57,6km/h=16m/s
m=2 tấn =2000kg
a)cầu võng xuống
\(F_{ht}=N-P\Leftrightarrow N=F_{ht}+P\)
\(\Leftrightarrow N=\dfrac{v^2}{R}.m+m.g=\)\(\dfrac{85600}{3}N\)
b) cầu võng lên
\(F_{ht}=P-N\Leftrightarrow N=P-F_{ht}\)
\(\Leftrightarrow N=m.g-\dfrac{v^2}{R}.m\)=\(\dfrac{34400}{3}N\)