Using a simply pulley/rope system, a crewman on an Arctic expedition is trying t

Question

Using a simply pulley/rope system, a crewman on an Arctic expedition is trying to lower a 6.30 kg crate to the bottom of a steep ravine of height 28.4 meters. The 50.3 kg crewman is being careful to lower the crate at a constant speed of 1.50 m/s. Unfortunately, when the crate reaches a point 15.8 meters above the ground, the crewman slips and the crate immediately accelerates toward the ground, dragging the hapless crewman across the ice and toward the edge of the cliff. If we assume the ice is perfectly slick (that is, no friction between the crewman and the ice once he slips and falls down), at what speed will the crate hit the ground? Assume also that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff.At what speed will the crewman hit the bottom of the ravine? (Assume no air friction.)

Explanation / Answer

The crate is travelling at 5.54 m/s when it hits the ground.

COMPREHENSIVE EXPLANATION BELOW:
In solving the problem, I'd suggest using kinematics equations and application of Newton's Laws, rather than Work-Energy methods. It'll make life a lot easier.
Most of the information given is unnecessary in solving the problem, so the tricky part is picking out the useful stuff. In order to solve the problem, I'm going to provide formulas to begin with and substitute all known data at the end, so you can see the relationship of the variables in the system. I will provide step-by-step algebraic working.

ASSUMPTIONS:
-The rope is inextensible and massless (no need to overcomplicate things!)
-g = 9.81 m/s^2
We can disregard any other drag force (air resistance, etc.) acting on the objects.

Firstly, you want to detail all the forces acting on the objects in the system after the crewman slips. YA doesn't really allow me to draw free body diagrams here, so I hope a description will suffice. We can disregard the forces acting on the rope as we have assumed it to be massless.

Therefore, we have firstly the forces acting on the crewman. There are THREE forces at play here.
The vertical forces are: the weight of the crewman and the normal force exerted by the ground on the crewman. The net vertical force acting on the crewman is zero, and since we can disregard friction, are largely superfluous in order to solve the problem.
The lone horizontal force acting on the crewman is the TENSION force in the rope pulling the crewman across the ice.

There are TWO forces acting on the crate: both vertical, they are the TENSION force pulling up on the crate and the WEIGHT of the crate pulling the crate down towards the earth. They are opposite in direction, and we know that the weight of the crate must be greater than the tension force pulling up on it as it's accelerating downwards.
It's also necessary to note that the tension force acting on the crate here is equal in magnitude to the tension force pulling on the crewman, as the string is inextensible.

As the string is inextensible, BOTH the crewman and crate have accelerations of equal magnitude.

Thus, according to Newton's Second Law, we can derive equations from knowledge of the forces acting upon the objects.

Let:
m1 = mass of crewman = 45.0kg
m2 = mass of crate = 5.00kg
a = acceleration of objects = UNKNOWN
?s = displacement of object = 14.5m
vi = inital velocity = 1.50m/s
vf = final velocity = UNKNOWN
T = Tension force in rope = UNKNOWN (Unnecessary, however, as will be shown later.)
W2 = Weight of the crate

To solve the problem, we have to find vf.

Therefore we know have, according to Newton's Second Law:

m1*a = T...........(1) (Since we know tension is the only horizontal force acting on the crewman)
m2*a = W2 - T...(2) (Since we know that the weight of the crate > tension force acting upon it)
W2 = m2*g........(3) (Weight of crate)

Thus, substituting (1) into (2), and making 'a' the subject of the equation we have:

m2*a = W2 - m1*a
m2*a +m1*a = W2
(m2+m1)*a = W2
a = W2/(m2 +m1)..........(4)

Substituting (3) into (4):

a = (m2*g)/(m2 + m1)....(5)

From our equations of motion we have:

vf^2 = vi^2 + 2*a*?s.......(6)

Substituting (5) into (6):

vf^2 = vi^2 +2*((m2*g)/(m2 + m1))*?s...... (7)

Now we can substitute our known values into equation (7) to find vf, which is our goal. Therefore we have:

vf^2 = 1.50^2 +2*((5.00*9.81)/(5.00 + 45.0))*14.5...... (7)
vf^2 = 2.25 + 2*14.5*((49.05)/50.0)
vf^2 = 2.25 + 29*(0.981)
vf^2 = 2.25 + 28.449
vf^2 = 30.699
Taking the square root of both sides of the equation:

vf = sqrt(30.699)
vf = 5.54 m/s (Rounded to 3 significant figures)

Therefore the crate is travelling at 5.54 m/s when it hits the ground.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.