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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
W03D1-2 Group Problem: Tension in a Suspended Rope Solution
Suppose a uniform rope of mass M and length L is suspended from a ceiling. The
magnitude of the acceleration due to gravity is g .
a) Find the tension in the rope at the upper end where the rope is fixed to the ceiling.
b) Find the tension in the rope as a function of the distance from the ceiling.
c) Find an equation for the rate of change of the tension with respect to distance
from the ceiling in terms of M , L , and g .
Be sure you show any free body force diagrams or sketches that you plan to use.
Solution: We begin by choosing a coordinate system with the origin at the ceiling
and the positive y -direction pointing downward.
a) Find the tension in the rope at the upper end where the rope is fixed to the ceiling.
In order to find the tension at then upper end of the rope, we draw a free body
diagram of the entire rope. The forces acting on the rope are the force at y  0
holding the rope up, (we refer to that force as T ( y  0) , the tension at the upper end)
j . The free body force diagram is
and the gravitational force on the entire rope Mg φ
shown in the figure below.
We now apply Newton’s Second Law noting that the acceleration is zero
Mg  T ( y  0)  0 .
Thus we can solve for T ( y  0) , the tension at the upper end,
T ( y  0)  Mg .
b) Find the tension in the rope as a function of the distance from the ceiling.
Recall that the tension at a point is the magnitude of the action-reaction pair of forces
acting at that point. So we make an imaginary cut in the rope a distance y from the
ceiling separating the rope into an upper and lower piece.
We choose the upper piece as our system with mass m  ( M / L) y . The forces acting on
j  ( M / L) ygφ
j , the force at y  0 holding
the upper rope are the gravitational force mg φ
the rope up, (we refer to that force as T ( y  0) , the tension at the upper end), and the
tension at the point y , T ( y) that is pulling the upper piece down. The free body force
diagram is shown in the figure below.
We now apply Newton’s Second Law noting that the acceleration is zero
mg  T ( y)  T ( y  0)  0 .
Thus we can solve for T ( y) , the tension at a distance y from the ceiling,
T ( y)  T ( y  0)  mg .
Using our results for the mass of the upper piece and the tension at the upper end we have
that
T ( y)  Mg(1  y / L)
(1)
As a check, we note that when y  L , the tension T ( y  L)  0 which is what we expect
because there is no force acting at lower end of the rope.
c) Find an equation for the rate of change of the tension with respect to distance
from the ceiling in terms of M , L , and g .
We can differentiate Eq. (1) with respect to y and find that
dT
( y)  ( M / L)g .
dy
So the rate that the tension is changing is constant.
(2)
Second Approach:
We can arrive at this result by an alternative method, a technique that will generalize to
many types of “continuous systems”. We consider as our system an imaginary segment of
the rope between the points y and y  y . This small element has length y and mass
m  ( M / L)y .
The rope has now been divided into three pieces, an upper piece, the small element, and a
lower piece. The forces acting on it are the tension T ( y) at y directed upward, (the force
of the upper piece holding the element up), the tension T ( y  y) at y  y directed
downward (the force of the lower piece pulling the element down), and the gravitational
j  ( M / L)ygφ
j . The free body force diagram is shown in the figure below.
force mg φ
We now apply Newton’s Second Law to the small element noting that the acceleration is
zero
mg  T ( y)  T ( y  y)  0 .
We now solve for the difference in the tension
T ( y  y)  T ( y)  mg
We now substitute our result for the mass of the element m  ( M / L)y , and find that
that
T ( y  y)  T ( y)  ( M / L)yg .
We now divide through by y yielding
T ( y  y)  T ( y)
 ( M / L)g
y
Now here comes the crucial step, the limiting argument. We consider the limit in which
the length of the small element goes to zero, y  0 .
T ( y  y)  T ( y)
 ( M / L)g .
y0
y
lim
Recall that the left hand side is the definition of the derivative of the tension with respect
to y , and so we arrive at
dT
( y)  ( M / L)g
dy
in agreement with Eq. (2).
We can solve this differential equation by a technique called separation of variables. We
rewrite the equation as
dT  ( M / L)gdy
and integrate both sides. Our integral will be a definite integral, the limits of the right
hand side are from y  0 to y , and the corresponding limits on the left hand side are
from T ( y  0) to T ( y) :
T ( y)

T ( y 0)
dT  ( M / L)g 
y y
y  0
dy .
Notice that we have introduced a “dummy” integration variable y  to distinguish the
integration variable from the endpoint of the integral y , the point that we are trying to
find the tension T ( y) . We now integrate and find that
T ( y)  T ( y  0)  ( M / L)gy .
We use our earlier result that T ( y  0)  Mg and find that
T ( y)  Mg(1  y / L)
in agreement with our earlier result.