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Tension
Tension
• Tension is a pulling force that arises when a rope,
string, or other long thin material resists being pulled
apart without stretching significantly.
• Tension always pulls away from a body attached to a
rope or string and toward the center of the rope or
string.
A physical picture of tension
Imagine tension to be the internal force preventing a rope or string
from being pulled apart. Tension as such arises from the center of
the rope or string. It creates an equal and opposite force on objects
attached to opposite ends of the rope or string.
Tension examples
Note that the
pulleys shown are
magic! They don’t
affect the tension
in any way, and
serve only to bend
the line of action of
the force.
Sample problem
A.
B.
A 1,500 kg crate hangs motionless from a crane cable. What
is the tension in the cable? Ignore the mass of the cable.
Suppose the crane accelerates the crate upward at 1.2 m/s2.
What is the tension in the cable now?
Springs (Hooke’s Law)
• The magnitude of the force exerted by a spring is
proportional to the amount it is stretched.
• F = kx
– F: force exerted by the spring (N)
– k: force constant of the spring (N/m or N/cm)
– x: displacement from equilibrium (unstretched and
uncompressed) position (m or cm)
• The direction of the force is back toward the
equilibrium (or unstretched) position.
Sample problem
• A 1.50 kg object hangs motionless from a spring with a
force constant of k = 250 N/m. How far is the spring
stretched from its equilibrium length?
Sample problem
• A 1.80 kg object is connected to a spring of force constant 120 N/m.
How far is the spring stretched if it is used to drag the object across a
floor at constant velocity? Assume the coefficient of kinetic friction is
0.60.
Laboratory
• Using the ramp at an angle, determine the
coefficient of kinetic friction between the felt
side of the block and the ramp by allowing the
block to accelerate down the ramp. Tomorrow,
you will turn in the appropriate free body
diagrams and calculations.
Sample problem
A 5.0 kg object (m1) is connected to a 10.0 kg object (m2) by a string. If a pulling
force F of 20 N is applied to the 5.0 kg object as shown,
A) what is the acceleration of the system?
B) what is the tension in the string connecting the objects?
(Assume a frictionless surface.)
Gravity
A very common accelerating force is
gravity. Here is gravity in action. The
acceleration is g.
Slowing gravity down
The pulley lets us use gravity as
our accelerating force… but a lot
slower than free fall. Acceleration
here is a lot lower than g.
Magic pulleys on a flat table
• Magic pulleys bend the line of action of the force
without affecting tension.
N
m1g -x
T
SF = ma
m2g + T – T = (m1 + m2)a
a = m2g/(m1+m2)
m1
T
Frictionless table
m2
x
m2g
Sample problem
Mass 1 (10 kg) rests on a frictionless table
connected by a string to Mass 2 (5 kg). Find
(a)the acceleration of each block.
(b)the tension in the connecting string.
m1
m2
Sample problem
Mass 1 (10 kg) rests on a table connected by a string to Mass
2 (5 kg). Find the minimum coefficient of static friction for
which the blocks remain stationary.
m1
m2
Sample problem - solution
fs
m1g
N
SF = 0
m2g - T + T – fs = 0
fs = m2g
msN = m2g
msm1g = m2g
ms = m2/m1 = 0.50
T
m1
T
m2
m2g
Note: we know from previous problem that the static
friction is not enough to hold the blocks in place!
Sample problem
Mass 1 (10 kg) rests on a table connected by a string
to Mass 2 (5 kg). If ms = 0.30 and mk = 0.20, what is
(a)the acceleration of each block?
(b)the tension in the connecting string?
m1
m2
Sample problem – solution (a)
fk
m1g
N
SF = ma
m2g - T + T – fk = ma
m2g - mkm1g = (m1 + m2)a
a = (m2- mkm1) g/(m1 + m2)
a = 2.0 m/s2
T
m1
T
m2
m2g
Using that the acceleration is 2.0 m/s2 from part a)
Sample problem – solution (b)
fk
m1g
N
Using block 2
SF = ma
m2g - T = m2a
T = m2(g – a)
T = 40 N
Using block 1
SF = ma
T - fk = m1a
T = m1(a + mkg)
T = 40 N
T
m1
T
m2
m2g
Magic pulleys on a ramp
• It’s a little more complicated when a magic pulley is installed on a
ramp.
N
SF = ma
m2g -T + T – m1gsinq = (m1+m2)a
T m2g – m1gsinq = (m1+m2)a
a = (m2 – m1sinq)g/(m1+m2)
T
m1gsinq
m1g
m1gcosq
m2
q
m2g
Sample problem
• Two blocks are connected by a string as shown in the
figure. What is the acceleration, assuming there is no
friction?
5 kg
45o
Sample problem - solution
N
T
m1gsinq
m1g
m1gcosq
SF = ma
m2g -T + T – m1gsinq = (m1+m2)a
m2g – m1gsinq = (m1+m2)a
T a = (m2 – m1sinq)g/(m1+m2)
a = [(5 – 10sin45o)(9.8)]/15
a = - 1.35 m/s2
5 kg
m2g
45o
Sample problem - solution
N
T
SF = ma
m2g -T + T – m1gsinq = (m1+m2)a
m2g – m1gsinq = (m1+m2)a
a = (m2 – m1sinq)g/(m1+m2)
a = [(5 – 10sin45o)(9.8)]/15
a = - 1.35 m/s2
T
m1gsinq
m1g
m1gcosq
5 kg
m2g
45o
How would this change
if there is friction on
the ramp?