Respuesta :
Answer:2.737 kN
Explanation:
Given
mass of log(m)=205 kg
ramp inclination[tex]=30^{\circ}[/tex]
coefficient of kinetic friction between log and ramp is ([tex]\mu _k[/tex])=0.9
log has an acceleration of 0.8 m/s^2
Let T be the tension in the rope
[tex]T-mgsin\theta -f_r=ma[/tex]
Where [tex]mgsin\theta [/tex]=Sin component of weight
[tex]f_r=friction\ Force=\mu _KN[/tex](Where N is Normal reaction)
[tex]T-mgsin\theta -\mu _k\left ( mgcos\theta \right )=ma[/tex]
[tex]T=m\left ( gsin\theta +\mu _kcos\theta +a\right )[/tex]
sin30=0.5
cos30=0.866
[tex]T=205\times \left ( 9.81\times 0.5+0.9\times 9.81\times 0.866+0.8\right )[/tex]
[tex]T=205\times 13.35=2.737 kN[/tex]
