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Home / Doubts and Solutions/JEE/Physics

q 15

Asked by chandanbr6004 25th December 2017, 9:02 AM
Answered by Expert
Answer:
Refer the above diagram for different forces acting on the wooden block as explained below
 
Fp = force due to average hydrostatic pressure = average pressure × area =  (1/2)ρgh × (b×h) = (1/2)×ρ×b×h2 ×g
 
Fpv = vertical component of Fp = (1/2)×ρ×b×h2 ×g sinθ
 
Fph = horizontal component of Fp = (1/2)×ρ×b×h2 ×g cosθ
 
Fg = weight = mg
 
Fr = normal reaction force = Fph = (1/2)×ρ×b×h2 ×g cosθ
 
Ff = friction force = μ×Fr = μ (1/2)×ρ×b×h2 ×g cosθ
 
By balancing the vertical forces we can write
begin mathsize 12px style F subscript p v end subscript space equals space F subscript f plus F subscript g F subscript p v end subscript minus space F subscript f space equals space F subscript g 1 half rho b h squared g space sin open parentheses theta close parentheses space minus space 1 half rho b h squared g space cos open parentheses theta close parentheses space equals space m g s o l v i n g space f o r space h space w e space w i l l space g e t space h space equals space square root of fraction numerator 2 m over denominator rho b space open parentheses space sin space theta space minus space mu cos space theta space close parentheses end fraction end root end style

 

 

Answered by Expert 25th December 2017, 5:01 PM
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