Newton’s second law:
FwΔt=m(v−0)F_w \Delta t = m(v-0)\\FwΔt=m(v−0)
FwΔt=ρVvF_w\Delta t = \rho V vFwΔt=ρVv
FwΔt=ρSvΔtvF_w\Delta t = \rho S v \Delta t vFwΔt=ρSvΔtv
Fw=ρπd24v2=1000kgm3π⋅25⋅10−4m24⋅400m2s2=250πNF_w = \rho\frac{\pi d^2}{4} v^2 = 1000 \frac{kg}{m^3} \frac{\pi \cdot 25 \cdot 10^{-4}m^2}{4} \cdot 400 \frac{m^2}{s^2} = 250\pi NFw=ρ4πd2v2=1000m3kg4π⋅25⋅10−4m2⋅400s2m2=250πN
P=sin(ϕ)Fw=sin(30∘)Fw=250π⋅0.5N=125πN=392.7NP = \sin (\phi) F_w = sin(30^\circ) F_w = 250\pi \cdot 0.5N = 125\pi N = 392.7NP=sin(ϕ)Fw=sin(30∘)Fw=250π⋅0.5N=125πN=392.7N
Answer: 392.7N
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