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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

509 Points
10 years ago

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

Xyz
13 Points
4 years ago
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

SIDDHARTH
15 Points
3 years ago
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