1 Ch5 5.1 a) A quickly applied jerk will break the upper string. The large mass will have a small acceleration. If the force is applied quickly mass will not move much. Correspondingly, the upper string will not elongate enough to break. b) The upper string initially has much more tension than the lower string. By slowly increasing the tension, the large mass has ample time to move. The force added to the bottom string directly adds to the upper string and causes the upper string to break. 5.2 Larger masses will tend to stay in place better by Newton’s 1 st law. You might be concerned about the elevated center of mass of the wine glass and it possibly tipping over. Tipping will be discussed much later. For now, try experimenting. I recommend a plastic wine glass for practice. This link shows one of my former students in action: https://www.youtube.com/watch?v=9NV1E0Myx4o 5.3 a) According to the coordinates −=−. Here a is the magnitude of acceleration; the minus sign appears on a because (the acceleration vector) points opposite the positive y-direction. This gives =− b) When the acceleration is zero the scale reads the true weight of the mass. This occurs when the mass is either at rest or moving with constant velocity!!! Vid 1 (accelerating upwards): https://youtu.be/vuldes602TA Vid 2 (accelerating downwards): https://youtu.be/tsdo5XLXyo4 Vid 3 (accelerating downwards in freefall): https://youtu.be/L7BGrov1HyU 5.4 a) Now the normal force points opposite the positive y-direction. We find −= which gives =− Compare this result to the previous problem. Hopefully you notice the only significant difference is how the coordinates were chosen. Notice the minus signs in the FBD equations all flip (including a) depending on the choice of coordinate system. In the end, both choices give identical results for the magnitudes. b) When elevator moves with constant velocity the scale reads the true weight of the mass. 5.5 10 N. Consider if the string on the right were connected to a wall. You’d have no trouble believing the result. But the scale can’t tell the difference between these two scenarios…the scale must read the same in either case. The point is to emphasize the scale reads tension in the sting…not the weight of a mass (of masses) causing tension.
Transcript
1
Ch5
5.1
a) A quickly applied jerk will break the upper string. The large mass will have a small acceleration. If the
force is applied quickly mass will not move much. Correspondingly, the upper string will not elongate
enough to break.
b) The upper string initially has much more tension than the lower string. By slowly increasing the tension,
the large mass has ample time to move. The force added to the bottom string directly adds to the upper
string and causes the upper string to break.
5.2 Larger masses will tend to stay in place better by Newton’s 1st law. You might be concerned about the elevated
center of mass of the wine glass and it possibly tipping over. Tipping will be discussed much later. For now, try
experimenting. I recommend a plastic wine glass for practice.
This link shows one of my former students in action: https://www.youtube.com/watch?v=9NV1E0Myx4o
5.3
a) According to the coordinates � − �� = ��−��. Here a is the magnitude of acceleration; the minus sign
appears on a because � (the acceleration vector) points opposite the positive y-direction. This gives � = ��� − �� b) When the acceleration is zero the scale reads the true weight of the mass. This occurs when the mass is
either at rest or moving with constant velocity!!!
Vid 1 (accelerating upwards): https://youtu.be/vuldes602TA
Vid 2 (accelerating downwards): https://youtu.be/tsdo5XLXyo4
Vid 3 (accelerating downwards in freefall): https://youtu.be/L7BGrov1HyU
5.4
a) Now the normal force points opposite the positive y-direction. We find �� − = �� which gives = ��� − �� Compare this result to the previous problem. Hopefully you notice the only significant difference is how
the coordinates were chosen. Notice the minus signs in the FBD equations all flip (including a) depending
on the choice of coordinate system. In the end, both choices give identical results for the magnitudes.
b) When elevator moves with constant velocity the scale reads the true weight of the mass.
5.5 10 N. Consider if the string on the right were connected to a wall. You’d have no trouble believing the result.
But the scale can’t tell the difference between these two scenarios…the scale must read the same in either case.
The point is to emphasize the scale reads tension in the sting…not the weight of
a mass (of masses) causing tension.
2
5.6
a) If the mass is not accelerating we expect �� = ��.
b) See figure at right.
c) For the x-direction we get �� sin � −�� cos � = 0
For the y-direction we get �� cos � ��� sin� − �� = 0
d) The quickest way is to first set up the equations as seen below: �� sin � = �� cos� �� cos � = �� − �� sin�
Notice �� doesn’t cancel. The result is ugly…so what. Verify the units check for practice.
e) The quickest way is to first set up the equations as seen below:�� sin � = �� cos� �� cos � = �� − �� sin�
Now square both sides of both equations ��� sin� � = ��� cos ��� ��� cos� � = ��� − �� sin���
Add the equations giving ����sin� � � cos� �� = ��� cos ��� � ��� − �� sin���
Using sin� � � cos� � = 1 and taking a square root gives �� = !��� cos ��� � ��� − �� sin��� In this case, because �� is the magnitude of tension 2, we should use the positive root. �� = !��� cos ��� � ��� − �� sin���
T3 = mg only when a = 0
�� cos� �� sin �
�� sin�
�� cos� � = 0 " #
3
5.7
a) The FBD changes to the one shown at right. Notice the following:
There is no normal force upwards on the hanging mass.
A lot of students initially think that do that.
I can see why you might think that but it is not a good model.
Since you have a single string and negligible friction, the angles are the same and
the tension throughout that string is the same everywhere.
The force equations are
Σ%&:� cos � − � cos� = 0 Σ%(:� sin� � � sin � − �� = 0
b) Using the vertical force equation gives � = ��2 sin�
c) For � = 90° the strings are vertical. We expect the tension should be half of the weight. � = ��2 sin 90° � = ��2�1� � = ��2
Answer is half the weight as we expect.
d) For � , 30° we find � . ��. Why care? For small angles we see the tension in the string will tend to be
very large. The string is likely to break if � is small.
e) Think about a clothesline. Typically � is small and you hang a lot of heavy, wet laundry on it. The tension
on a fully loaded clothesline must be huge � = /0� 1234. Think about the force this must exert on the points
where the clothesline connects to a wall or pipe. Because these forces are huge, clotheslines probably need
to use very sturdy supports, bolts, wires, etc.
Braces for teeth end up with a similar FBD. In this case the tension in the arch wire gradually realigns the
teeth. The connection points for the arch wire come from brackets on adjacent teeth. Again the angles tend
to be somewhat small. This implies the tension in the wire connecting adjacent teeth is probably pretty
large. Also, the brackets are basically bonded onto your teeth with adhesive. That adhesive had better be
pretty awesome to work well.
WATCH OUT! I usually do this during a classroom demo and sometimes there is enough friction between the
hanging mass and string to cause asymmetry in the angles.
Real life does not always match our simplified models!
��
� �
� �
4
5.8
a)
Σ%&:� cos � = ��& Σ%(:� sin � � − �� = ��5 b) One finds = ��� � �5� − � sin �. Do some reality checks: when the angle is zero the zombie is not
lifting up on the block at all and we get back a familiar result (if �6 = 0 then = ��). As the angle
increases the normal force should decrease. Also, the units check.
c) Beware. If a problem says “acceleration” you are to assume this implies acceleration vector �. One way to describe this is
� = �� cos � 7̂ � �59̂ 5.9
a)
Σ%&:� cos � − �� sin � = 0 Σ%(:� sin� � − �� cos � = 0 b) Notice the only unknown in Σ%& is �. This gives : = ;< =>? @AB=C.
c) One finds = �� cos � − � sin �
Remember, in this problem you were told � is unknown…that means it shouldn’t be in your final answer!
Plug in the result from part b for the tension to find
= �� cos � − �;< =>? @AB=C� sin �
= �� cos � − �� sin � sin�cos �
= �� cos � − �� sin � tan�
D = ;<�AB= @ − =>? @ EF?C�
� cos �
� sin �
��
� "
#
�& �6 �
� cos�
� sin�
��
Constant velocity � = 0
"
#
�
�
�� sin �
�� cos �
5
5.10
a)
Σ%&:�� sin � − % cos � = �� Σ%( :− % sin � � − �� cos � = 0
b) Use the Σ%&. This gives % cos � = �� sin � − ��
% = �� sin �cos � − ��cos �
% = �� tan � − ��cos �
% = �� tan � − �� ∙ ��cos �
H = ;<�EF?@ − I< AB= @� c) One finds = �� cos � � % sin � = �� cos � �;<�EF?@ − I<AB= @� sin �
= �� Jcos � � tan � �sin � − ���K
% cos � % sin �
��
�
"
#
�
�� sin �
�� cos �
6
5.11 For each case think �& = LMN/ and �( = LMO/ . We also know sin 45° = √�� = √�√�√� = �√� = cos 45°. In doing part D, the force components of the diagonal force are smaller than the purely horizontal or vertical forces!
To get the third column for part B directly, a trick is to realize the vertical and horizontal forces cancel. Then rotate
the coordinates to line up with the diagonal force and VOILA!
Side note: perhaps you are wondering about part C…A common physics trick
for two identical forces like part C is to rotate the coordinates such that axis
splits the angle between the two forces (see at right). From this we see the net
force is simply twice the component parallel to the blue axis. Said another way
� = %S5T� = 2%� cos 22.5° One can use a half angle formula from trig (notice cos 22.5° = cos VWX°� Y) to
show this result simplifies to
� = !2 � √2%� Z 1.85 %�
Evidently this result cannot be “denested”. The cleanest way to simplify it is to
write a decimal using a calculator as shown above. You might read about “denesting radicals” using a web search. I
found it pretty darn fun.
Case I\ I] I = ^I\_ � I]_
A %� 0
%�
B %√2� Z 0.707 %�
%√2� Z 0.707 %� %�
C %√2� Z 0.707 %� �1 � 1√2� %� Z 1.707 %� 1.85 %�
D �1 − 1√2� %� Z 0.293 %� �1 − 1√2� %� Z 0.293 %� 0.414 %�
FBD Case D
%
%
%
M√� %√2
FBD Case C
%
%
%√2
%√2
FBD Case B
%
%
%√2 %
FBD Case A
%
%
%
FBD Case C - Rotated
%
%
22.5°
7
5.12
a) After the push the masses should move with constant speed (acceleration is zero). � = 10N.
b) During the push the masses are both accelerating and the scale reading should increase.
If it helps, consider an FBD of the right mass.
The tension (magnitude) must be greater than �� for it to accelerate upwards.
5.13 An object in motion tends to stay in motion (in a straight line) unless acted upon by an external force. The tube
exerted a force on the marble (towards the center of the circle) to cause it travel in a circle (instead of a straight line
path). With the tube no longer touching the marble, the marble follows the straight line path C.
8
5.13¼ By Newton’s SECOND law Σ% = �� %� � %� = �� %� = �� − %�
The first force was given as %� = �07̂ � 12.509�̂N.
The object has acceleration � = �−3.007̂ � 09�̂ b1c. The object has mass � = 2.50kg. The object has �� = �−7.507̂ � 09�̂N
Notice the units of ��& and ��( are kg∙ b1c = N. %� = �� − %�
I will leave off units for clarity but remember to put them back in on the final answer. %� = �−7.507̂ � 09�̂ − �07̂ � 12.509�̂ %� = �−7.507̂ − 12.509�̂N
Immediately sketch this to get a feeling for it. See my sketch at right.
The magnitude is %� = ^%�&� � %�(� = 14.58N.
The angle is found using � = f��� VMcOMcNY. In practice, I ignore the minus signs and use � = tan�� gMcOMcNg in tandem with my sketch.
Angle is � = 59.0° below the negative "-axis as shown in the figure.
The second force has magnitude %� = 20.0N pointing 30.0° above the negative "-axis.
See figure at right. %� = �−17.327̂ � 10.009�̂N
The object has acceleration � = �2.597̂ − 8.759�̂ b1c. The object has mass � = 2.00kg. Notice the units of ��& and ��( are kg∙ b1c = N.
Since everything has units of Newton’s, I will leave off units for now but include them on the final answer. %�& = �2.00��2.59� − �0� − �−17.32�jkl%�( = �2.00��−8.75� − �−19.6� − �10.0� %�& = 22.5jkl%�( = −7.9 Immediately sketch this to get a feeling for it. See my sketch at right.
The magnitude is %� = ^%�&� � %�(� = 23.7N.
The angle is found using � = tan�� VM}OM}NY. In practice, I ignore the minus signs and use � = tan�� gM}OM}Ng in tandem with my sketch.
Angle is � = 19.3° below the positive "-axis as shown in the figure.
5.13rqb) Assume each division represents 5N. The figure looks like the one shown below.
Hmn
20.0Nsin
30.0°
20.0N cos30.0°
7.9N
22.5N
11
5.14 The sun pulls the earth towards the center of the sun with gravitational force. The earth pulls the sun towards
the center of the earth with gravitational force.
5.15 The earth pulls the block downwards using gravity. The block pulls the earth upwards using gravity. We don’t
notice the force on the earth because the its mass is so large we cannot discern the effects of the tiny force mg.
5.16 The earth pulls down on the block using gravity; the block pulls up on earth using gravity.
The zombie pushes to the right on the block; the block pushes to the left on the zombie.
The incline pushes up & left on block using normal force; block pushes down & right on incline using normal force.
5.17
a) The seat pushes the driver forward using a normal force; driver pushes seat backwards using normal force.
Note: the seat also provides a normal force directed upwards to oppose the weight of the driver.
By Newton’s 3rd law the driver also pushes downwards on the seat with a normal force.
The net force of the car seat on the driver is forwards and up.
The net force of the driver on the car seat is backwards and down.
b) Friction between tires & road pushes car forwards; friction between road & tires pushes road backwards.
The road also exerts a normal force upwards on the car (on the tires) to support the weight of the vehicle.
By Newton’s 3rd law, the vehicle (the tires) exerts a normal force downwards on the road.
The net force of the road on the car (tires) is forwards and up.
The net force of the car (tires) on the road is backwards and down.
5.18 Your body presses outwards against the car door; the door presses inwards against your body. Notice the force
on your body is towards the center of the circle. Think about the marble in the tube…more in chapter 6.
5.19 Friction between the floor and your feet pushes you forwards; you exert a frictional force backwards on the
floor (and thus the earth). Again, we don’t notice the effect on the earth due to the earth’s enormous mass.
5.20
a) Between stages 2 & 3 the person is moving downwards but slowing down. Therefore person is
accelerating upwards. Must have �~� . �� by Newton’s second law (% = ��).
b) The two forces described are an action-reaction pair. Must have ��� = ��� by Newton’s third law.
c) The action-reaction pair associated with the weight force is best summarized as follows:
Action: The earth exerts a gravitational force directed downwards on the person.
Reaction: The person exerts a gravitational force directed upwards on the earth.
Video of medicine ball being dropped on scale:
The video shows the force versus time for the scale as the ball impacts the scale and comes to rest.
Notice the massive upwards scale reading recorded when the ball first impacts the scale (much larger than the 27 N
weight of the ball). The scale reading then oscillates a bit because the ball actually bounces off the scale twice
before coming to rest. Notice the force reading drops to zero twice after the initial impact as the ball is in the air
during each bounce.
https://youtu.be/2aP974_uoDA
Video of me jumping off the scale:
Tip: you just analyzed what should happen when I jump onto a scale.
Predict what should happen if I jump off of it before watching the video.
The video shows a plot of force versus time during the jump, starting from a crouched position.
https://youtu.be/JSX0lNgrBN0
12
5.20½
Parts a&b) The two blocks are in direct contact with each other.
We are asked to compare the forces two objects exert on each other (forces between two blocks).
Use Newton’s THIRD law: m1�2 = −m2�1 (force vectors) regardless of velocity or acceleration values!
Since the question asked about force magnitudes, we know Dn�D_ = D_�Dn.
Notice the masses have nothing to do with the reasoning required!
This is true regardless of the frictional forces acting on the blocks!
Note: this problem only makes sense if friction is non-negligible. Why?
If a force is applied horizontally AND the objects move with constant speed, there must be additional horizontal
force to balance the applied force. FBDs are shown below. Think: if the system moves at constant speed you could
set acceleration equal to zero.The relative sizes of the arrows in this figure are approximately correct.
We are asked to compare two forces acting on a single object.
Use Newton’s SECOND law to analyze the effects of m1�3&m2�3 ON BLOCK 3.
An FBD for object three only is shown at right.
Sum of forces in the horizontal direction gives
��S� − ��S� = ���
If the blocks move with constant velocity, acceleration is zero.
a) If acceleration is zero, ��S� = ��S�
b) If � . 0, ��S� . ��S�.
Notice the masses have nothing to do with the reasoning required!
��S�
���
� �
2�3
14
5.21 The fan pushes air backwards. By Newton’s third law the air pushes forwards on the fan (and thus the cart).
5.22
a) The fan will not move significantly.
b) The fan pushes air to the left which implies the air pushes to the right on the fan (and thus the cart).
Unfortunately, the sail put will exert a nearly identical force to the right on the air to stop its motion.
The net force exerted by the cart (fan & sail) on the air is zero!
By Newton’s third law, the net force exerted by the air on the cart (fan & sail) is also zero!
5.23
a) It should move.
b) In this example we still have the same force exerted by the fan on the air.
The funky tube, however, exerts approximately twice as much force as before.
It takes more force to reverse the direction than it does to stop the motion.
We will learn more about this in a chapter on momentum.
The fan pushes air to the right but the tube pushes almost twice as hard on the air to the left.
Accounting for both the fan & tube, the net force exerted by the cart on the air is directed to the left!
By Newton’s 3rd law, net force exerted by the air on the cart (including fan & tube) is directed to the right!
Note: there will be losses due to drag. The cart is now heavier than in the cart in 5.21 (fan without a tube).
As such, this contraption is less effective at producing thrust than the simple case of a fan without a tube.
5.24
a) "�f� = ���f� b) "�f� = V���/Y�/� f�/�
c) One way is to plot " vs f� and also plot " vs f�/�.
If the first graph is linear:
• the cart has constant acceleration
• ����6 = �� �
• If mass is known, the slope can be used to determine the average force.
If the second graph is linear:
• the constant power model is better
• ����6 = V���/Y�/�.
• If mass is known, the slope can be used to determine the average power.
A second method is to determine the position at two points in time.
The constant power and constant force equations will predict different ratios.
Interesting to note: taking the ratio causes the unknown terms (i.e. % or �) to drop out.
The first method probably is more reliable but the second method is probably faster.
5.25
a) Only the normal force between 4 & 5.
b) The normal force between 2 & 3, the normal force between 1 & 2, and the frictional force between 1 & 2.
c) The 1-2-3-4-5 system would work. In fact, any system with both 2 & 5 in it will work.
d) The blocks do not move together as a single unit so I wouldn’t treat it as a system. They don’t have the
same accelerations.
15
5.26 Notice that the force % does not touch �2 and, as such, it should not appear in the �2 FBD.
I typically label the normal force between two blocks with both subscripts. If you do this, you will hopefully
remember that normal force must appear on the FBD of both blocks. Notice the subtle difference when you write
the normal force between the blocks as vectors vs. scalars: m�� = −m�� 12 = 21 The magnitudes are equal but the vectors are opposite directions. In the FBDs we indicate direction with arrows and
list the magnitude next to it. This is why, in both the �� and �� FBDs, the magnitude of the force is written as 12…you could have instead written 21 since it is the same thing.
Notice the weight of each scale on each side cancels out…seems reasonable. The difference in mass between the
two sides is what contributes to the net force making the system accelerate.
a) I found the following results for the system acceleration and the tensions � = � �� −���� ��� � 2�� �� = 2����� ������� ��� � 2��
�� = 2����� ������� ��� � 2��
�� = 2����� ������ ��� �������� ��� � 2��
b) Notice the above results all reduce back to the results of the previous problem when �� = 0. Also notice
that all three tensions are the same. This is one indicator the solution is correct. The units also check. The
acceleration still makes sense when �� = �� or �� , ��. We can feel confident in this result. Please
note, it is important to find as many ways to check your results as possible. Sometimes you won’t find a
mistake until your 4th or 5th check…
5.30 Suppose the lighter mass is shoved downwards and released. The instant after your hand is removed, it no
longer exerts a force on the masses. The lighter mass will accelerate upwards. Since it is moving downwards it is
being lowered while accelerating upwards. Similarly the heavier mass, on the other end of the string, must be
moving upwards and accelerating downwards.
�� ���
� ��
�2�
�
FBD Left
Block (;n� ��
���
�
FBD Left
Scale
�1 �3
���
FBD Right
Scale
�2
�
FBD Right
Block (;_)
20
5.31 the FBDs are drawn below
Force Equations for p; Force Equations for ; #: = 3�� cos � ":� = 3�� sin � ":� = ��
Setting the two results for � equal to each other gives �� = 3�� sin � 13 = sin �
@ = =>?�n �np� Z n�. �r° In the spirit of preparing for the final, recall we usually write final answers with three sig figs. Engineers often write
four sig figs for numbers whose first digit is a 1.
3�� sin �
3�� cos � 3��
�
��
�
� = 0 � = 0
21
5.32 To save time, these forces were not drawn to scale. The FBD for each block is shown below. Get the original
picture and hold out your hand in the shape of the coordinate system. I set it up using my left hand and made my
thumb line up with the string. Let your hand follow the string around making sure your thumb is always lining up
with the string. Notice you get the coordinate systems shown in my figure below.
Note: other choices of coordinate systems are equally valid. We know the final result for acceleration should not
depend on the choice of coordinate system. That said, I often try to line up the coordinates such that the acceleration
is parallel to one of the coordinates. By doing this, one avoids the need to split the acceleration into components.
Force equations
for ;n
Force equations
for ;_
Force equations
for ;p
Force equations
for ;�
X: �� −��� = ���
Y: none
X: �� − �� = ���
Y: � = ���
X: �� ���� sin � − �� = ���
Y: � = ��� cos �
X: % cos� − �� = �W�
Y: W � % sin � = �W�
Now we will see the advantage of aligning the coordinate systems with acceleration always aligned with the same
axis (in this case the "-axis). Stack the "-axis force equations and add them.
• Internal forces (the tensions shown in different colors below) will drop out.
• On the left side we get total mass times acceleration! �� − ��� = ��� �� − �� = ��� �� ���� sin � − �� = ��� �% cos � − �� = �W� % cos � � ��� sin � − ��� = ��� ��� ��� ��W��
I = HAB=C �;p<=>? @ −;n<;n �;_ �;p �;�
b) If the blocks are to remain at rest, the acceleration must be zero. This implies % cos� � ��� sin � − ��� = 0 % cos� = ��� − ��� sin �
% = ��� − ��� sin �cos�
We were also told all masses are �.
% = ��1 − sin �cos�
Solution continues on the next page…
�� �� �2
�3
�1 �3 % % sin� % cos�
2 3 4
��� �2� �4�
�
� � �
22
c) Our model fails if the rightmost block in the chain (�W) lifts off the surface.
If the lead block no longer touches the ground we know W = 0.
This allows us to say our model fails whenever % sin� . �W�
If the magnitude of the force H is known, this constraint tells us the largest angle � the zombie can use.
If the angle C of the force is known, this constraint gives a limit on the largest % the zombie can apply.
This constraint may tell us it is physically impossible to design such a system of blocks.
Alternatively, the system accelerates opposite the direction drawn if �� is really large.
23
5.33 Parts a, b, and c are shown below.
FBD ;n&_ FBD ;n FBD Pulley Σ%(:2� −�T�T� = �T�T�
Note: �T�T = �T�T�� = �� ���
2� − �T�T� = �T�T �10
� = 1120�T�T�
Σ%(: − ��� � � = ���
− ��� � � = �� �10
= 1110��� − �
= 1110��� − 1120 ��� �����
= 2220��� − 1120��� − 1120���
= 1120 ��� −����
Σ%(:% − 2� = 0
Note: I am assuming the pulley has
negligible mass compared to �T�T��.
% = 2� = 1110�T�T�
Notice the ceiling must apply a force
greater than the total weight when the
system center of mass accelerates
upwards…this is reasonable.
When the masses are equal the normal force goes to zero! That result surprised me. Then I tried to reason through
it to see if the result makes sense. To make life simpler, first consider the two masses equal but with zero
acceleration…it is just like a balanced Atwood’s machine! This made me feel a lot better about this result. Also,
the ceiling must exert a force greater than the total mass if the system is accelerating upwards. This checks out. We
expect if the system is not accelerating, the tension should be half the total mass…this checks out.
�
�
�T�T���
�
���
� �
% � � � = 0
24
5.34
a) I found ���� = % −�� sin �% cos � −�� sin �
Since cos � , 1, the numerator is larger and the ratio is greater than 1. In case 1 the zombie is pushing in
the direction of acceleration. In case 2 the zombie is wasting some force pushing the block into the ramp
instead of up the ramp. Seems reasonable.
b) I found �� = �� cos �% sin � � �� cos �
Notice, since sin � . 0, the denominator is larger and the ratio is less than 1. In case 2 the zombie is
wasting some force pushing the block into the ramp instead of up the ramp. Therefore, in case 2, the ramp
must exert more force on the block. Seems reasonable.
c) As � → 0, in each case the system and applied force is horizontal. We expect the ratios should go to 1.
As � → 0 we know sin � → 0 while cos � → 1. It all checks out.
As the angle approaches 90° the block is being pushed, essentially, straight up. Case 2 makes no sense. How could
a zombie pushing horizontally on a block cause it to go upwards? In fact, if friction is negligible as the problem
stated, we expect the block to fall no matter how hard the zombie pushes…
5.35
a) Between 8 and 10 seconds.
b) Between 6 and 8 seconds AND at 1 second.
c) If block accelerates upwards with magnitude � find � = �� ���. If the block accelerates downwards
with magnitude � find � = �� −��. Note: � is less than �� when acceleration is downwards.
Acceleration is downwards when slope of ~f is negative. Between 4 and 6 seconds AND between 8 and 10
seconds.
d) Tension in the cable is equal to the weight only when acceleration is zero. Notice velocity could be non-
zero as long as acceleration is zero. Zero acceleration implies zero slope (flat line) on ~f graph. Between 2
and 4 seconds AND between 6 and 8 seconds.
e) The initial position is 20.0m from problem statement. The initial velocity is −4.0b1 from graph. Note:
probably unrealistic to get three sig figs from reading this graph. Notice slope of ~f plot is constant for the
first 2.0 seconds which implies � is constant. Use rise over run to determine the slope as � = �4.0 b1c. Since � is constant we know
"� = "� � ~�&f � 12 �&f� = 20.0m − V4.0ms Y f � V2.0 ms�Y f�
Note: it is common for us to assume "� = "�f�. You may use this same procedure to determine an
equation for the motion for the next two seconds. Be careful thought! Remember the start time is now 2.0s instead of 0!!! In all equations shift the time by 2.0s using f → f − 2.0s. You can keep working
your way across the graph to determine an equation of motion for any straight line segment.
25
5.36
Action Reaction
String pulls up and to the right on cheese using tension
force.
Cheese pulls down and to the left on string using
tension force.
Floor pulls left on cheese using friction force. Cheese pulls right on floor using friction force.
Floor pushes up on cheese using normal force. Cheese pushes down on floor using normal force.
Earth pulls down on cheese using gravitational force. Cheese pulls up on earth using gravitational force.
5.37 Imagine a mass connected to a spring or a rubber band. It will move downwards but eventually slow down.
• Moving down implies velocity is negative.
• If it is slowing down, the signs of signs of � and ~ must be opposite.
• Slowing down, then, must imply acceleration is positive.
A bungee jumper is a specific case of the above example.
Another thought: imagine a space ship landing on the moon. Perhaps it is initially moving downwards at high
speed. As it nears the surface it fires a thruster so that it might slow down. Again we have negative velocity
(moving down) and upwards acceleration.
As a funky example, consider an Atwood’s machine where the string is too long. The heavy mass sits on the ground
with ample extra string. The light mass is released from rest. Initially there is zero tension in the string and the light
mass falls freely to the ground. Just before hitting the ground the string goes taut. The light mass will lift the heavy
mass off the ground. At that point we have an Atwood’s machine where the lighter mass is still moving downwards
(negative velocity) but accelerating upwards (slowing down).
26
5.38 WATCH OUT! Only use the system FBD if all objects are moving in unison with identical acceleration.
If the system FBD is applicable, ignore internal forces on the system FBD.
The system FBD is awesome for this problem because the situation is not that complicated.
Doing separate FBD’s for each object will ALWAYS solve the problem.
Doing the system FBD will SOMETIMES solve the problem
Most often I use the system FBD to determine the total acceleration (or external applied force).
Then I use the separate FBDs to find the internal forces.
Force Equations for � Force Equations for 2� Force Equations for 3� System Force Equation
�� − � = ��
Σ%&:�� = �2���
Σ%(:� − �2��� = 0
Σ%&:� − �� = �3���
Σ%(:� − �3��� = 0
�� = ��T�T����
Notice �� is the only unbalanced force!!!
Notice the internal forces � & �� do not
appear in the system FBD!!!
a) I assumed motion to the left or down was positive. I found � = /�/��/c�/} � = 0 . Then I found � = X ��.
b) �� = ����
FBD � FBD 2� FBD 3� FBD system
�
��
3��
� � ��
�
2��
��
� " # � "
#
� " # �
p _
n 3��
2��
��
� �
27
5.39
a) Fill in the table for the three special values of �� shown below.
���kg) � V���Y �(N)
0 −� = −10 0
1 0 �� = 10
20 1921 � ≈ 9.0 ≈ 19.0
b) See below.
c) See below. Notice I used a variable step size on �� to ensure a smooth plot. Also, remember the problem
statement said to use � = 10 b1c.
d) Whenever one of the masses is very small (compared to the other one) the larger mass is essentially in
freefall. The smaller mass is thus accelerating upwards with rate �. This can only happen if the tension is
twice as large as the smaller mass. From � = �/�/c0/c�/�
we find ¡
/c0= �/�0
/c�/�. As �� → 0,
¡/c0
→ 2.
m1 (kg) g (m/s2)
1 10
m2 (kg) a (m/s2) T (N)
0.0 -10.0 0.0
0.2 -6.7 3.3
0.4 -4.3 5.7
0.6 -2.5 7.5
0.8 -1.1 8.9
1.0 0.0 10.0
1.2 0.9 10.9
1.4 1.7 11.7
1.6 2.3 12.3
1.8 2.9 12.9
2.0 3.3 13.3
3.0 5.0 15.0
4.0 6.0 16.0
5.0 6.7 16.7
6.0 7.1 17.1
7.0 7.5 17.5
8.0 7.8 17.8
9.0 8.0 18.0
10.0 8.2 18.2
12.0 8.5 18.5
14.0 8.7 18.7
16.0 8.8 18.8
18.0 8.9 18.9
20.0 9.0 19.0
-10.0
-5.0
0.0
5.0
10.0
0.0 5.0 10.0 15.0 20.0
a (m/s2)
m2 (kg)
a vs m2
0.0
5.0
10.0
15.0
20.0
0.0 5.0 10.0 15.0 20.0
T (N)
m2 (kg)
T vs m2
28
5.40
a) The answer is in the problem statement.
b) Use 90°.
c) The component of weight down the plane must equal the hanging mass (�� = �� sin �). Use 30°.
d) See the plots below.
e) See the plots below.
f) I used Excel because that is what I happen to have installed at the time of writing this book. The formatting
in Excel is brutal for this type of plot but, since this is just for fun, I’m ok with having crappy formatting
here. The two darker shades of grey are negative values of acceleration while the two lighter shades
indicate positive acceleration. Having colors would help a lot.
Note: As you move into more technical fields, contour plots can become more useful. Think about how
much information is contained in that plot. You can see acceleration trends as �� is increased for any
fixed angle or as the angle is increased for any fixed �� in one quick glance. Powerful. You are familiar
with contour plots already if you have read a topographic map or watched a weather report. All branches
of science use contour plots to express relationships between three variables instead of the usual two.
-4.0
-2.0
0.0
2.0
4.0
0 15 30 45 60 75 90
a (m/s2)
θ (deg)
a vs θ
0.0
2.0
4.0
6.0
8.0
10.0
12.0
14.0
0 15 30 45 60 75 90
T (N)
θ (deg)
T vs θ
0
30
60
90
0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
θ (deg)
m2 (kg)
-10--5 -5-0 0-5 5-10
29
5.41
a) Both the string above and below the left pulley must move distance �. Expect �� moves 2�.
b) Since, from the previous part, "� = 2"� we expect ~� = 2~� and �� = 2��. c)
Σ%&:�� = ����
Σ%&:�� − 2�� = ��¢��5(��
since ��¢��5( ≈ 0
Σ%&:�� = 2��
Σ%&:��� − �� = ����
Σ%(: = ���
In reality ��¢��5(� downwards is countered by
angled tensions. Since ��¢��5( ≈ 0 we can
neglect this angle issue. This assumption is valid
when ��¢��5( is small compared to �� and ��.
d) We find
�� = � ���� + 4��
�� = 2�� = � 2���� + 4��
�� =2������� + 4��
�� = 2�� =4������� + 4��
Recall �� = 2�� due to the constrained motion. Furthermore, the FBD of the pulley tells us �� = 2��.
e) In my opinion, this is part is the most fun.
i. When �� → 0 there is essentially nothing impeding the motion of m2. We expect a2 should have
magnitude � and it does (verify this yourself). We also expect �� ≈ 0 which checks out.
ii. When �� → ∞ we expect nothing should accelerate. We also expect �� = ���. Verify.
iii. When �� → 0 we expect nothing should accelerate. We also expect �� ≈ 0. Verify.
iv. When �� → ∞ we expect �� ≈ �, �� ≈ 2� and �� ≈ 2���. Verify.
n
m1g
T1
T1
T1
T2
T2
m2g
a1
x
y
a2
x
y
a2
x
y
FBD m1 FBD pulley FBD m2
30
5.42
a) We see "� = 2"� which implies ~� = 2~� and �� = 2��. b)
� −��� = ���� �¤ = 2� ��� − �¥ = ����
�¥ − 2� = ����
If /¦/�
≪ 1 and /¦/c
≪ 1
Then �¥ ≈ 2�
c) We find �� = � /c��/��c/c��/�
= � �/c�W/�/c�W/�
and = 3 /c/�0/c�W/�
. Recall �� = 2�� due to the constrained motion.
Furthermore, the FBD of the pulley tells us �¤ = 2�. Lastly, note that the total downward force on the
ceiling is given by �¤ + � = 3�.
d) In my opinion, this is where you finally get to have some fun.
i. When �� → 0 there is essentially nothing impeding the motion of m2. We expect �� should have
magnitude � and it does (verify this yourself). We also expect � ≈ 0 which checks out.
ii. When �� → ∞ we expect �� ≈ −�. Note: the minus sign comes from the fact that we did all our
math assuming a1 was upwards. We also expect � ≈ 0. Verify.
iii. When �� → 0 the results should be the same as when �� → ∞ . Verify.
iv. When �� → ∞ the results should be the same as when �� → 0. Verify.
v. When �� = 2�� the accelerations are zero, � = ���, �¥ = ���, and the downward force on the
ceiling is 3� = �T�T���.
m1g
T
T T
TA TB
m2g
a1
a2
FBD m1 FBD pulley A FBD m2
T
TB
FBD pulley B
T
a2
a = 0
31
5.43
a) See the table below.
Assumption Why?
Rope is massless. Ensures constant mass on each side of pulley which keeps acceleration constant and
reduces number of forces in FBDs.
Rope is inextensible. Not really needed here but usually this assumption is made. This is important in an
Atwood’s machine because we want to ensure both masses have the same acceleration.
Pulleys are massless
with no axle friction.
Both of these assumptions are required to ensure the tension in a rope does not change
when going from one side of a pulley to the other.
All rope sections are
purely vertical.
Avoids having to split up the forces into components. As long as angles are less than
8° from vertical this assumption introduces less than 1% error.
Mass of the chair is
negligible.
A simple chair made from a plank of wood probably weighs a lot less than a human. If
the human is a small child and the chair is large and metal, this assumption fails.
b) The ceiling supports Chuey (and chair); Chuey is not accelerating. Force on ceiling is �� where � is the
mass of Chuey (and chair).
c) The middle pulley tells us �� = 2��. Vertical forces on Chuey and swing give �� + 2�� = ��.
Combining you should find �� = ���� and �� = �
W��. A 60-lbs kid need only apply 15 lbs to hold himself
up. How cool!
d) If Zuel holds the rope the force increases! When Chuey holds the rope �� is applied to the Chuey-swing
system twice but when Zuel holds the rope only once. The forces from the middle pulley do not change.
The forces on Chuey and the swing become �� + �� = ��. One finds �� = ���� and �� = �
���. This is a
33% increase in the amount of force required to hold up Chuey…silly Zuel, should’ve let him do it himself.
e) Trick question. More info is needed. If lowered at a constant rate the acceleration is still zero and nothing
changes. If lowered with acceleration, the net force on the ceiling should be less than mg. All the tensions
would be slightly lowered. The question would need to be solved similar to problem 5.23.
f) While the problem still has ropes connected everywhere, our assumptions simplify the problem
dramatically. The tensions drop to zero and Chuey is in free fall (acceleration magnitude �). This
demonstrates the need for safety precautions built into any pulley chair demonstration. Simplest method is
to have Zuel back up Chuey on the rope. Another style is to limit the amount of rope that can be pulled by
keeping a short distance between the upper pulleys and the middle one. Some people will put the chair on
guided tracks to reduce swinging or attach the bottom of the swing to the ground using springs or ropes.
