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Physics 11 - Unit 4 FORCES
Lesson 6- Atwood machine Testing Newton’s Laws of motion
George Atwood (1746-1807) developed a simple construction known as the Atwood machine
which gives you an exceptionally simple but excellent tool for testing and verifying Newton’s
Laws of Motion. The idea of the Atwood Machine is explained on Fig.1
A pulley is mounted about 2.0m from the ground on a stand with a scale on it to measure the
positions of attached objects.
 Two weights are attached to both ends of a long string which passes over the pulley. In the
free body diagram of the Atwood's machine.
 T is the tension in the string, m is the smaller mass, M is the larger mass, and g is gravity’s
acceleration. Assuming that the pulley and the string are massless and the string doesn't
stretch, and that there is no friction, the resultant force on M is the difference between
the tension and mg . The net force on M is the difference between the tension and Mg.
The resultant force exerted on both masses is
If the difference M-m is relatively small, acceleration a is easy to measure with a simple
stopwatch. If we assume the knowledge of gravitational acceleration g, the Atwood Machine
allows us to check Newton’s Second Law of Motion.
When M = m we can verify Newton’s First Law of Motion. If we assumed Newton’s Second
Law of Motion as true, we can then determine, from Equation (3), the value of gravitational
acceleration g. The applications of the Atwood Machine are quite universal, especially for
educational purposes.
Example: Refer to the following information for the next six questions
In our example the two masses are attached to the ends of a single cord that passes over a mass
less, frictionless pulley suspended from the ceiling.
This situation is called an Atwood Machine.
a) Write the equation for net Fy = may for the 2 kg mass.
b) Write the equation for net Fy = may for the 5 kg mass.
c) Why did we not need to subscript the tension variables?
d) Why did we not need to write the equations for net Fx = max for these two bodies?
e) What is the acceleration of the system?
f) What is the tension in the cord?
Hanging Masses
Now suppose that the right mass is hanging off a frictionless table and the cord connecting it to the
left mass passes over a "mass less, frictionless" pulley.
How would this change the free body diagrams for each mass and their equations of motion?
Table Surface
Let's consider as our first example, two identical objects being dragged across the surface of a
frictionless table by two cords.
The two free body diagrams would look like:
A man lowers himself with the help of a pulley
A man with a mass of m1=80kg lowers himself by h=10m along a wall, while fixed to a rope that
runs over a frictionless pulley to a m2=70kg sandbag.
(a) What is his final speed if he started from a state of rest?
(b) How long does it take to “travel” this distance?
Solution.
The Figure below illustrates the problem schematically. We can write following equations.
Block On an Incline (after the Christmas break)
A block of mass m1=2.0kg on a frictionless inclined plane of angle 20 deg is connected by a rope
over a pulley to another block of mass m2=1.0kg. What are the magnitude and direction of the
acceleration of the second block?
The figure below will help us to solve this problem.
The force m1g, caused by gravity, can be decomposed into two forces: F1a and F1b.
Elementary geometry and the definitions of trigonometric functions allow us to write
The force F1a, represented by the red vector is compensated, according to Newton’s Third Law of
Motion, by a force represented by the green vector. So we are left with forces F1b and F2 acting
one against the other trough a rope connecting blocks.
Let’s assume that F1b > F2. This means that block m2 will move with acceleration a directed
upwards. If our assumptions are wrong, the calculated value of acceleration will be negative.
The minus sign tells us, that acceleration has direction opposite to the one chosen by us for
writing equations leading to the solution of the problem. This is a general rule in all kinds of
problems. Negative numerical value means the direction of the parameter found is opposite to the
one which was assumed for writing equations.
Name:________________________
Physics 11
Block: _________
Forces: The Atwood Machine Assignment
Solve the following in your paper –
Please note that this assignment is due Friday Jan. 6, 2012(After the Christmas break)
1. If m1= 8.3 kg and m2= 12.4 kg
a) Draw a free body diagram for each mass, labeling each force.
b) Find the magnitude of the acceleration of both masses. Which mass is mass is
up? Which mass is accelerating down?
c) Find the tension in the rope.
accelerating
2. If m1= 9.3 kg and m2= 16.4 kg
a) Draw a free body diagram for each mass, labeling each force.
b) Find the magnitude of the acceleration of both masses. Which mass is mass is accelerating
up? Which mass is accelerating down?
c) Find the tension in the rope.
3. If m1= 18.3 kg and m2= 12.4 kg
a) Draw a free body diagram for each mass, labeling each force.
b) Find the magnitude of the acceleration of both masses. Which mass is mass is accelerating
up? Which mass is accelerating down?
c) Find the tension in the rope.
4. If m1= 8.3 kg and m2= 8.3 kg
a) Draw a free body diagram for each mass, labeling each force.
b) Find the magnitude of the acceleration of both masses. Which mass is mass is accelerating
up? Which mass is accelerating down?
c) Find the tension in the rope.
5. Suppose the tension in the rope is 75 N, and m1 is accelerating upwards at 2.3 m/s2.
What is m1? What is m2?
6. Suppose the tension in the rope is 105 N, and m1 is accelerating downwards at 1.9 m/s2.
What is m1? What is m2?
7. Suppose the tension in the rope is 98 N, and m1 is accelerating upwards at 5.3 m/s2. What is m1?
What is m2?
8. Suppose m2 = 18 kg, and is accelerating downwards at 2.4 m/s2.
a) What is the tension in the rope?
b) What is the value of m1?
9. Suppose m2 = 5.0 kg, and is accelerating upwards at 3.4 m/s2.
a) What is the tension in the rope?
b) What is the value of m1?
10. Nasty Question. This is from an old Physics 12 scholarship exam.
A 32.0 kg box is placed on a scale inside an elevator. The total mass of the elevator plus
contents is 725 kg. Suddenly, the supporting cable breaks and the elevator accelerates
downwards. However, it is slowed slightly by the emergency brakes, which drag along the
walls of the shaft. While the elevator is falling, the scale reads 210 N.
a) Draw a free body diagram showing the forces acting on the box.
b) Draw a free body diagram showing the forces acting on the elevator.
c) What is the acceleration of the elevator?
d) What is the magnitude of the frictional force of the brakes against the walls of the shaft?
11. From the June 2004 Provincial Exam:
Two masses are connected by a light cord passing over frictionless pulleys as shown
diagram below.
What is m2 if the system accelerates as shown?
in the
The Tension Force
Draw FBDs for all problems!
1. A 550 kg elevator is at rest and is being held by a cable. Determine the tension forces in the
cable.
2.
A 25 kg sign hangs outside a restaurant and is being held by two cables. Determine the
tension forces in each cable. (hint: the weight is distributed evenly between the two cables!)
3. A 12.0 kg bucket is lifted upwards with constant velocity by a rope. What is the tension in the
rope?
4. A 12.0 kg bucket is accelerated upwards at 0.60 m/s2 by pulling a rope. What is the tension in
the rope?
5. A 12.0 kg bucket held by a rope is accelerated downwards at 0.80 m/s2. What is the tension in
the rope?
6. A 30.0 kg block is pulled along a table surface by a rope. If the table surface has a µ = 0.18,
what applied force (the tension force) in the rope necessary to accelerate the block at 1.5 m/s2?
7. A 20.0 kg block is pulled along a table surface by a rope. If the table surface has a µ = 0.10,
what applied force (the tension force) in the rope necessary to accelerate the block at 2.5 m/s2?
8. Two buckets are at rest as arranged in the diagram below. Determine the tension in each rope.
Rope #1
10 kg
Rope #2
20 kg
9. What is the tension in the rope that joins the two blocks that are pulled by a constant force of
500 N in the diagram below:
10 kg
20 kg
µ = 0.12
10. Safety engineers estimate that an elevator can hold 15 persons of 80 kg average mass. The
elevator itself has a mass of 500 kg. Tensile strengths show that the cable supporting the
elevator can withstand a maximum force of 2.85 x 104 N. What is the greatest acceleration that
the elevator can attain without breaking the cable?
The Tension ForceAnswers
Draw FBDs for all problems!
1. A 550 kg elevator is at rest and is being held by a cable. Determine the tension forces in the cable.
Fnet = Fg - FT
0 = mg - FT since it not accelerating Fnet is zero
FT = mg
FT = (550)(9.80) = 5390 N
2. A 25 kg sign hangs outside a restaurant and is being held by two cables. Determine the tension forces
in each cable. (hint: the weight is distributed evenly between the two cables!)
Fnet = Fg - FT
0 = mg - FT since it not accelerating Fnet is zero
FT = mg
FT = (25)(9.80) = 245 N (divide this by 2 to find Tension in one cable)
FT = 123 N
3. A 12.0 kg bucket is lifted upwards with constant velocity by a rope. What is the tension in the rope?
Fnet = FT – Fg
0 = FT - mg
since bucket is not accelerating Fnet is zero
FT = mg
FT = (12)(9.80) = 118 N
4. A 12.0 kg bucket is accelerated upwards at 0.60 m/s2 by pulling a rope. What is the tension in the
rope?
Fnet = FT – Fg
ma = FT - mg since bucket is accelerating Fnet is included
(12)(0.60) = FT - (12)(9.80)
FT = 125 N (note that there is more tension in the rope as the bucket is pulled upwards!)
5. A 12.0 kg bucket held by a rope is accelerated downwards at 0.80 m/s2. What is the tension in the
rope?
Fnet = Fg – FT
ma = mg - FT since bucket is accelerating Fnet is included
(12)(0.80) = (12)(9.80) - FT
FT = 108 N (note that there is less tension in the rope as the bucket is dropped downwards!)
6. A 30.0 kg block is pulled along a table surface by a rope. If the table surface has a µ = 0.18, what
applied force (the tension force) in the rope necessary to accelerate the block at 1.5 m/s2?
Fnet = FT – Ffr
ma = FT – µFN
since block is accelerating Fnet is not zero!
(30)(1.5) = FT - µmg
(30)(1.5) = FT – (0.18)(30.0)(9.80)
FT = 98N
7. A 20.0 kg block is pulled along a table surface by a rope. If the table surface has a µ = 0.10, what
applied force (the tension force) in the rope necessary to accelerate the block at 2.5 m/s2?
Fnet = FT – Ffr
ma = FT – µFN
since block is accelerating Fnet is not zero!
(20)(2.5) = FT - µmg
(20)(2.5) = FT – (0.10)(20.0)(9.80)
FT = 70 N
8. Two buckets are at rest as arranged in the diagram below. Determine the tension in each rope.
Rope #1
10 kg
Rope #2
20 kg
Fnet = FT – Fg
the FT of the other rope cancels out!
0 = FT - mg10 – mg20
since bucket is not accelerating Fnet is zero
FT = mg10 + mg20
FT = (10)(9.80) + (20)(9.80) = 294 N
Fnet = FTrope 1 – mg10 - FTrope 2
0 = 294 - (10)(9.80) - FTrope 2
FTrope 2 = 196 N
9. What is the tension in the rope that joins the two blocks that are pulled by a constant force of 500 N
in the diagram below:
10 kg
20 kg
µ = 0.12
Fnetsys = Fapp – Ffr20 – FT + FT – Ffr10
msysa = Fapp – Ffr20 – Ffr10
since block is accelerating Fnet is not zero!
(30)(a) = Fapp - µm20g - µm10g
(30)(a) = 500 – (0.12)(20.0)(9.80) – (0.12)(10.0)(9.80)
a = 15.5 m/s2
Fnet10 = FT – Ffr10
m10a = FT – Ffr10
since block is accelerating Fnet is not zero!
(10)(15.5) = FT - µm10g
155 = FT - (0.12)(10.0)(9.80)
FT = 167 N
10. Safety engineers estimate that an elevator can hold 15 persons of 80 kg average mass. The
elevator itself has a mass of 500 kg. Tensile strengths show that the cable supporting the elevator can
withstand a maximum force of 2.85 x 104 N. What is the greatest acceleration that the elevator can
attain without breaking the cable?
accelerated the object upwards, we should consider this when constructing our FBD. We should
use FT = 2.85 x 104 N
Fnet = FT – Fg
ma = FT - mg
total mass is 1700 kg (15 x 80 + 500)
(1700)(a) = 2.85 x 104 – (1700)(9.80)
a = 6.96 m/s2