Someone is able to pull the 1800 kg shack at a constant velocity with a force of 190 N. Calculate the coefficient of friction between the shack and the ice.
4 years ago
$\left(F_{net}\right)_x=m.a_x=0$(Fnet)x=m.ax=0 , $F_{app}-F_f=0$Fapp−Fƒ =0 , $F_{app}=F_f$Fapp=Fƒ , $F_f=\mu.F_N$Fƒ =μ.FN , $F_N=m.g$FN=m.g , $F_f=\mu.m.g$Fƒ =μ.m.g
$F_{app}=\mu.m.g$Fapp=μ.m.g , $\mu=\frac{F_{app}}{m.g}=\frac{190N}{\left(1800kg\right)\left(9.81m.s^{-2}\right)}=0.0108$μ=Fappm.g =190N(1800kg)(9.81m.s−2) =0.0108
4 years ago
Answered By Majid B
$\left(F_{net}\right)_x=m.a_x=0$(Fnet)x=m.ax=0 , $F_{app}-F_f=0$Fapp−Fƒ =0 , $F_{app}=F_f$Fapp=Fƒ , $F_f=\mu.F_N$Fƒ =μ.FN , $F_N=m.g$FN=m.g , $F_f=\mu.m.g$Fƒ =μ.m.g
$F_{app}=\mu.m.g$Fapp=μ.m.g , $\mu=\frac{F_{app}}{m.g}=\frac{190N}{\left(1800kg\right)\left(9.81m.s^{-2}\right)}=0.0108$μ=Fappm.g =190N(1800kg)(9.81m.s−2) =0.0108