Question #133030

Two smooth spheres P,Q each of radius 25 cm and weighing 500N, rests in a horizontal channel having vertical walls. if the distance between the walls is 90 cm, make calculations for the pressure exerted on the wall and floor at points A,B and C.

Expert's answer

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


Nc=wtanθ=500×43=20003NN_c=\frac{w}{tan\theta}=\frac{500\times4}{3}=\frac{2000}{3}N

contact force at contact of sphere

N=wsinθ=500×53=25003NN=\frac{w}{sin\theta}=\frac{500\times5}{3}=\frac{2500}3N


contact force at point A


NA=Ncosθ=20003NN_A=Ncos\theta =\frac{2000}{3}N


contact force at point B


NB=Nsinθ+W=500+25003=40003NN_B=Nsin\theta+W=500+\frac{2500}{3}=\frac{4000}3N



effective area for pressure


A=πr2=3.14×25×10−4=0.2m2A=\pi r^2=3.14\times25\times10^{-4}=0.2m^2



pressure at point A , B , and C is as following


PA=NAA=20000.6=3.33KPaP_A=\frac{N_A}{A}=\frac{2000}{0.6}=3.33KPa


PB=NBA=40000.6=6.66KPaP_B=\frac{N_B}{A}=\frac{4000}{0.6}=6.66KPa


PC=NcA=20000.6=3.33KPaP_C=\frac{N_c}{A}=\frac{2000}{0.6}=3.33KPa




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