Exactly. Although he could save time by just letting go, as I noticed his belay technique is pretty bad.
And Paul has it coming, as clearly it is his plan to drop Jay later by untieing.
Paul you are a bastard, dropping Jay after he held you. I will never climb with you.

Sorry, the smiley was intended to be implied!

it would be interesting if we added USN as the third party falling on crevasse so now one person is holding two on a 40%
edit: add 10 meter between second and third

there'd be two different accelerations due to two different tensions: one while both USN and Paul are falling, then another after USN craters, then get's landed on!

@adatesman: sucka!!!

Um, no, I have no such thing. I just happen to be an engineer and know how to use sine... and the equations of motion.

p.s. Majid, I'm not going to let you cheat.

but, find the vector of force pulling on the rope from jay's weight, add that to paul's force (hint F=ma), there's your tension. subtract those two values to get the force paul is accelerating toward the ground with. calculate paul's acceleration and plug it into the EOM. voila time it takes paul to crate, thus you can back calculate his final velocity.

now, calculate the length of the slope, knowing it's 25 m tall at a 40deg incline (with sine). subtract 5 m from it, since there is still 5 m of rope left when paul crater's, now that's how far jay will slide. change that back into the vertical he will fall (again sine), since it's a frictionless slope. he will fall at 1G, again since it's a frictionless slope. now you know jay's cratering acceleration, EOM's, voila, dun.

Remember when you'd post sets of climbing related problems for people to calculate the answer to? Well, I was flipping through one of my old physics textbooks last night and happened to notice this:



Obviously it needs more Red and Green Arrows, but not bad otherwise!

-a.

I do not believe you are correct on the first part.

First off, in a static situation, tension is only calculated with the force on one object. IE, if you have two objects hanging dead off of a frictionless pulley, the total tension in your system will only be that of the weight of one of the objects, not both of the objects combined.

What I did (and there are probably errors in this, so bear with me:

Force on Jay from Gravity = 52*9.81=510.1
Force on Paul from Gravity = 74*9.81=725.94
Force directed downhill from Gravity on Paul = 725.94*Cos(50)=466.1N.

The net force on our system is 510.1-466.1=44N


Calculate the net acceleration:

F(net) = m(net) * a
44N = (52+74)*a
a = 0.349 m/s^2


Calculate the tension on the rope:

F(due to acceleration) = m(paul) * a

F(due to acceleration) = 74*.349 = 25.8N

466.1 + 25.8 = 491.9N

The tension on the rope is 491.9N


Calculate the speed at which Jay decks:


x= 1/2*a*t^2
25=.5*0.349^2
t = 11.97 seconds

V = a * t = 0.349 * 11.97 = 4.18 m/s

I could have made an error in there somewhere. If I did, please let me know!

Oooh, this looks like a fun one.

a) The tension in the rope can be figured by adding the opposing forces on the rope, in this case, Paul and Jay. Pauls force on the rope is his full 52kg, but because Jay is on a slope, his will only be 47.565kg (74*sin40), Giving us 99.565kg of tension in the rope.

b) Because there is no friction, Paul will still fall at a rate of 9.81m/s^2, even though his effective mass is only 4.435kg. So if he falls the 20m, he will hit the ground at 19.809m/s. Jay is a very different case, because when Paul hits and unties (assuming instantaneous untying of a figure 8 that has just taken a fatty fall :p) then Jay will still be 5m from the top. At that moment, he will have the same velocity as Paul, 19.809m/s. Because he is going uphill, the force of gravity (Fg) will slow him down before he goes over the edge, with a force of 47.565kg. Using the equation F=ma (Force=Mass*Accelaration) we solve for a, giving us F/m=a, or 47.565/74=.643m/s^2. So Jay will have an initial velocity of 19.809m/s, and slow down with an accelaration of -.643m/s^2, leaving him going 19.646m/s at the point of the mountain. Now, when he takes off, he will be traveling upward at an angle of 40 degrees. With a little trig, we find his upward velocity to be 12.628m/s, and his horizontal to be 15.050m/s. Gravity will slow him down at a rate of 9.81m/s^2, which using the equation Vf^2= Vi^2 + 2ax, solved for x, being Vf^2-Vi^2/2a= x, he will travel a full 8.12m upward before he stops midair. This will put him a total of 33.12m above the ground, so when he hits, the vertical component of his velocity will be 25.491m/s. Because of his horizontal velocity, and since he spent a full 3.877 seconds in the air, he will be 58.5 meters away from Paul when he hits the ground, and his resultant velocity will be 29.602m/s, striking at an angle of 59.4 degrees. So Paul will be going 19.809m/s (44.3mi/h), and Jay will be going 29.602m/s (66.2 mi/h)

I hope that made sense and is correct. Either way, their both gonna be pretty messed up, if not dead, after this "predicament"

kg of tension? sweet jebus!

Oooh, this looks like a fun one.

a) The tension in the rope can be figured by adding the opposing forces on the rope, in this case, Paul and Jay. Pauls force on the rope is his full 52kg, but because Jay is on a slope, his will only be 47.565kg (74*sin40), Giving us 99.565kg of tension in the rope.

b) Because there is no friction, Paul will still fall at a rate of 9.81m/s^2, even though his effective mass is only 4.435kg. So if he falls the 20m, he will hit the ground at 19.809m/s. Jay is a very different case, because when Paul hits and unties (assuming instantaneous untying of a figure 8 that has just taken a fatty fall :p) then Jay will still be 5m from the top. At that moment, he will have the same velocity as Paul, 19.809m/s. Because he is going uphill, the force of gravity (Fg) will slow him down before he goes over the edge, with a force of 47.565kg. Using the equation F=ma (Force=Mass*Accelaration) we solve for a, giving us F/m=a, or 47.565/74=.643m/s^2. So Jay will have an initial velocity of 19.809m/s, and slow down with an accelaration of -.643m/s^2, leaving him going 19.646m/s at the point of the mountain. Now, when he takes off, he will be traveling upward at an angle of 40 degrees. With a little trig, we find his upward velocity to be 12.628m/s, and his horizontal to be 15.050m/s. Gravity will slow him down at a rate of 9.81m/s^2, which using the equation Vf^2= Vi^2 + 2ax, solved for x, being Vf^2-Vi^2/2a= x, he will travel a full 8.12m upward before he stops midair. This will put him a total of 33.12m above the ground, so when he hits, the vertical component of his velocity will be 25.491m/s. Because of his horizontal velocity, and since he spent a full 3.877 seconds in the air, he will be 58.5 meters away from Paul when he hits the ground, and his resultant velocity will be 29.602m/s, striking at an angle of 59.4 degrees. So Paul will be going 19.809m/s (44.3mi/h), and Jay will be going 29.602m/s (66.2 mi/h)

I hope that made sense and is correct. Either way, their both gonna be pretty messed up, if not dead, after this "predicament"

And what would that be? If there is no friction, then masses dont matter, they will both accelarate at 9.81m/s^2...

um, jay has an opposing force.... so yes mass does matter. however as jay falls once the rope is cut, you are correct, his mass does not matter due to the frictionless surface.

for bonus points we should consider that a person has a drag coefficient of Cd ~ 1.3

GO!!

go check out the simple force diagram on any climbing pulley and report back.

thanks.

you got the net force paul is falling with correct however. thus you can get paul's acceleration.

I realize he has an opposing force, but since Pauls is larger, and there is no friction, they will both be taken Pauls direction with the accelaration of gravity. And ive done enough math today to deal with adding in more crap into this problem... :p

um, jay has an opposing force.... so yes mass does matter. however as jay falls once the rope is cut, you are correct, his mass does not matter due to the frictionless surface.

for bonus points we should consider that a person has a drag coefficient of Cd ~ 1.3

GO!!

Except that you have to look at the vector components of Paul's weight. The force along the slope is less than Jay's weight.

Exactly. Although he could save time by just letting go, as I noticed his belay technique is pretty bad.
And Paul has it coming, as clearly it is his plan to drop Jay later by untieing.
Paul you are a bastard, dropping Jay after he held you. I will never climb with you.

There are no components of Paul's weight, he is falling straight down. And the opposing component of Jay's weight is still less than Paul's total weight.

Except that you have to look at the vector components of Jay's weight. The force along the slope is less than Paul's weight.