NEW L4

Lecture 4
Application of Newton’s Law

Outlines
1. Newton’s third law
2. Free body diagram
3. Equilibrium In
Translational Motion
4. Non - Equilibrium In
Translational Motion

1.0 Newton’s 3rd Law
For every action there is an equal and opposite re-action.

1.0 Newton’s 3rd Law Cont.
 If object 1 and object 2 interact, the force exerted by object 1 on object 2 is equal in

magnitude but opposite in direction to the force exerted by object 2 on object 1.

 Equivalent to saying a single isolated force cannot exist

• F12 may be called the action force and F21 the reaction force.
• Actually, either force can be the action or the reaction force

Some Action-Reaction Pairs



is the normal force, the force the table exerts on the TV and always perpendicular to the surface.

is the reaction – the TV on the table.





is the force the Earth exerts on the object.



is the force the object exerts on the earth.



Object
 Newton’s Law uses the forces acting on an

object.



are acting on the object.



are acting on other objects.

Example 1:
A man of mass M = 75.0 kg and woman of mass m = 55.0 kg stand facing each other on an ice rink, both wearing ice skates. The woman pushes the man with a horizontal force of
F = 85.0 N in the positive x-direction. Assume the ice is frictionless.
(a) What is the man’s acceleration?
(b) What is the reaction force acting on the woman?
(c) Calculate the woman’s acceleration.

Solution:
(a) What is the man’s acceleration?

F Ma M  aM 

aM 

F
M

85.0
1.13 m/s 2
75.0

Solution:
(b) What is the reaction force acting on the woman?

F12  F21
R  F  85.0 N
(c) Calculate the woman’s acceleration.

R
R maW  aW  m  85.0 aW 
 1.55 m/s 2
55.0

Free Body Diagram
 A diagram of the forces acting on an object.
 Must identify all the forces acting on the object of interest.
 Choose an appropriate coordinate system.
 If the free body diagram is incorrect, the solution will likely be incorrect.
 Only forces acting directly on the object are included in the free body

diagram.
 Reaction forces act on other objects and so are not included
 The reaction forces do not directly influence the object’s motion

Free Body Diagram,
Example

 The magnitude of force T is the tension acting on the box.
 The tension is the same at all points along the rope.


are the forces exerted by the earth and the ground.

Solving Newton’s Second Law Problems
 Read the problem at least once.
 Draw a picture of the system.
 Identify the object of primary interest
 Indicate forces with arrows
 Label each force.
 Use labels that bring to mind the physical quantity involved
 Draw a free body diagram.
 If additional objects are involved, draw separate free body diagrams for each object
 Choose a convenient coordinate system for each object
 Apply Newton’s Second Law.
 The x- and y-components should be taken from the vector equation and written

separately
 Solve for the unknown(s).

Equilibrium
 An object either at rest or moving with a constant velocity is said to be in equilibrium.
 The net force acting on the object is zero (since the acceleration is zero).

 Easier to work with the equation in terms of its components:

 This could be extended to three dimensions.
 A zero net force does not mean the object is not moving, but that it is not accelerating.

Equilibrium Example – Free Body
Diagrams

Example 2:
A traffic light weighing 1.00×102 N hangs from a vertical cable tied to two other cables that are fastened to a support, as in figure. The upper cables make angles of 37.0° and 53.0° with the horizontal. Find the tension in each of the three cables.

F

y

0

T3  Fg 0
T3 Fg 1.00 10 2 N

F

x

0

 T1 cos 37.0  T2 cos 53.0 0
 cos 37.0 
 1.33 T1
T2 T1 
 
 cos 53.0 

F

y

0

T1 sin 37.0  T2 sin 53.0  1.00 10 2 0
T1 sin 37.0  1.33 T1  sin 53.0  1.00 10 2 0
T1 60.1 N
T2 1.33 60.1 79.9 N

Inclined Planes
 Choose the coordinate system with x along the incline and y perpendicular to the

incline.
 Replace the force of gravity with its components.

 n 
- mg cos


- mg sin

mg

Example 3:
A sled is tied to a tree on a frictionless, snow-covered hill, as shown in figure. If the sled weighs 77.0 N, find the magnitude of the tension force T : exerted by the rope on the sled and that of the normal force n: exerted by the hill on the sled.

Assume parallel to the slope is x-axis and perpendicular to the slope is y-axis

 F T  n  F 0
 F T  0  mg sin  0 g x

T  77.0  sin 30.0 38.5 N

F

y

0  n  mg cos  0

n  77.0  cos 30.0 66.7 N

Multiple Objects – Example
 When you have more than one object, the

problem-solving strategy is applied to each object.  Draw free body diagrams for each object.
 Apply Newton’s Laws to each object.
 Solve the equations.

Forces of Friction

 When an object is in motion on a surface or through a viscous medium, there will be a

resistance to the motion.
 This is due to the interactions between the object and its environment
 This is resistance is called friction.
 Friction is proportional to the normal force.
 The force of static friction is generally greater than the force of kinetic friction.
 The coefficient of friction (µ) depends on the surfaces in contact.
 The direction of the frictional force is opposite the direction of motion.
 The coefficients of friction are nearly independent of the area of contact.

Friction

Static friction, ƒs acts to

keep the object from moving. If F increases, so does ƒs.
If F decreases, so does ƒs.
ƒs  µs n.
Use = sign for impending

motion only
The

force of kinetic friction, ƒk acts when the object is in motion.
ƒk = µk n.
Variations

of

the

Some Coefficients of
Friction

Block on a Ramp, Example with Friction
 Axes are rotated as usual on an incline.
 The direction of impending motion

would be down the plane.
 Friction acts up the plane.
 Opposes the motion
 Apply Newton’s Laws and solve equations. Example 4:
Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down?

F

x

0

mg sin    s n 0

F

y

   (1)

0

n  mg cos  0

   ( 2)

From Equation (1) & (2),

mg sin    s  mg cos   0 tan   s 0.350

 19.3

Connected Objects
 Apply Newton’s Laws separately to

each object.
 The magnitude of the acceleration of both objects will be the same.
 The tension is the same in each diagram.  Solve the simultaneous equations.

ibrium In Translational Motion
An object is in equilibrium when the net force acting on the object is equals to zero (or balanced).



Fx 0



Fy 0

Non-Equilibrium In
An object is notTranslational in equilibrium when the net force acting onMotion the object is not equals to zero (or not balanced).
According to Newton’s second law of motion, the object will accelerates.

F
F


x

max

y

ma y

Example 5:
The combined weight of the crate and dolly in figure is 3.00×102 N. If the man pulls on the rope with a constant force of 20.0 N, what is the acceleration of the system (crate plus dolly), and how far will it move in 2.00 s? Assume the system starts from rest and that there are no friction forces opposing the motion.

W mg
W 3.00 10 2 m 
30.6 kg g 9.80

F

x

ma x

Fx 20.0 ax  
0.654 m/s 2 m 30.6

x v x 0t  12 a x t 2
x 0 

1
2

 0.654 2.00

2

1.31 m

Example 6:
(a) A car of mass m is on an icy driveway inclined at an angle
 = 20.0°, as in figure. Determine the acceleration of the car, assuming the incline is frictionless.
(b) If the length of the driveway is 25.0 m and the car starts from rest at the top, how long does it take to travel to the bottom? (c) What is the car’s speed at the bottom?

(a)

F

x

max

mg sin  max a x  g sin   9.80  sin 20.0 3.35 m/s 2
(b) x v x 0t  12 a x t 2
25.0 0  12  3.35t 2 t 3.86 s

(c) v x v x 0  a x t 0   3.35 3.86 12.9 m/s

Example 7:
A woman weighs a fish with a spring scale attached to the ceiling of an elevator, as shown in Figures a and b. While the elevator is at rest, she measures a weight of 40.0 N.
(a) What weight does the scale read if the elevator accelerates upward at 2.00 m/s2? (b) What does the scale read if the elevator accelerates downward at 2.00 m/s2, as in Figure b?
(c) If the elevator cable breaks, what does the scale read?

Assume upward positive.

(a)

F

y

ma y

T  mg ma y

T m a y  g   4.08 2.00  9.80  48.1 N
(b) T m a y  g  where a y  2.00 m/s 2

T  4.08  2.00  9.80  31.8 N
(c) T m a y  g  where a y  9.80 m/s 2

T  4.08  9.80  9.80  0 N

Example 8:
Two objects of mass m1 and m2, with m2 > m1, are connected by a light, inextensible cord and hung over a frictionless pulley, as in figure. Both cord and pulley have negligible mass.
Find the magnitude of the acceleration of the system and the tension in the cord.

F

y

ma y

T  m1 g m1a1  T m1  a1  g 

(1)

T  m2 g m2 a2  T m2  a2  g 

( 2)

a2  a1
From Eq (2),

T m2   a1  g 

(3)

Eq (1) = Eq (3),

From Eq (1),

 m1  m2  a1  m2  m1  g

 m2  m1 
 g  m1 g
T m1 
 m1  m2 
 2m1m2 
 g
T 
 m1  m2 

 m2  m1 
 g a1 
 m1  m2 

Example 9:
The hockey puck in figure, struck by a hockey stick, is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 1.20×102 m, slowing down steadily until it comes to rest. Determine the coefficient of kinetic friction between the puck and the ice.

v 2 v02  2ax

v 2  v02
0   20.0
2
a



1
.
67 m/s 2 x
21.20 10 2 
2

F

y

0

n  mg 0  n mg f k  k n  k mg

F

x

max

 f k max    k mg max ax 
 1.67 
 k 

0.170 g 9.80

Example 10:
(a) A block with mass m1 = 4.00 kg and a ball with mass m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley, as shown in figure. The coefficient of kinetic friction between the block and the surface is 0.300. Find the acceleration of the two objects and the tension in the string.
(b) Check the answer for the acceleration by using the system approach.

F

x

T  f k m1a

F

y

n  m1 g 0

T   k m1 g m1a  T m1  a   k g 

F

y

(1)

T  m2 g  m2 a

T m2  g  a 

(2)

Eq (1) = Eq (2),

m2 g   k m1 g  7.00  9.80    0.300 4.00  9.80 a 
 4.00  7.00 m1  m2 a 5.17 m/s 2

From Eq (1),

T  4.00 5.17   0.300  9.80 

T 32.4 N

 F Ma m2 g    f k   m1  m2  a

m2 g   k m1 g a m1  m2

Example 11:
A block of mass m = 5.00 kg rides on top of a second block of mass M = 10.0 kg. A person attaches a string to the bottom block and pulls the system horizontally across a frictionless surface, as in figure. Friction between the two blocks keeps the 5.00-kg block from slipping off. If the coefficient of static friction is 0.350,
(a) what maximum force can be exerted by the string on the 10.0-kg block without causing the 5.00-kg block to slip?
(b) Use the system approach to calculate the acceleration.

(a) Write the two components of Newton’s second law for the top (5.00 kg) block:

x - component : s n1 ma (1) y - component : n1  mg 0  n1 mg

( 2)

From Eq (1) & Eq (2),

a  s g
Write the x-component of Newton’s second law for the bottom (10.0 kg) block:

T  f s Ma
T   s mg M s g  T  m  M   s g
T  5.00  10.0  0.350  9.80  51.5 N

(b) Write the second law for the x-component of the force on the system:  F m

a

system

T  M  m  a
T
51.5 a 
3.43 m/s 2 m  M 5.00  10.0