Download 4-2 Force, Mass and Newton`s 2nd Law

4-4 Forces In Nature
There are many types of forces:
The Fundamental Forces / Field Forces
The fundamental forces are all types of field forces which means “action at a
distance” forces, or no direct contact forces. What that means is that as objects
enter a field, they are subject to the effects of the field. In order of decreasing
strength, they are:
1. Strong Nuclear Force = the attractive force that holds the subatomic
particles of a nucleus together. You probably remember this one from
AP Chemistry!
2. Electromagnetic Force = the force between electrical charges. May be a
force of attraction or repulsion. It is assumed that even though the
masses of an electron and proton are different (mass of an e- = 1/1837 of
mass of a proton), their charges are equal in magnitude, but opposite in
direction.
3. Weak Nuclear Force = the force between subatomic particles during
certain types of radioactive decay.
4. Gravitational Force = the force of mutual attraction between two
objects. Whether viewed from a Newtonian or an Einsteinian
perspective, there is a force of gravity between any two objects.
Both the strong and weak nuclear forces operate only on a nuclear scale, and
are thus only viable within a range of 10-15 m. Beyond this range, they have no
effect. Classical physics deals only with gravitational and electromagnetic
forces which have, for intensive purposes, infinite range.
Contact Forces
Any force that involves direct contact between two or morr objects is a contact
force. Some examples of contact forces are:
A.) Solids
1. Push / pull – may be parallel, perpendicular, or anti-parallel (at an angle)
to the contact surface.
2. Normal Force = any force perpendicular to the contact surface. These are
particularly relevant in drawing free-body diagrams (FBD’s).
3. Frictional Force = the force (or component of the force) that is parallel to
the contact surface and oppositely directed to the sliding motion of the
object. Frictional forces are response forces. Their magnitudes increase
to a certain point in direct response to the push / pull exerted on the
object. We will cover this in greater detail next chapter.
B) Springs
When a spring is either compressed or stretched by a certain amount, x, the
force it exerts back is given by:
where k is the force constant measuring the stiffness of the spring.
Q: Why is there a negative sign in the Hooke’s Law equation?
A: The force the spring exerts back is in the opposite direction of the original
force. Because it acts in the opposite direction, the effect is to return, or
restore, the object to its “equilibrium” position, and is thus called a restoring
force.
Q: What does equilibrium mean? Describe the types of equilibrium.
A: Equilibrium is when the sum of the forces, or the net force, = zero. The
two types of equilibrium are:
 Static equilibrium – the object is at rest and any forces that exist are not
enough to accelerate the object.
 Dynamic equilibrium – the sum of the net forces = 0; the object is not at
rest, but it is not accelerating. In other words, it is moving at constant
velocity.
The natural resting place of a spring is called its stable equilibrium.
In a molecule or solid, the forces between atoms varies approximately linearly
with the change in separation (for small changes), like the forces in a spring. If
you push or pull the object greatly, you can compress, warp, stretch, or even
break it, depending on its internal structure.
Clamping down on a plastic block
produces stress patterns that are
visible when viewed with polarized
light.
C) Strings
Strings are used to pull things. Think of a string as being like a spring with
such a large force constant that the extension of the string is negligible.
Though strings can flex (out), or bend, they can’t be used to push things. The
amount of force that, technically, one segment of string exerts on the next is
called tension.
Really, tension is the amount of force in a spring, string, wire, or rope, et
cetera. My daddy taught me that! 
D) Constraints
A condition that restricts or controls the motion of another object in any way is
called a constraint. For example, a railroad car moving along a track; a
wooden pony on a carousel; a bobsled on a luge track.
Before delving fully into the next example and the next section, let’s review
some notes from 10th grade on free-body diagrams (FBD’s).
If ∑ F = 0, the object is not accelerating. Which is / are accelerating below?
Applied Forces
Net Force
DEF: Free-body Diagram (FBD) = a basic diagram that:
 Considers only one object at a time
 Represents the object with a simple sketch or dot
 uses vectors to represent all relevant
forces acting on that object
 Superimposes a coordinate system to give direction
and magnitude to the vectors
 NOTE: All forces must be “like” in that they
contribute to the net motion of the object
 NOTE: As a matter of convention, start all vectors
from the center of the object
Draw Free-Body Diagrams (FBD’s) for all of the following:
Ex 1) A block sitting on a table
Ex 2) A piece of rope being pulled on by two equally strong children
Ex 3) A child on a stationary sled pointed down a hill (300 incline)
Ex 4) A child sledding at constant velocity down a hill (300 incline)
Ex 5) This guy hanging from the ceiling wires. (C’mon, it could happen!)
You try it.
1. A person sliding down a slide (450 incline)
2. A crate being pulled across the floor
3. A hockey puck sliding across the ice
4. Two people successfully pushing a stalled car
5. A ball rolling down a frictionless ramp
See Ex 4-5, p 94.
A 110 kg basketball player hangs on the rim following a slam dunk. Prior to
dropping to the floor, he hangs motionless with the front of the rim deflected
down a distance of 15 cm. Assume the rim can be approximated by a spring
and calculate the force constant, k.
This diagram is appropriate because it says he hangs motionle
means he is not accelerating in any direction. Because of that, t
are all balanced. To say he is hanging there implies the forces
the y-direction. Also, they told us to assume the rim acts like
which means that force will be calculated using Hooke’s Law.
The sum of the forces will be the weight, in Newtons, of the basketball player +
the spring force of the rim.
Q: What units will you end up with if you divide N by kg? 
Don’t forget to convert cm to m.
∑Fy = wy + Fy-rim = ma
mg + -kΔy = m  0
mg = kΔy
mg = k
Δy
k = (110 kg)(9.81 N/kg) = 7.19 x 103 N/m
0.15 m
4-5 Problem Solving: FBD’s
From your text:
See Ex 4-6, p 96
During your winter break, you enter a dogsled race in which students replace
the dogs. Wearing cleats for traction, you begin the race by pulling on a rope
attached to the sled with a force of 150N at 250 to the horizontal. The mass of
the sled-passenger-rope object is 80 kg and there is negligible friction between
the sled runners and the ice. Find:
a) the acceleration of the sled
b) the normal force, Fn, exerted by the surface on the sled.
What did they do to the
vectors? What does the
black vector, m a , signify?
There are three vectors to consider here. Their sum will be equal to ma.
Conceptually: The normal force has to cancel out the difference between the y
component of the F and the weight, because Fnet is in the x-direction only!
a) Because ma is in the x-direction only, and weight and normal are in the ydirection only, we have:
Fn + w + F = ma
This gives us two sets of equations to deal with: x and y directions
Fn,x + wx + Fx = max
AND
0 + 0 + (150 N)(cos 250) = 80 kg  ax
m0
ax = (150 N)(cos 250)
80 kg
0
– (150 N)(sin 25 )
ax = 1.70 m/s2
b) Fn,y + wy + Fy = may
Fn,y + -mg + Fsin =
Fn = mg - Fsin
= (80 kg)(9.81 N/kg)
= 721 N
See Ex 4-7, p 97
You are working for a big delivery company, and must unload a large, fragile
package from the truck using a delivery ramp. If the downward component of
the velocity of the package when it reaches the bottom of the ramp is greater
than 2.5 m/s (the speed it would have if you dropped it from a height of 1 ft),
the package would break. What is the largest angle at which you can safely
unload? The ramp is 1 m high, has rollers (i.e., is frictionless), and is inclined
at an angle θ.
This FBD shows the x-axis along th
direction of natural motion, o
acceleration, as recommended. Note tha
the weight vector always points straigh
down, and that the Fn is always  th
contact surface, or x-axis, in this case.
Some very important “tricks” for solving these types of problems:
1. The weight vector is straight down and = - mg. Does that make sense in
terms of N2? (F = ma → F = m(-g)).
2. All of the θ’s are equal. (Do you know why? θ1 = θ2 Alt Int ’s are equal; θ2
= θ3 complements of the same  are equal ) Because that’s true, θ will
actually be resolved along the y-axis. Look at it! For that reason, the weight
vector, which needs to be resolved in both x and y will switch wrt x and y. So,
the weight vector resolved in x will = mgsin θ, and resolved in y will be mgcos θ.
3. Drawn this way, the normal force will always be only in the + y-direction.
Now for the problem-solving process:
1st: Find the sum of the forces in x, and if the object will accelerate in x, set
them equal to ma. Fn,x = 0.
Fx :
0 + mgsin θ = max 
ax = gsin θ
The m’s cancel
We don’t know ax, but we do know vf and vi, so we can use those to set up and
solve for vf in terms of ax
vf2 = vi2 + 2axx
vf2 = 2axx
Since vi = 0
vf2 = 2 gsin θ x
Substituting ax from above
For this next part draw a couple of right triangles, and use right-triangle
geometry.
x
h
Ө
When x = the length of the
ramp (as it’s moving along the
x-axis), h = x sin Ө. Thus, vf2
= 2gsin  becomes
vf2 = 2gh
vf
vd
Ө
Looking at the velocity in the downward direction, vd, will give:
vd = vf sin Ө
Solve the equation in the box above for vf,
and you get vf = 2 gh . Substitute, and you get
= 2 gh sin Ө
equation for vd.
Note: There is an additional sin Ө from the
2.5 m/s =
2(9.81m / s 2 )(1.0m)
sin Ө
Ө = 34.40
See Ex 4-8, p 98
A picture weighing 8N is suspended by two wires with tensions T1 and T2 as
shown. Find each tension.
Remember that tension just means the force in a string, spring, rope or wire.
The 600 angle on the top = the 600 angle on the bottom because alt int ’s are
equal. Same is true for the 300 ’s.
The picture is stationary; that means there is no acceleration in any direction.
∑Fx:
- T2cos 600 + T1 cos 300 = 0
T2 = T1 cos 300 = 1.73 T1
cos 600
∑Fy:
T2sin 600 + T1 sin 300 – w = 0
1.73 T1 sin 600 + T1 sin 300 – w = 0
evaluate
substitute for T2
add w to both sides,
the products, sin’s, and
cos’
1.5 T1 + 0.5 T1 = 8N
2 T1 = 8N
T1 = 4N 
T2 = 6.93N (7N)
See Ex 4-9, p 99
As your jet plane speeds down the runway on takeoff, you decide to determine
its acceleration, so you take out your yo-yo and note that when you suspend it,
the string makes an angle of 220 with the vertical. a) What is the acceleration
of the plane?
b) If the mass of the yo-yo is 40 g, what is the tension in the string?
Ө
Ө
Draw the y-axis, and then
another line parallel to the yaxis. Then, these two Ө’s are
equal.
(Alt int ’s are )
So, they used the fact that alt int ’s are , and
then moved the tension vector up to the origin.
The way this is drawn, T will be resolved along
the y-axis (not the x-axis). This means x will be
in terms of sin Ө, and y. will be in terms of cos
Ө.
Because of how the acceleration vector is drawn, we know there is only
acceleration in the x-direction.
∑Fx:
T sin Ө = max
There is an acceleration in x
∑Fy:
2nd eq
T cos Ө - mg = 0
There is not an acceleration in y. Use this
to solve for T.
T cos Ө = mg
T = mg
= (0.040 g)(9.81 m/s2) = 0.423 N
Now, subst. into the
st
1 eq
cos Ө
cos 220
to solve for ax
ax = (0.423 N)(sin 220) = 3.96 m/s2
0.040 kg
See Ex 4-10, p 100
Suppose that your mass is 80 kg, and you are standing on a scale fastened to
the floor of an elevator. The scale measures force and is calibrated in N’s.
What does the scale read when (a) The elevator is rising with upward
acceleration of magnitude a; (b) the elevator is descending with a downward
acceleration of magnitude a’; (c) the elevator is rising at 20 m/s and its speed is
decreasing at a rate of 8 m/s2?
Remember this from 11th grade? 
Draw FBD’s for each
scenario. All the motion is in
the y-direction only, and is
some derivation of the FBD
below.
(a) ∑Fy: Fn,y + wy = may
Fn,y – mg = may
Fn,y = mg + may
Fn,y = m(g + ay)
This will make you weigh morr because
you are
adding something to “g”
(b) ∑Fy: Fn,y + wy = ma’
Fn,y – mg = m(- a’)
Fn,y = mg - ma’
Fn,y = m(g – a’)
This will make you weigh less because
you are
subtracting something from “g”
(c) Apply ∑Fy:
Fn,y + wy = may
Fn,y – mg = may
Fn,y = mg + may
Fn,y = m(g + ay) =
= (80 kg)(9.81 m/s2 + - 8.00 m/s2)
= 145 N
the
So, you weigh much less, as you should;
acceleration is almost g. 80 kg = 784.8 N.
145 / 784.8  18.5 %
9.81 – 8.00 = 1.81
1.81 / 9.81  18.5 %. The percentages are
the
same, as they should be! 