Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

I am in summer school for physics and I got this question for home work. A rock climber of mass 55 kg is hanging suspended from a rope tied to another climber of mass 65 kg on a horizontal cliff ledge. If the coefficient of kinetic friction between the climber on the ledge and the ledge 0.45, what is the net acceleration of the climber on the ledge?
I have been stuck on this question for 2 hours and I need it for tomorrow can anyone help me. Thanks


Sagot :

AL2006
First of all, we have to sort out how this situation is put together. You have 55 kg that's hanging suspended, but it's tied to another 65 kg on a horizontal ledge.  Somewhere in there, the rope has to turn a corner.  Obviously, it has to go around the edge of the ledge, and we have top assume that there's absolutely no friction against the rope at that point,  because we're not told anything different.  So we have to treat the edge like a frictionless pulley. And we're also going to ignore the weight of the rope.

OK. The weight of the 55-kg guy who is hanging suspended is (mass) x (gravity) =
(55 x 9.8) = 539 newtons. That force soaks upward through the rope, makes the turn
at the edge, and is exerted horizontally against the 65-kg guy on the ledge. 

Hopefully, the suspended guy can hold on for a few minutes longer, while we analyze
the forces around the heavier guy up on the ledge.

The weight of the guy on the ledge is (mass) x (gravity) = (65 x 9.8) = 637 newtons.
That's his weight, pointing downward, against the ledge.
As his boots slip along the ledge, the friction force against his motion is
(his weight) x (the coefficient of kinetic friction between him and the ledge) =
(637 newtons) x (0.45) = 286.65 newtons.

The man on the ledge has the rope pulling him toward the edge with 539N of force,
and 286.65N of friction force holding him back.  You can see that he's slipping toward
the edge, because the friction force isn't enough to hold him.

The net force on him is (539N forward) + (286.65N backward) = 252.35N forward.

Since the man on the ledge has a net force pulling him forward toward the edge,
he accelerates in that direction.

Force = (mass) x (acceleration)

Acceleration = (force) / (mass) = (252.35N) / (65kg) = 3.88 meters per second²

He's sliding toward the edge with an acceleration of about 0.396 G ... his speed is increasing about 39 or 40% as fast as it will after he falls over the edge, and the both of them proceed toward their ultimate and apparently unavoidable 'splut' below.

I've totally terrified myself answering this one.