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1206 - Concepts of
Physics
Wednesday, October 14th
Demonstrations
• The spinning chair, etc.
• Thank you Mark !
• Remember the ice skater example?
An ice skater is spinning with both arms and a leg
outstretched (a) and then she pulls her arms and leg
inward (b). Does her angular velocity increase or
decrease?
Notes
• Midterm: November 02nd 16:30 - 18:00
• For those who have class: 19:30 - 21:00
(we meet at my office F511)
• If you still have conflict, you need to come
see me in person today or Friday !!!
Notes
• Friday, we will go over the rules for the
midterm
• Friday, we will also go over the topics for
the midterm
• Friday, we will go over the formula sheet
for the midterm
A little more about forces
Accelerating blocks. Block 1 (mass m1 = 8.00 kg) is moving on a frictionless 30° incline. This block
is connected to block 2 (mass m2 = 22.0 kg) by a massless cord that passes over a massless and
frictionless pulley (see figure below - left). Find the acceleration of each block and the tensions in
the cord.
Free body diagrams separate incline and vertical motion
+y
+y
+x
FN T
a
FN
T
m1
T’
30.0°
W1
m2
30°
W2
W1 sin30°
a
W1
T
W1 cos30°
30°
Block 1
+x
W2
Block 2
Since both blocks accelerate, there must be a net force acting on each one. The key to solving is to
realize that Newton’s second law can be used separately for each block to relate the net force to the
acceleration. Note also that both blocks have accelerations of the same magnitude a, since they move
as a unit. We assume that block 1 accelerates up the incline and choose this direction to be the +x
axis. If block 1 in reality accelerates down the incline, then the value obtained for the acceleration
will be a negative number.
Let’s look at block 1 first. Three forces act on the block:
(a) W1 is the weight of block 1 (W1 = m1 * g = 8.00 kg * 9.80 m/s2 = 78.4 N)
(b) T is the force applied due to the tension in the cord
(c) FN is the normal force that the incline exerts
Free body diagrams separate incline and vertical motion
+y
+y
+x
FN T
W1 sin30°
W1
T
W1 cos30°
30°
Block 1
+x
W2
Block 2
The Weight is the only force that does not point along the
x or y axes and its x and y components are given in the
diagram. Applying Newton’s second law (ΣFx = m1 ax ):
ΣFx = -W1 sin30° + T = m1 a
This equation cannot be solved directly, since both T and a
are unknown. So, we have to consider block 2 to complete
our solution.
Two forces act on block 2:
(a) W2 is the weight of block 2
W2 = m2 * g = 22.0 kg * 9.80 m/s2 = 216 N)
(b) T’ is exerted as a result of block 1 pulling back on the
connecting cord. Since the cord and the pulley are
frictionless, the magnitude of T and T’ are the same.
Applying Newton’s second law (ΣFy = m2 ay) to block 2 reveals:
ΣFy = -W2 + T = - m2 a
Now we have two equations with two unknowns and can solve the system. Starting from the second
equation we obtain: T = W2 - m2 a (*)
Substitute in equation above: -W1 sin30° + W2 - m2 a = m1 a
And find a as: a = (-W1 sin30° + W2)/(m1 + m2) = (-78.4 N * 0.5 + 216 N)/30 kg = 5.89 m/s2
Now we know a and can put it back into equation (*):
T = W2 - m2 a = 216 N - 22.0 kg * 5.89 m/s2 = 86.4 N
Example: hoisting scaffold
A window washer on a scaffold is hoisting the scaffold up the side of a building by pulling downward
on a rope, as shown in the figure. The magnitude of the pulling force is 540 N, and the combined
mass of the worker and the scaffold is 155 kg. Find the upward acceleration of the unit.
P
T
T
T
W = mg
+y
TTT
+x
W = mg
The worker and the scaffold form a single unit, on which the rope
exerts a force in three places. The left end of the rope exerts an
upward force T on the worker’s hands. This force arises because he
pulls downward with a 540 N force, and the rope exerts an
oppositely directed force of equal magnitude on him, in accord with
Newton’s third law. Thus the magnitude of the upward force T is
540 N and is the magnitude of the tension of the rope. We assume
the masses of the pulley and each rope are negligible and the
pulleys are frictionless. Therefore a 540 N tension force T acts
upward on the right side of the scaffold pulley. A tension force is
also applied to the point P, where the rope attaches to the roof.
The roof pulls back on the rope in accord with the third law, and
this pull leads to the 540 N tension force T that acts on the left
side of the scaffold pulley. In addition the weight of the unit must be
taken into account:
W = mg = 155 kg * 9.80 m/s2
Newton’s second law can be applied to find a: ΣFy = m * ay
ΣFy = T + T + T - W = m ay
ay = (3T - W)/m = (3*540 N - 155 kg * 9.80 m/s2)/155 kg = 0.65 m/s2
Pulleys - a few more things
Pulleys are used on many occasions - usually to make the necessary force to lift something smaller.
A Pulley is a simple machine made with a rope, belt or chain wrapped around a wheel. A pulley
changes the direction of a force. There are three kinds of pulleys:
(a) a fixed pulley
(b) a movable pulley
(c) a combined pulley
A single pulley changes the direction of the lifting force. For example, if you are lifting a heavy
object with a single pulley anchored to the ceiling, you can pull down on the rope to lift the object
instead of pushing up. The same amount of effort is needed as without a pulley, but it feels easier
because you are pulling down. A movable pulley is a pulley that moves with the load. The movable
pulley allows the effort to be less than the weight of the load. The movable pulley also acts as a
second class lever. The load is between the fulcrum and the effort.
If you add a second pulley, the amount of effort to lift the
heavy object is much less. For example, to lift a box weighing
150 N, one would need to exert 150 N of force without the
help of pulleys. However, by using just two pulleys, the person
would only need to use 50 N of force.
A combined pulley (one fixed and one movable pulley)
makes life easier as the effort needed to lift the load is less
than half the weight of the load.
This figure shows a single pulley with a weight on one end of the
rope. The other end is held by a person who must apply a force to
keep the weight hanging in the air (in equilibrium). There is a force (tension) on the rope that is equal to the weight of the object.
This force or tension is the same all along the rope. In order for the weight
and pulley (the system) to remain in equilibrium, the person holding the end
of the rope must pull down with a force that is equal in magnitude to the
tension in the rope. For this pulley system, the force is equal to the weight, as
shown in the picture. The mechanical advantage of this system is 1! In the second figure, the pulley is moveable. As the rope is pulled up, it
can also move up. The weight is attached to this moveable pulley.
Now the weight is supported by both the rope end attached to the upper bar
and the end held by the person! Each side of the rope is supporting the
weight, so each side carries only half the weight (2 upward tensions are
equal and opposite to the downward weight, so each tension is equal to 1/2
the weight). So the force needed to hold up the pulley in this example is 1/2
the weight! The mechanical advantage of this system is 2; it is the weight
(output force) divided by 1/2 the weight (input force).
The mechanical advantage of each system is easy to determine. Count the number of rope/
cable segments on each side of the pulleys, including the free end. If the free end is to be
pulled down, subtract 1 from this number. This number is the mechanical advantage of
the system! To compute the amount of force necessary to hold the weight in equilibrium,
divide the weight by the mechanical advantage!
Here there are 3 sections of rope. Since the applied force is downward,
we subtract 1 for a mechanical advantage of 2.
It will take a force equal to 1/2 the weight to hold the weight steady.
This figure has the same two pulleys, but the rope is applied differently
and it is pulled upwards. The mechanical advantage is 3, and the force to
hold the weight in equilibrium is 1/3 the weight.
YOUR TURN:
When designing the system you will utilize to raise and lower a TV, etc., think about the load
you plan to lift.
1. How heavy is it?
2. Can pulley positions be placed to take advantage of leverage?
Particularly important to remember is that both sides of your platform MUST be raised with
identical pulley layouts. If this isn't done, one side will move faster than the other --
Typical pulley problem for assignment (and midterm)
This here is the simplest system - it can get far more complex. But the physics doesn’t change ...
This is a pulley system where masses m and M are connected by a
rope over a massless and frictionless pulley. Note that M > m and
both masses are at the same height above the ground. The system is
initially held at rest, and is then released.
Calculate the acceleration of the masses, the velocity of mass m when
it moves a distance h, and the work done by the tension force on
mass m as it moves a distance h.
Before we start calculating, let’s go through the three steps for
problem solving:
1. Ask yourself how the system will move
2. Choose a coordinate system
3. Draw free-body diagrams
YOUR TURN
+y
T
T
mg
Mg
1. Ask yourself how the system will move: From
experience, we know that the heavy mass, M, will fall, lifting
the smaller mass, m. Because the masses are connected, we
know that the velocity of mass m is equal in magnitude to the
velocity of mass M, but opposite in direction. Likewise, the
acceleration of mass m is equal in magnitude to the
acceleration of mass M, but opposite in direction.
2. Choose a coordinate system: You should choose the
coordinate system that it simplifies your calculation. In this
case, let’s follow the standard convention of saying that up is
the positive y direction and down is the negative y direction.
3. Draw free-body diagrams: We know that this pulley
system will accelerate when released, so we shouldn’t expect
the net forces acting on the bodies in the system to be zero.
1. What is the acceleration of mass M?
YOUR TURN
+y
T
T
mg
Mg
1. What is the acceleration of mass M?
Because the acceleration of the rope is of the same magnitude
at every point in the rope, the acceleration of the two masses
will also be of equal magnitude. If we label the acceleration of
mass m as a, then the acceleration of mass M is –a. Using
Newton’s Second Law we find:
for mass m: T - mg = ma
for mass M: T - Mg = -Ma
By subtracting the first equation from the second, we find
(M – m)g = (M + m)a or a = (M – m)g/(M + m).
Because M – m > 0, a is positive and mass m accelerates
upward as anticipated. This result gives us a general formula
for the acceleration of any pulley system with unequal
masses, M and m. Remember, the acceleration is positive for
m and negative for M, since m is moving up and M is going
down.
g (M - m)
a=
(M + m)
2. What is the velocity of mass m after it travelled a
distance h?
YOUR TURN
+y
T
T
mg
Mg
2. What is the velocity of mass m after it travelled
a distance h?
We can tackle this problem in terms of energy. Because the masses in
the pulley system are moving up and down, their movement
corresponds to a change in gravitational potential energy. Because
mechanical energy E is conserved, we know that any change in the
potential energy PE of the system will be accompanied by an equal but
opposite change in the kinetic energy KE of the system: ΔKE = -ΔPE
Remember that since the system begins at rest, KE0=0.
As the masses move, mass M loses Mgh joules of potential energy,
whereas mass m gains mgh joules of potential energy. Applying the law
of conservation of mechanical energy, we find:
1/2Mv2 + 1/2mv2 = -(mgh - Mgh) --> 1/2(M+m)v2 = gh(M-m)
v2 = 2gh (M-m)/(M+m)
Mass m is moving in the positive y direction.
3. What is the work done by the force of tension in
lifting mass m a distance h?
YOUR TURN
+y
T
T
mg
3. What is the work done by the force of tension in
lifting mass m a distance h?
Since the tension force T is in the same direction as the displacement
h we know that the work done is equal to hT. But what is the
magnitude of the tension force? We know that the sum of forces
acting on m is T – mg which is equal to ma. Therefore, T = m(g – a).
From the solution to question 1, we know that a = g(M – m)/(M + m),
so substituting in for a, we get:
W = hT = h m(g-a) = mgh - mha = mgh - mh[g(M-m)/(M+m)]
Mg
W = mgh (1 -
M-m
)
M+m
Many examples ....
Pulleys are used everywhere: industry, climbing, sailing, etc.