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Rain falls into a roof of 1.5m2 inclined to horizontal at an angle of 30 degree. The drops of rain strikes the vertical roof with the vertical velocity 3m/s where 2.5*10-2 m3 water is collected in one minute. If the velocity of the raindrop after impact is 0, Find the vertical force exerted by the rain drop on the roof, and the normal pressure applied on the roof.
Rain falls vertical on the roof. See attached figure below. The initial vertical speed `V` of drops decomposes into a normal to the roof component `V_n` and a parallel to the roof component `V_p` . Only the normal component of the speed created pressure on roof (pressure is always perpendicular to the surface).This happens because of the change in the vertical component of the rain momentum.
`P =F_n/S =(m*V_n)/S `
where `V_n =V*cos(alpha) =3*cos(30) =2.598 m/s`
Because in the problem it is given only the volume of water collected in one minute, only the medium pressure over one minute (or one second) can be computed.
`m =rho*V =1000*2.5*10^-2 =25 (kg)/min`
`P =(25*2.598)/1.5 =43.3 N/(m^2*min) =43.3/60 N/(m^2*s) =0.72 N/(m^2*s)`
The normal force on the roof is just
`F_n = P*S =0.72*1.5 = 1.08 N/s =1.08*60 N/min =64.8 N/min`
Since the normal force on the roof is just the normal component of the initial vertical force we have for the vertical force the value
`F =F_n/cos(alpha) =1.08/cos(30) =1.247 N/s =74.8 N/min`
The pressure on the roof is 0.72 Pa*s and the vertical force on the roof is 1.247 N/s