Answer:3062.37 N
Explanation:
Given
log weight =2250 N
[tex]\theta =30[/tex]
coefficient of kinetic friction [tex]\mu =0.9 [/tex]
[tex]a_{max}=0.80 m/s^2[/tex]
During Pulling Forces on cable is log sin component and friction and tension (T)
[tex]T-mg\sin \theta -f_r=ma[/tex]
[tex]T=2250\sin (30)+\mu 2250 \cos (30)+\frac{2250}{g}\times 0.8[/tex]
T=1125+1753.70+183.67=3062.37 N
so Rope must be able to withstand a force of 3062.37 N