# Find the tension in the rope.

A 212-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 29.1° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.757, and the log has an acceleration of 0.720 m/s2. Find the tension in the rope.

Fx=ma(of x)
Fy=ma(of y)

i know i have a normal force

this is what i have but i am stuck

how do i set this up?

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black02 said:
Fx=ma(of x)
Fy=ma(of y)

i know i have a normal force

this is what i have but i am stuck

how do i set this up?
You have a block on a ramp. <--[BLOCK]-->

Those (<--) are vectors. They are pointing in opposite directions. One of them is a pulling force and one is a resistive force. The resistive force and the normal force have a relationship which you can use.

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• PhanthomJay
i have come back to it. this is what i have so far.

force of friction = .757*185.24
f=140.227
i got this using this formula

f = mkn

f = Force of Friction.
mk = Coefficient of kinetic friction.
n = The normal force pressing the surfaces together.

however the object is still accelerating so the tension force is greater just not sure how i get the tension force from this.

black02 said:
i have come back to it. this is what i have so far.

force of friction = .757*185.24
This is close. But the forces in the Y direction are N - mg cos 29.1 = 0. You have them as N - m cos 29.1 = 0, so your value for the normal force (and therefore the frictional force) are too low.

ihowever the object is still accelerating so the tension force is greater just not sure how i get the tension force from this.
Have you drawn a free-body diagram? Those help me a lot. You will have an arrow up the incline for T, the tension in the rope. And you will have arrows downward, along the incline for the friction and gravity. These represent vectors, which add up to the total force created by the accelerating mass of the log.

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• PhanthomJay