Title: For Midterm
1For Midterm 2
- Ch. 4 (2D Motion) General Problems (Problems
50-76 p. 101 - 102) - Ch. 5 (Newtons Laws) General Problems
(Problems 45 -67 p. 135 - 136)
2Physics 1020Applications of Newtons Laws(Ch. 6
of Walker)
- Frictional Forces
- Strings and Springs
- Translational Equillibrium
- Connected Objects
- Circular Motion
3Applications of Newtons Law
- Assumptions
- Objects can be modeled as particles
- Masses of strings or ropes are negligible
- When an object is being pulled by a rope, the
rope exerts a force on the object. Its
direction is along the rope, away from the
object. The magnitude of this force is called
the tension. - Interested only in the external forces acting ON
the object do not include reaction forces
4Problem-Solving Hints Newtons Laws
- Conceptualize the problem
- Draw a diagram (include convenient coordinate
axes) - Categorize the problem
- Equilibrium (SF 0) or Newtons Second Law (SF
m a) - Analyze
- Draw free-body diagrams for EACH object
- Include only forces acting ON the object
5Equilibrium Example
- A traffic light weighing 122 N hangs from a
cable tied to two other cables fastened to a
support, as shown. The upper cables make angles
of 37 and 53 with the horizontal. These upper
cables are not as strong as the vertical cable
and will break if the tension in them exceeds 100
N. Does the traffic light remain in this
situation, or will one of the cables break? - Conceptualize the traffic light
- Categorize as an equilibrium problem
- No movement, so acceleration is zero
6Equilibrium, Example, cont
- Analyze
- Need two free-body diagrams
- Apply equilibrium equation to the light and find
7Equilibrium, Example, cont
- Analyze, cont.
- Apply equilibrium equations to the knot and find
and
8Equilibrium, Example, cont
- Analyze, cont
- Solve the 1st of these for T2
- Substitute into the 2nd equation to give
Both less than 100 N, so cables do not break!
9Objects Experiencing a Net Force
- If an object that can be modeled as a particle
experiences an acceleration, there must be a
nonzero net force acting on it. - Draw a free-body diagram
- Apply Newtons Second Law in component form
10Nonzero Net Force - Example
- Forces acting on the crate
- A tension, the magnitude of force
- The gravitational force,
- The normal force, , exerted by the floor
- Apply Newtons Second Law in component form
- Solve for the unknown(s)
11Inclined Plane Recipes
Memorize!!!!
- Forces acting on the object
- The normal force acts perpendicular to the plane
- The gravitational force acts straight down
- Choose the coordinate system with x along the
incline and y perpendicular to the incline - Replace the force of gravity with its components
12Inclined Plane Recipes, cont.
- Newtons 2nd law
- From 1st equation
- Check special cases to see that answer makes
sense
What is acceleration in each of these cases?
13Forces of Friction (Ch. 6 Sect. 6-1)
- 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 resistance is called the force of friction
14Static Friction
Static friction acts to keep the object from
moving If increases, so does If
decreases, so does
15Static Friction
- 0 ? s ? fs, max
- fs, max µs N
- µs is called the coefficient of static friction
- µs - dimensionless
- s ? µs N where the equality holds when the
surfaces are on the verge of slipping - Called impending motion
- The force of static friction between 2 surfaces
is parallel to the surface of contact and has the
direction opposite to the motion (Does not act in
the direction of the normal force!!!!)
16Static Friction (more magic!)
- How does the static friction know how big it has
to be? - Same thing as for the normal force!
- As you apply a force to move an object, you start
to distort the stuff at the contact points of
the object - Floor, table, air, water whatever
- The more force you apply, the more distortion
- If you push hard enough you break the contact
points - At this point, the kinetic friction model comes
into play
17Kinetic Friction
- The force of kinetic friction acts when the
object is in motion - Although µk can vary with speed, we shall neglect
any such variations - k µk N
- µk - the coefficient of kinetic friction
- µk dimensionless
- Typical values are in the range between 0 and 1
- Simpler to work with than the static case because
of the equality
18Static vs. Kinetic
19Forces of Friction, Summary
- The direction of the frictional force is (in
general) opposite the direction of motion (or
impending motion) and parallel to the surfaces in
contact - The coefficients of friction are nearly
independent of the area of contact - The coefficient of kinetic friction (mk) it
typically smaller than the coefficient of static
friction (ms) - The maximum magnitude of static friction, s, is
generally greater than the magnitude of kinetic
friction, k - The coefficient of friction (µ) depends on the
surfaces in contact - fs, max µs N fk µk N
- These equations relate the magnitudes of the
forces, they are not vector equations
!
!
20Friction in Newtons Laws Problems
- Friction is a force, so it simply is included in
the SF in Newtons Laws - The rules of friction allow you to determine the
direction and magnitude of the force of friction
21Example (simple)Problem 4 p. 165
- When you push a 1.80-kg book resting on a
tabletop, it takes 2.25 N to start the book
sliding. Once it is sliding, however, it takes
only 1.50 N to keep the book moving with constant
speed. What are the coefficients of static and
kinetic friction between the book and the
tabletop? - P. 5 p. 165. What is the frictional force exerted
on the book when you push on it with a force of
0.75 N?
22Example (more difficult) Problem 7 p. 166
- To move a large crate across a rough floor, you
push down on it at an angle of 21, as shown.
Find the force necessary to start the crate
moving, given that the mass of the crate is 32 kg
and the coefficient of static friction between
the crate and the floor is 0.57.
23HOMEWORK !!!!
- Example 6-2 page 141 (Making a big splash)
24Check you understanding
- A physics student is pulling upon a rope which is
attached to a wall. In the bottom picture, the
physics student is pulling upon a rope which is
held by the Strongman. In each case, the force
scale reads 500 Newtons. The physics student is
pulling - with more force when the rope is attached to the
wall. - with more force when the rope is attached to the
Strongman. - the same force in each case.
25Check you understanding
- Consider the addition of two forces, both having
a magnitude of 10 Newtons. What is their sum?
26(No Transcript)
27Multiple Objects (Sect. from 6-2 to 6-4)
- When two or more objects are connected or in
contact, Newtons laws may be applied to the
system as a whole and/or to each individual
object. - Whichever you use to solve the problem, the other
approach can be used as a check.
28Strings and Springs (Sect. 6-2)
- Assumptions
- Masses of strings or ropes are negligible
- Pulleys have no mass, no friction
- A pulley changes the direction of the tension but
does not change its magnitude
When an object is being pulled by a rope, the
rope exerts a force on the object. Its
direction is along the rope, away from the
object. The magnitude of this force is called
the tension.
29Conceptual Checkpoint 6-2a p.148How do the scale
readings compare?
- The scale at left reads 9.81 N. Is the reading of
the scale at right - greater that 9.81N
- equal to 9.81N
- less that 9.81 N
30Multiple Objects
31Multiple Objects
- Forces acting on the objects
- Tension (same for both objects since one string)
- Gravitational force
- Each object has the same acceleration since they
are connected - Draw the free-body diagrams
- Apply Newtons Laws
- Solve for the unknown(s) - Here a and T)
32Atwoods Machine (Ex. 6-7 p. 156)
Masses m1 and m2 are attached to an ideal
massless string and hung as shown around an ideal
massless pulley.
Fixed Pulley
- Find the accelerations, a1 and a2, of the masses.
- What is the tension in the string T ?
y
T1
T2
m1
a1
m2
a2
33Atwoods Machine...
- Draw free body diagrams for each object
- Applying Newtons Second Law ( y -components)
- T1 - m1g m1a1
- T2 - m2g m2a2
- But T1 T2 T since pulley is ideal
- and a2 -a1 -a.since the masses are
connected by the string
Free Body Diagrams
T1
T2
y
a1
a2
m2g
m1g
34Atwoods Machine...
- T - m1g m1 a (a)
- T - m2g -m2 a (b)
- Two equations two unknowns
- we can solve for both unknowns (T and a).
- subtract (a) - (b)
- g(m2 m1 ) a(m1 m2 )
- a
35Atwoods Machine
- T - m1g m1 a (a)
- T - m2g -m2 a (b)
- divide (a) by (b)
36Atwoods Machine...
37Is the result reasonable? Check limiting
cases!
- Special cases
- i.) m1 m2 m a 0 and T mg. OK!
- ii.) m2 or m1 0 a g and T 0.
OK! - Atwoods machine can be used to determine g (by
measuring the acceleration a for given masses). -
-
-
38Example (Pr. 31 p. 167)
- After a skiing accident, your leg is in a cast
and supported in a traction device, as shown.
Find the magnitude of the force exerted by the
leg on the small pulley. (By Newtons third law,
the small pulley exerts an equal and opposite
force on the leg.) Let the mass m be 2.50 kg.
39Multiple Objects, Example 2
- Three blocks of mass 3m, 2m, and m are connected
by strings and pulled with constant acceleration
a. What is the relationship between the tension
in each of the strings?
(a) T1 gt T2 gt T3 (b) T3 gt T2 gt T1
(c) T1 T2 T3
40Solution
- Draw free body diagrams!!
T3 3ma
T1 gt T2 gt T3
41Multiple Objects - Example 3
- First treat the system as a whole
- Apply Newtons Laws to the individual blocks
- Solve for unknown(s) Here a and P12
- Check P21 P12
42Multiple Objects - Example 2, cont
P21 P12 P12 -P21 a1 a2 a
- First treat the system as a whole
- Apply Newtons 2nd law to m2
- Check. Apply Newtons 2nd law to m1
as before.
43Example (Pr. 36 and 39 p. 168)
- Find the acceleration of the masses shown in
Figure, given that m11.0 kg, m22.0 kg and
m33.0 kg - Find the tension in each of the strings in Figure.
44Next lecture