Question how to find acceleration with kinetic friction involved.

[baird.lindsay]

Homework Statement

[URL=http://imageshack.us/photo/my-images/132/81100769.png/][PLAIN]http://img132.imageshack.us/img132/2433/81100769.png[/URL] Uploaded with ImageShack.us[/PLAIN]

I have a question on part of this problem.

Paul accidentally falls off the edge of a glacier as shown in Fig 4-21 (p 103). He
is tied by a long rope to Steve, who has a climbing ax. Before Steve sets his ax
to stop them, he slides without friction along the ice, attached by the rope to
Paul. Assume no friction between the rope and the glacier. Find the
acceleration of each person and the tension in the rope. (figured this part out already)

Questions: The two climbers each have masses of 110kg each, the coefficient of kinetic friction between steve and the rock is .5 and theta is 15 degrees. If steve has 3.2 meters to reach the edge of the cliff how long till he reaches the cliffs edge. I think I use Δx=1/2at^2?

question. after steve goes over the cliffs edge what is the tension in the rope (have no idea how to do this part)

Homework Equations

these are the solutions for frictionless.
Steve:
ƩS: Fx : T + msg sin = msax
ƩFy : Fn – msg cos = ms · 0

Paul:
ƩP: Fy : T – mpg = may

g(mp + ms sin)
__________________=ax
(ms + mp)

The Attempt at a Solution

I know how to figure out acceleration when its frictionless , but I don't know how with friction. I am thinking I add the kinetic friction vector opposite direction of tension (to pic above) and when I sum the forces I am subtracting it off (ƩFx: T + mgsintheta - fk) for Steve in the y direction. So this:
g(mp + ms sin)
__________________=ax
(ms + mp)

would be this:
g(mp + ms sin)
__________________-0.5 (kinetic friction)=ax
(ms + mp)

The line is a divison line...

 

[haruspex]

0.5 is the coefficient of friction. So what is the frictional force opposing motion?

After going over the edge, suppose the tension in the rope is T. What then would be the acceleration of each climber?

 

[baird.lindsay]

0.5 is the coefficient of friction. So what is the frictional force opposing motion?

kinetic friction is opposing the motion. which is 0.5. So steve in the x direction is: ƩFx: T + mgsintheta - fk) =ma ?
[STRIKE][/STRIKE]

 

[haruspex]

You don't seem to know what is meant by a 'coefficient of friction'. It's the ratio between the magnitude of the frictional force and that of the normal force between the surfaces.
What is the magnitude of the normal force between Steve and the ground?

 

[Nickie]

Your approach is correct. To find the acceleration with friction involved, you need to take into account the kinetic friction force in the equation. So your equation for Steve would be:

ƩS: Fx : T + mgsinθ - μkN = msax

Where μk is the coefficient of kinetic friction and N is the normal force. You can then solve for ax using this equation.

For the second part of the question, after Steve goes over the cliff, the tension in the rope will be equal to the weight of Paul. This is because there is no longer any force acting on the rope, so it will be in equilibrium. So the equation to find the tension would be:

ƩP: Fy: T - mg = 0

Solving for T, you would get:

T = mg

Hope this helps!