A 217 kg log is pulled up a ramp by means of a rope that is parallel to the surf

Question

A 217 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.830, and the log has an acceleration of 0.800 m/s2. Find the tension in the rope.
I've followed the example given in the textbook as well as on this website, yet I still dont get the correct answer. The equation I used: T=ma+mgsin30+µmgcos30 T=(217)(.8)+(217)(9.8)(sin30)+(.83)(217)(9.8)(cos30) T=? A 217 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.830, and the log has an acceleration of 0.800 m/s2. Find the tension in the rope.
I've followed the example given in the textbook as well as on this website, yet I still dont get the correct answer. The equation I used: T=ma+mgsin30+µmgcos30 T=(217)(.8)+(217)(9.8)(sin30)+(.83)(217)(9.8)(cos30) T=?

Explanation / Answer

we know that the frictional force is equal to the coefficient of friction times the normal force.

the normal force is the force perpendicular to the ramp due to gravity.

Fn=mg*cos(theta) where theta is the angle that the ramp makes with the ground. You can get this formula by using similar triangles.

Thus;

Fn=217*9.8*cos(30) = 1841.69N

Now, Ff = Fn * u (u is coefficient listed in your question)

Thus:

Ff= 1841.69 * 0.860 = 1583.85
That is the frictional force opposing upward motion.

Now, the force applied on the rope is equal to the forces opposing motion plus the motivational force;

F(total) = Tension = 0.8 * 217 + 1583.85
______________
| T = 1757.45N |

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