anonymous
  • 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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1343893542262:dw|
anonymous
  • anonymous
x > 45 and friction acts towards Q
anonymous
  • anonymous
i will explain

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More answers

anonymous
  • anonymous
|dw:1343894068439:dw|
anonymous
  • anonymous
@thushananth01 Plzz Procede
anonymous
  • anonymous
give me a min
anonymous
  • anonymous
OK
anonymous
  • anonymous
x = \[45^{o}\]
anonymous
  • anonymous
hw
anonymous
  • anonymous
|dw:1343894414685:dw| mgsinx=1Ncosx g=10m/s
anonymous
  • anonymous
ok.......... This is a multiple choice correct answer there is one more option which is correct..
anonymous
  • anonymous
x=45
anonymous
  • anonymous
|dw:1343894824349:dw|
anonymous
  • anonymous
friction?
anonymous
  • 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
  • anonymous
If I take g=10. sinx = cosx only when x=45.
anonymous
  • anonymous
ok.....1 st opt ion is correct .......i1 more we want
anonymous
  • anonymous
If g=9.8. sinx < cos x. So to make it stationary I would need friction towards P.
anonymous
  • 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
  • anonymous
@Ishaan94 in the question it is said that g =10m/s^2 i forgot to mention
anonymous
  • 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
  • anonymous
thxxx
ghazi
  • ghazi
m*g*sin(x)= 1* cos (x)substituting the value of m and g will give x= 45

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