solution
given data
radius of sphere P and Q =25cm
weight of sphere P and Q =500N
diagram for this question can be drawn as
whole system is in equilibrium so all force in x direction and y direction are equal
contact force at point C
"N_c=\\frac{w}{tan\\theta}=\\frac{500\\times4}{3}=\\frac{2000}{3}N"
contact force at contact of sphere
"N=\\frac{w}{sin\\theta}=\\frac{500\\times5}{3}=\\frac{2500}3N"
contact force at point A
"N_A=Ncos\\theta =\\frac{2000}{3}N"
contact force at point B
"N_B=Nsin\\theta+W=500+\\frac{2500}{3}=\\frac{4000}3N"
effective area for pressure
"A=\\pi r^2=3.14\\times25\\times10^{-4}=0.2m^2"
pressure at point A , B , and C is as following
"P_A=\\frac{N_A}{A}=\\frac{2000}{0.6}=3.33KPa"
"P_B=\\frac{N_B}{A}=\\frac{4000}{0.6}=6.66KPa"
"P_C=\\frac{N_c}{A}=\\frac{2000}{0.6}=3.33KPa"
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