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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

aditya bhardwaj , 15 Years ago
Grade 11
anser 3 Answers
vikas askiitian expert

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

 

ApprovedApproved
Last Activity: 15 Years ago
Xyz
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
 
Last Activity: 9 Years ago
SIDDHARTH
F=20N Ans
Frictional Force = Fcos60 (begins slide)
u.n= F* ½
1/2root3*(mg+F sin60) = F*1/2
F/4 + 5 = F*1/2
Hence F= 20 N
 
Last Activity: 8 Years ago
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