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aditya bhardwaj Grade: 11
        

a bloack of mass under root 3kg is resting on a horizontal plane (coefficient of static friction =1/2under root3 ) . A foce is applied to the block at an angle of 60 . The minimum magnitude of force for which the block begins to slide is (g =10m/s square)      please explain

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

taking two components of force seperatly fcos@ and f sin@  


 fcos@ will become driving force and


  (mg-fsin@) will be the normal reaction acting on the body


  we know, friction force= u.normal reaction ,u=cofficient of friction


for minimum force fcos@=friction force 


    f=20/3 N


 

6 years ago
Xyz
13 Points
										
Normal reaction:Fsin60+mg
Root3(F/2+10). 
As frictional force= u.N
u=coefficient of friction),N = normal reaction
 
Fcos60>(or equal)(F+20)/4
On solving, min F =20N
 
yesterday
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