Frictionless tension/pulley problem.

[CandyApples]

Homework Statement

Two people are trying to climb an iced-over mountain. They are connected by a rope which passes smoothly over a rock where the terrain bends (acting as a pulley). One climber (110kg) is on top on a horizontal surface while the other (125kg) is on a 22.56 degree incline. Assume the ice is frictionless and the climbers have no tools to aid their climb. What is the tension in the rope, and what would be their acceleration?

Homework Equations

F = m*a and the variation T - mg = ma.

The Attempt at a Solution

\SigmaF_x = -T + mgsin(22.56)
\SigmaF_y = N - mgcos(22.56)

Now I am unsure what to do next, as all previous problems assumed the object was at rest. Should I set both of these equal to ma then solve for the acceleration vector? The issue I run into with doing so is that there are two different masses to be concerned with in this problem.
-Thanks for any assistance in advance.

 

[Doc Al]

The Attempt at a Solution

\SigmaF_x = -T + mgsin(22.56)
\SigmaF_y = N - mgcos(22.56)

Those apply to the person on the slope. What about the person on the horizontal surface?

(Since the only motion is parallel to the surfaces, all you care about are forces parallel to the surfaces.)

Now I am unsure what to do next, as all previous problems assumed the object was at rest. Should I set both of these equal to ma then solve for the acceleration vector? The issue I run into with doing so is that there are two different masses to be concerned with in this problem.

You need to apply Newton's 2nd law to each mass, then combine the two equations.

 

[CandyApples]

the person on the slope would have the forces:
\SigmaF_x = T
\SigmaF_y = 0

Can I solve using T = 110(a) in my x force component for the other person, seeing as tension is constant throughout the entire rope?

Doing so gives:
T = 110a
F = -t + mgsin(22.56)
125a = -110a +125(9.8)(sin(22.56))
225a = 125*9.8*sin(22.56)
a = 2.09 m/s/s

then:
T = ma
T = 110*2.09
T = 229.9N

 

[Doc Al]

the person on the slope would have the forces:
\SigmaF_x = T
\SigmaF_y = 0

That's for the person on the horizontal surface.

Can I solve using T = 110(a) in my x force component for the other person, seeing as tension is constant throughout the entire rope?

Sure.

Doing so gives:
T = 110a
F = -t + mgsin(22.56)
125a = -110a +125(9.8)(sin(22.56))
225a = 125*9.8*sin(22.56)
a = 2.09 m/s/s

You made an error when adding 125 and 110. Redo the calculation.

 

[CandyApples]

oh my, it was 235. The new a is 2 and hence the tension is 220. Got too enthused that I solved the problem and made a silly mistake. Thank you very much for the assistance and thank you for saving me some embarrassment :).