## Essential University Physics: Volume 1 (3rd Edition)

We know that the force of friction is given by: $F_f=F_n\mu$. We know that the given force is 35 degrees downward, so we see that the normal force is equal to: $F_n=mg+Fsin35$. Thus, the overall force against the motion of the mower is: $F_f=(mg+Fsin35)\mu=((22)(9.81)+Fsin(35))(.68)=146.76+.39F$. We set this equal to the force pushing the mower forward to find: $146.76+.39F=Fcos35 \\ 146.76+.39F=.82F \\ \fbox{F=342 Newtons}$ We know that the mower's weight is: $mg=22(9.81)=215.82$ Thus, this force is $\frac{342 N}{215.82 N}=1.585$ times the weight of the mower.