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A small block of mass 0.1kg lies on a fixed inclined plane PQ which makes an angle x with horizontal.A horizontal force of 1N on the block through its center of mass as shown The block remains stationary if
x = 45
x > 45 and friction acts towards p
x > 45 and friction acts towards Q
x < 45 and friction acts towards Q
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best.shakir
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x > 45 and friction acts towards Q
best.shakir
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i will explain
thushananth01
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|dw:1343894068439:dw|
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@thushananth01 Plzz Procede
thushananth01
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give me a min
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OK
celos
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x = \[45^{o}\]
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hw
celos
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|dw:1343894414685:dw|
mgsinx=1Ncosx
g=10m/s
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ok.......... This is a multiple choice correct answer there is one more option which is correct..
Ishaan94
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x=45
Ishaan94
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Ishaan94
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friction?
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x = 45
x > 45 and friction acts towards p
x > 45 and friction acts towards Q
x < 45 and friction acts towards Q
Ishaan94
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If I take g=10. sinx = cosx only when x=45.
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ok.....1 st opt ion is correct .......i1 more we want
Ishaan94
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If g=9.8. sinx < cos x. So to make it stationary I would need friction towards P.
Ishaan94
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okay. so g must be 10. other option should be the third one.
if x>45. sinx>cosx. and for block to be stationary we need friction towards Q.
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@Ishaan94 in the question it is said that g =10m/s^2 i forgot to mention
celos
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if g =10 m/s^{2} then there is no friction because it arises when there is any tendency of motion. but at x =\[45^{o}\] the net force is zero.
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thxxx
ghazi
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m*g*sin(x)= 1* cos (x)substituting the value of m and g will give x= 45