Answer
The tension force is $2734.4N$
Work Step by Step
There are 3 forces affecting the movement of the log:
- The pulling force, or the tension in the pulling rope $T$
- Kinetic friction $f_k$ and the weight of the log $m_{log}g\sin30$, both of which oppose the upward motion
We can write $$\sum F=T-f_k-mg\sin30$$
As the log has $a=0.8m/s^2$, according to Newton's 2nd Law: $$T-f_k-mg\sin30=ma$$ $$T-\mu_kN-(205\times9.8\sin30)=205\times0.8$$ $$T-0.9N-1004.5=164$$ $$T-0.9N=1168.5$$
On the vertical, normal force $N$ opposes the weight $mg\cos30$. Since there is no vertical acceleration, $$N=mg\cos30=1739.8N$$
Therefore, $$T=0.9N+1168.5=2734.4N$$
