Question

A 205 kg log is pulled up a ramp by means of a rope that

Answer 1

Answer:2.737 kN

Explanation:

Given

mass of log(m)=205 kg

ramp inclination$=30^{\circ}$

coefficient of kinetic friction between log and ramp is ($\mu _k$)=0.9

log has an acceleration of 0.8 m/s^2

Let T be the tension in the rope

$T-mgsin\theta -f_r=ma$

Where $mgsin\theta$=Sin component of weight

$f_r=friction\ Force=\mu _KN$(Where N is Normal reaction)

$T-mgsin\theta -\mu _k\left ( mgcos\theta \right )=ma$

$T=m\left ( gsin\theta +\mu _kcos\theta +a\right )$

sin30=0.5

cos30=0.866

$T=205\times \left ( 9.81\times 0.5+0.9\times 9.81\times 0.866+0.8\right )$

$T=205\times 13.35=2.737 kN$