Answer:3062.37 N
Step-by-step explanation:
Given
log weight =2250 N
![\theta =30](https://img.qammunity.org/2020/formulas/physics/college/9q58z72jr9593y4ewdhdoiqqnisbanj6au.png)
coefficient of kinetic friction
![\mu =0.9](https://img.qammunity.org/2020/formulas/physics/college/3wke89791nza56u3kxyp2gh5wn3fbrqgtl.png)
![a_(max)=0.80 m/s^2](https://img.qammunity.org/2020/formulas/physics/college/tu5ikm25iuhxuedtj2wxtbp71v429jlfjw.png)
During Pulling Forces on cable is log sin component and friction and tension (T)
![T-mg\sin \theta -f_r=ma](https://img.qammunity.org/2020/formulas/physics/college/emoc20mywt2cox1rlwiw72l25evj2ll9hf.png)
![T=2250\sin (30)+\mu 2250 \cos (30)+(2250)/(g)* 0.8](https://img.qammunity.org/2020/formulas/physics/college/2lj0wxl5uida2kd1avyu1w2j1qwdw1wn3l.png)
T=1125+1753.70+183.67=3062.37 N
so Rope must be able to withstand a force of 3062.37 N