Engineering Mechanics - PKRB: Force and Animation - Discussion
Discussion Forum : PKRB: Force and Animation - General Questions (Q.No. 2)
2.
A clown, mounted on stilts, loses his balance and falls backward from the position, where it is assumed the 
= 0 when 
= 07deg;. Paralyzed with fear, he remains rigid as he falls. His mass including the stilts is 80 kg, the mass center is at G, and the radius of gyration about G is kG = 1.2 m. Determine the coefficient of friction between his shoes and the ground at A if it is observed that slipping occurs when = 30°.
= 0.833 = 0.468 = 0.243 = 0.400Discussion:
1 comments Page 1 of 1.
Radhika said:
1 decade ago
We know that, F=coefficient of friction * R
R= normal force = (80*2.5*9.81) =1962 N
F = Force = (80*9.81) N
hence,
coefficient of friction = F/R = (80*9.81)/(80*2.5*9.81)
= 0.4 ans
R= normal force = (80*2.5*9.81) =1962 N
F = Force = (80*9.81) N
hence,
coefficient of friction = F/R = (80*9.81)/(80*2.5*9.81)
= 0.4 ans
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