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- anonymous

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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- anonymous

- katieb

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- anonymous

|dw:1343893542262:dw|

- anonymous

x > 45 and friction acts towards Q

- anonymous

i will explain

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- anonymous

|dw:1343894068439:dw|

- anonymous

@thushananth01 Plzz Procede

- anonymous

give me a min

- anonymous

OK

- anonymous

x = \[45^{o}\]

- anonymous

hw

- anonymous

|dw:1343894414685:dw|
mgsinx=1Ncosx
g=10m/s

- anonymous

ok.......... This is a multiple choice correct answer there is one more option which is correct..

- anonymous

x=45

- anonymous

|dw:1343894824349:dw|

- anonymous

friction?

- anonymous

x = 45
x > 45 and friction acts towards p
x > 45 and friction acts towards Q
x < 45 and friction acts towards Q

- anonymous

If I take g=10. sinx = cosx only when x=45.

- anonymous

ok.....1 st opt ion is correct .......i1 more we want

- anonymous

If g=9.8. sinx < cos x. So to make it stationary I would need friction towards P.

- anonymous

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.

- anonymous

@Ishaan94 in the question it is said that g =10m/s^2 i forgot to mention

- anonymous

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.

- anonymous

thxxx

- ghazi

m*g*sin(x)= 1* cos (x)substituting the value of m and g will give x= 45

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