymous
 3 years ago
In the given figure all the surfaces are smooth. The ratio of forces exerted by the wedge on the mass M WHEN F is not applied and when the force F is applied such that M is at rest with respect to the wedge is  Answers options1,1:2,sec^2 theta,cos^2 theta.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350498047651:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350584039429:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0could you explain a bit how the FBD is drawn with Normal Reaction as mg/cos theta

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sure... there are only two forces on the block in that case... N and mg... the horizontal component of N must be F and the vertical component must be mg

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh . yes i have not got the logic clear.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah whoops, I see why you're asking..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I labelled the theta incorrectly...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350585942860:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there now it works...:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1350586072251:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But what about M at rest ? Is the Normal Reaction force mg/cos theta is sufficient to negate mg sin theta component. I am not so sure whether I have got the concept right. Could you please elaborate. I have got the values right.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0After a second thought it struck me that the wedge is pitted against the Mass M and there is as such no component as mg sin theta...am i correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0m is at rest with respect to the ramp when it's accelerating with the ramp...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I mean, you could decompose mg into vectors along the ramp or into/out of the ramp, but it would be pretty pointless...
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