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PHYS 100 (from 221)
Newton’s Laws
Week8
Exploring the Meaning of Equations
Exploring the meaning of the relevant ideas and equations introduced recently.
This week we’ll focus mostly on Newton’s second and third laws:
Kinematics describes the development of the motion – once velocities & accelerations
are specified, kinematics tells us how the system will move as a result of those velocities
and accelerations. Newton’s Laws describe the cause of the motion; where the
velocities and accelerations come from. When calculating the net force we’ll often use a
tool called a Free Body Diagram, which is a pictorial way of doing a vector sum:
This tells us how to sum different forces and what the total result will be. Newton’s
second law then relates that total result to the motion of the system. Today we’ll use
most of the forces you are currently familiar with. In particular, we will use the
constraint forces (e.g., T and N) that do not have formulas for them – you have to use
information in the problem to solve for them.
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PHYS 100 (from 221)
Newton’s Laws
Week8
Problem Solving Strategy
Practicing strategic analysis by going through the 3 steps 1) Read and translate to
physics, make drawing (FBD); 2) Conceptual analysis; 3) Connect 1 & 2 to choose a
strategy. Verify your answer with your TA for credit before you move on to the next
problem.
(Spring 04 Exam 1 #4) (74% correct)
Two blocks, of mass M1 = 8 kg and M2 = 3 kg are in contact with each other on a
frictionless floor. A horizontal force F = 72 Newtons is applied to block M1 as shown.
F
M1
M2
What is the force F2,1 of the block of mass M1 on the block of mass M2?
Strategy:
1. Use FBD 1 to determine acceleration
2. Use FBD 2 and acceleration to determine F12
Result:
F12 = F (M2/(M1+M2))
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PHYS 100 (from 221)
Newton’s Laws
Week8
(Fall 2003 Exam 1 #10)
A 211 student runs to catch a Frisbee. Just when the Frisbee passes over her head it is
moving at a speed of 4 m/s, and this is when she starts to run accelerating at a rate of 1
m/s2. The Frisbee is decelerating at a rate of 1.5 m/s2. (You may neglect the vertical
motion of the Frisbee). How far from her initial position does the student catch up with
the Frisbee?
Strategy:
1. Write down equations of motion for Frisbee and student (x as a function of
time)
2. Set positions of Frisbee and student equal to each other and solve for time
3. Substitute time into equation of motion for student to determine distance
Result:
t = 2vfrisbee/(astudent – afrisbee)
(note: afrisbee is negative)
x = (1/2)astudent t2
x = 5.12 m
(Spring 06 Exam 1 #24) (68% correct)
A ball of mass M = 0.2 kg is suspended from a string. The ball travels in a horizontal
circle of radius R = 0.5 m at a constant speed of v = 1.5 m/s.
What is the magnitude of the net force, Fnet, on the mass?
Strategy:
1. Determine acceleration from speed and radius of uniform circular motion
2. Use acceleration and Newton’s second law to determine Fnet.
Result:
Fnet = M v2/R
(note: Fnet is horizontal = vector sum of T + Mg = Tsinθ
θ)
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PHYS 100 (from 221)
Newton’s Laws
Week8
(Spring 05 #7) (97% correct)
A mass of 5 kg is suspended from the ceiling of an elevator by an ideal massless rope.
The rope will break if the tension T exceeds Tmax = 60 N.
g
a
T
The elevator moves upward with the acceleration a. What is the maximum vertical
acceleration the elevator can have without breaking the rope?
Strategy:
1. Write down Newton’s second law for M
2. Substitute T = Tmax to determine amax
Result:
amax = (Tmax – Mg)/M
Tmax= 60N
5kg
(Spring 07 Exam 1 #3) (85% correct)
A mass m = 8.1 kg is suspended by two wires hanging from a beam. The wires are set up
so that they hang at equal angles, θ, from the beam. The magnitude of the tension in
one wire, T1, is measured to be 109 N.
θ
θ
T1
T2
m = 8.1 kg
What is the angle, θ, in degrees?
Strategy:
1. Draw FBD and write down Newton’s second law (note: T1 = T2 from symmetry)
2. Solve for θ
Result:
sinθ
θ = mg/(2T)
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PHYS 100 (from 221)
Newton’s Laws
Week8
Problem Practice
Practicing strategic analysis by going through the 3 steps 1) Read and translate to
physics, make drawing (FBD); 2) Conceptual analysis; 3) Connect 1 & 2 to choose a
strategy; 4) Follow your strategy and solve the problem; 5) Reflect – did the strategy
work? Does the result make sense?
(Spring 05 #8) (81% correct)
A mass of 5 kg is suspended from the ceiling of an elevator by an ideal massless rope.
The rope will break if the tension T exceeds Tmax = 60 N.
Now the elevator is moving in horizontal direction with the acceleration a. What is the
maximum horizontal acceleration the elevator can have without breaking the rope?
Strategy:
1. Write down Newton’s second law for M in vertical and horizontal directions
2. Solve for angle θmax when T = Tmax in vertical equation
3. Substitute T = Tmax and θ = θmax in horizontal eqn to determine amax
Result:
cosθ
θmax = mg/Tmax
amax = Tmaxsinθ
θmax/m
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PHYS 100 (from 221)
Newton’s Laws
Week8
(Fall 03 Exam 1 #18) (65% correct)
A car, which weighs 2000 N, travels over a bumpy road with a constant speed. Gravity
acts. The road at point B is in the shape of a circle with a radius of 100 meters.
With what speed must the car travel such that the force that it exerts on the road at
point B is 600N?
B
100m
A
Strategy:
1. Draw FBD for car at B and write down Newton’s second law
2. Set acceleration at B = v2/R
3. Solve for v from vertical equation using N = 600N
Result:
v2 = R (g – (N/M))
(Fall 08 Exam 1 #10) (% correct)
m
What is the acceleration of the hanging mass?
Result:
a = g (m/(m + 2M))
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PHYS 100 (from 221)
Newton’s Laws
Week8
(Fall 03 Exam 1 #24) (67% correct)
A 1000 kg car is driving with constant velocity up a hill inclined at an angle of 5° with
respect to the horizontal. What is Fnet, the magnitude of the sum of all of the forces on
the car?
Result:
Fnet = 0 (acceleration = zero !!)
(Spring 06 Exam 1 #21) (56% correct)
A box is at rest on a table. What is the Newton’s Third Law reaction force corresponding
to the weight of the box?
Result:
The force the box exerts on the Earth (Universal gravitation: we haven’t done
this one yet). Note: it is not equal to the normal force since both the wieight and the
normal force are forces exerted ON THE BOX.
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PHYS 100 (from 221)
Newton’s Laws
Week8
(Spring 04 Exam 1 #17) (78% correct)
A student twirls a tennis ball of mass m at the end of a string of length l in a vertical
plane.
v
m
l
g
What is the minimum speed, v, required to just keep the string taut when the ball is at
the top of its travel?
Result:
v2 = Rg
You are lying on a hill that slopes at 45°. You throw a ball at a right angle to the hill’s
surface at 6m/s. When the ball lands on the hill below you, how far is it from you?
Geometrical challenge: Draw the picture !
Strategy:
1. determine the initial angle with respect to horizontal
2. write down equations of motion (constant velocity in horizontal and constant
acceleration = -g in the vertical)
3. solve for position of intersection with the y = x line (the hill)
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