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Forces February 18, 2013 - p. 1/9
February 18, Week 6
Today: Chapter 4, Forces
Exam #1 is in mailboxes
Homework Assignment #5 - Due March 1.
Mastering Physics: 10 problems from chapters 4 and 5.
Written Questions: 5.74
Help sessions with Jonathan:
M: 1000-1100, RH 111 T: 1000-1100, RH 114Th: 0900-1000, RH 114
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Weight - −→w
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Weight - −→w , the downward force on an object due to gravity.
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Weight - −→w , the downward force on an object due to gravity.
Normal Force - −→n
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Weight - −→w , the downward force on an object due to gravity.
Normal Force - −→n , the perpendicular force exerted by one solidobject onto another solid object.
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Weight - −→w , the downward force on an object due to gravity.
Normal Force - −→n , the perpendicular force exerted by one solidobject onto another solid object.
Friction -−→f
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Weight - −→w , the downward force on an object due to gravity.
Normal Force - −→n , the perpendicular force exerted by one solidobject onto another solid object.
Friction -−→f , force which slows a moving object, always
opposed to the motion ⇒ opposite to −→v .
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Weight - −→w , the downward force on an object due to gravity.
Normal Force - −→n , the perpendicular force exerted by one solidobject onto another solid object.
Friction -−→f , force which slows a moving object, always
opposed to the motion ⇒ opposite to −→v .
Tension -−→T
Forces February 18, 2013 - p. 2/9
Force Examples
Forces to be identified in any problem:
Weight - −→w , the downward force on an object due to gravity.
Normal Force - −→n , the perpendicular force exerted by one solidobject onto another solid object.
Friction -−→f , force which slows a moving object, always
opposed to the motion ⇒ opposite to −→v .
Tension -−→T , pulling force exerted by rope, chain, or spring,
always at same angle as rope.
Forces February 18, 2013 - p. 3/9
Newton’s First Law
First Law - The Law of Inertia
An object at rest stays at rest, an object in uniform motionstays if uniform motion if (and only if) the net force acting onthe object is zero.
Forces February 18, 2013 - p. 3/9
Newton’s First Law
First Law - The Law of Inertia
An object at rest stays at rest, an object in uniform motionstays if uniform motion if (and only if) the net force acting onthe object is zero.
Uniform motion - Straight line and constant speed, i.e, constantvelocity.
Forces February 18, 2013 - p. 3/9
Newton’s First Law
First Law - The Law of Inertia
An object at rest stays at rest, an object in uniform motionstays if uniform motion if (and only if) the net force acting onthe object is zero.
Uniform motion - Straight line and constant speed, i.e, constantvelocity.
Inertia - The property of all matter to stay in motion if already inmotion; to stay at rest if already at rest.
Forces February 18, 2013 - p. 4/9
First Law Example
Example: A 6860N car is traveling with a constant 30m/s
speed on a straight road. If the ground is exerting a forward350N force∗, what is the magnitude and direction of all forcesacting on the car? (∗ We’ll learn later that this is due to thecar’s engine.)
Forces February 18, 2013 - p. 4/9
First Law Example
Example: A 6860N car is traveling with a constant 30m/s
speed on a straight road. If the ground is exerting a forward350N force∗, what is the magnitude and direction of all forcesacting on the car? (∗ We’ll learn later that this is due to thecar’s engine.)
Free-Body Diagram - f. b. d. sketch of all the forces acting onan object using a convenient coordinate system.
Forces February 18, 2013 - p. 5/9
First Law Exercise
A 5 kg mass is hung from the ceiling using a "massless" rope.What is the magnitude of the tension force exerted by the ropeon the mass? Hint: A 5 kg mass has a weight of 49N on earthwhere this problem is taking place.
Forces February 18, 2013 - p. 5/9
First Law Exercise
A 5 kg mass is hung from the ceiling using a "massless" rope.What is the magnitude of the tension force exerted by the ropeon the mass? Hint: A 5 kg mass has a weight of 49N on earthwhere this problem is taking place.
(a) 0N
Forces February 18, 2013 - p. 5/9
First Law Exercise
A 5 kg mass is hung from the ceiling using a "massless" rope.What is the magnitude of the tension force exerted by the ropeon the mass? Hint: A 5 kg mass has a weight of 49N on earthwhere this problem is taking place.
(a) 0N
(b) 24.5N
Forces February 18, 2013 - p. 5/9
First Law Exercise
A 5 kg mass is hung from the ceiling using a "massless" rope.What is the magnitude of the tension force exerted by the ropeon the mass? Hint: A 5 kg mass has a weight of 49N on earthwhere this problem is taking place.
(a) 0N
(b) 24.5N
(c) 49N
Forces February 18, 2013 - p. 5/9
First Law Exercise
A 5 kg mass is hung from the ceiling using a "massless" rope.What is the magnitude of the tension force exerted by the ropeon the mass? Hint: A 5 kg mass has a weight of 49N on earthwhere this problem is taking place.
(a) 0N
(b) 24.5N
(c) 49N
(d) 98N
Forces February 18, 2013 - p. 5/9
First Law Exercise
A 5 kg mass is hung from the ceiling using a "massless" rope.What is the magnitude of the tension force exerted by the ropeon the mass? Hint: A 5 kg mass has a weight of 49N on earthwhere this problem is taking place.
(a) 0N
(b) 24.5N
(c) 49N
(d) 98N
(e) Not enough information to determine
Forces February 18, 2013 - p. 5/9
First Law Exercise
A 5 kg mass is hung from the ceiling using a "massless" rope.What is the magnitude of the tension force exerted by the ropeon the mass? Hint: A 5 kg mass has a weight of 49N on earthwhere this problem is taking place.
(a) 0N
(b) 24.5N
(c) 49N
(d) 98N
(e) Not enough information to determine
Forces February 18, 2013 - p. 6/9
First Law Exercise
Two 5 kg masses are connected to each over pulleys using arope. What is the tension force that the rope exerts on theright-hand mass if they are both at rest?
Forces February 18, 2013 - p. 6/9
First Law Exercise
Two 5 kg masses are connected to each over pulleys using arope. What is the tension force that the rope exerts on theright-hand mass if they are both at rest?
(a) 0N
Forces February 18, 2013 - p. 6/9
First Law Exercise
Two 5 kg masses are connected to each over pulleys using arope. What is the tension force that the rope exerts on theright-hand mass if they are both at rest?
(a) 0N (b) 24.5N
Forces February 18, 2013 - p. 6/9
First Law Exercise
Two 5 kg masses are connected to each over pulleys using arope. What is the tension force that the rope exerts on theright-hand mass if they are both at rest?
(a) 0N (b) 24.5N (c) 49N
Forces February 18, 2013 - p. 6/9
First Law Exercise
Two 5 kg masses are connected to each over pulleys using arope. What is the tension force that the rope exerts on theright-hand mass if they are both at rest?
(a) 0N (b) 24.5N (c) 49N (d) 98N
Forces February 18, 2013 - p. 6/9
First Law Exercise
Two 5 kg masses are connected to each over pulleys using arope. What is the tension force that the rope exerts on theright-hand mass if they are both at rest?
(a) 0N (b) 24.5N (c) 49N (d) 98N
(e) Not enough information to determine
Forces February 18, 2013 - p. 6/9
First Law Exercise
Two 5 kg masses are connected to each over pulleys using arope. What is the tension force that the rope exerts on theright-hand mass if they are both at rest?
(a) 0N (b) 24.5N (c) 49N (d) 98N
(e) Not enough information to determine
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
So if Σ−→F 6= 0 ⇒?
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
So if Σ−→F 6= 0 ⇒
−→a 6= 0.
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
So if Σ−→F 6= 0 ⇒
−→a 6= 0.
Forces cause acceleration
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
So if Σ−→F 6= 0 ⇒
−→a 6= 0.
Forces cause acceleration
Newton found that the acceleration is:
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
So if Σ−→F 6= 0 ⇒
−→a 6= 0.
Forces cause acceleration
Newton found that the acceleration is:
(a) In the same direction as the net force
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
So if Σ−→F 6= 0 ⇒
−→a 6= 0.
Forces cause acceleration
Newton found that the acceleration is:
(a) In the same direction as the net force(b) Directly proportional to the net force
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
So if Σ−→F 6= 0 ⇒
−→a 6= 0.
Forces cause acceleration
Newton found that the acceleration is:
(a) In the same direction as the net force(b) Directly proportional to the net force(c) Inversely proportional to the mass
Forces February 18, 2013 - p. 7/9
Newton’s Second Law
The first law tells us that if Σ−→F = 0 then we have a constant −→v
Constant −→v ⇒−→a = 0.
So if Σ−→F 6= 0 ⇒
−→a 6= 0.
Forces cause acceleration
Newton found that the acceleration is:
(a) In the same direction as the net force(b) Directly proportional to the net force(c) Inversely proportional to the mass Measure of the
amount of matterinside an object
Forces February 18, 2013 - p. 8/9
Second Law II
−→a =
Σ−→F
M
Forces February 18, 2013 - p. 8/9
Second Law II
−→a =
Σ−→F
M⇒ Σ
−→F = M−→
a
Forces February 18, 2013 - p. 8/9
Second Law II
−→a =
Σ−→F
M⇒ Σ
−→F = M−→
a
Units: Newton is a unit simplification.
Forces February 18, 2013 - p. 8/9
Second Law II
−→a =
Σ−→F
M⇒ Σ
−→F = M−→
a
Units: Newton is a unit simplification.
Ma
Forces February 18, 2013 - p. 8/9
Second Law II
−→a =
Σ−→F
M⇒ Σ
−→F = M−→
a
Units: Newton is a unit simplification.
Ma ⇒ kg ·m/s2
Forces February 18, 2013 - p. 8/9
Second Law II
−→a =
Σ−→F
M⇒ Σ
−→F = M−→
a
Units: Newton is a unit simplification.
Ma ⇒ kg ·m/s2
ΣF
Forces February 18, 2013 - p. 8/9
Second Law II
−→a =
Σ−→F
M⇒ Σ
−→F = M−→
a
Units: Newton is a unit simplification.
Ma ⇒ kg ·m/s2
ΣF ⇒ N
Forces February 18, 2013 - p. 8/9
Second Law II
−→a =
Σ−→F
M⇒ Σ
−→F = M−→
a
Units: Newton is a unit simplification.
Ma ⇒ kg ·m/s2
ΣF ⇒ N
N = kg ·m/s2
Forces February 18, 2013 - p. 9/9
Second Law Examples
Example: A 6860N car is in free-fall, what it its mass?
Forces February 18, 2013 - p. 9/9
Second Law Examples
Example: A 6860N car is in free-fall, what it its mass?
Example: A 6860N car is sitting stationary on the ground, whatis its mass?
Forces February 18, 2013 - p. 9/9
Second Law Examples
Example: A 6860N car is in free-fall, what it its mass?
Example: A 6860N car is sitting stationary on the ground, whatis its mass?
w = Mg
