Galileo discovered that an object will continue to move as it had in the past unless an
unbalanced force acts on it (for example, an unmoving object remains unmoving, or an
object moving at 4.25 m/s west continues to move at 4.25 m/s west unless something—a
force—causes its motion to change). So, the First Law tells us how to recognize a force—
through its action. If the motion of an object changes, it must have been acted on by a
force. Examples of changes in motion include
•the motion of a thrown football from the time it is in the quarterback’s hands
until it is caught by the receiver,
•the motion of a baseball from the time it is picked up by the pitcher until after it
has hit the ground and come to rest or been caught and come to rest, or
•the motion of a roller coaster car.
Newton’s Second Law of motion
Newton recognized that the changes in the positions of objects and their velocities could
be found if the object’s acceleration were known. The acceleration is a measure of the
change in the motion of the object seen when investigating Newton’s First Law. That
acceleration is the only knowledge required to determine the object’s motion (provided we
know at a specific time where it was and how fast it was going). The Second Law
recognizes that the more inertia (measured by mass) an object has, the harder it is for a
required to give an object of mass 2m the same acceleration as the mass m is twice as great
as the net force on the first object, the net force required to give an object on mass 3m the
same acceleration as m is three times as great, and so on; in other words,
acceleration = net force/mass.
This law is most often written in the form
F = ma
Let’s give some examples of the use of Newton’s Second Law of motion. An object of
mass 3.50 kg is changing its velocity in a specific direction by 2.00 m/s for every second.
What net force is being exerted on it that causes this change?
According to what we know from Extension 3.2 Average acceleration, the acceleration is
the change in velocity divided by the time required to change the velocity; here, that is
a = 2.00 m/s/1.00 s
= 2 m/s² in the direction in which the velocity changes.
Now, Newton’s Second Law of motion F = ma says that the net force F required is the product of the mass and acceleration,
F = ma (3.50 kg) x (2.00 m/s², in the direction in which the velocity changes)
= 7.00 N, in the direction in which the velocity changes.
acceleration is 0.100 m/s², west. What net force must be acting on it?
According to Newton’s Second Law of motion, the rope and all the other things the box is
in contact with (for example, the floor) must be exerting a net force on the box of
F = ma
= (6.75 kg) x (0.100 m/s², west) = 0.675 N, west.
A kite is subject to a net force of 0.389 N, northwest. Its acceleration is 2.35 m/s2,
northwest. According to Newton’s Second Law F = ma , which is a vector relation,
the directions of the net force on an object and acceleration of that object must be the
same. Luckily, this appears true for this problem. If this were not so, it would have meant
we had lost track of some of the forces that contribute to the change in motion. Any time
you come on a case for which the net force and acceleration have different directions, you
must have missed identifying at least one of the forces acting that contribute to the net
force. What must the kite’s mass be if these data are true?
If mass is not known but force and acceleration are, the mass may be determined from the
relation found as a consequence of F = ma ,
a = F/m = (0.389 N)/(2.35 m/s²) = 0.166 kg.
Note that we did not give the directions, because we knew that F and a were both in the
same direction. The mass m is a scalar; it has no direction associated with it.
Newton’s Third Law of motion
What causes a force? Forces do not just appear magically. Every force that is exerted on
an object (call it object A) is the result of some other material agent acting on it to cause
that force. Suppose object B touches object A or holds something that touches A.
Because of the presence of B, there is a force acting on A.Let’s suppose Ashley and Beth are each standing still holding the ends of a rope. Beth
pulls on Ashley. If Beth lets go, both Beth and Ashley will briefly struggle to regain their
balance—in other words, when they let the rope go, their motions will change. Letting go
of the rope changes the forces that act; for example, Beth is no longer pulling on Ashley,
who had braced herself to keep herself motionless. When Ashley’s force is removed, she
might briefly stagger before recovering.
The fact that Beth has also to struggle indicates that the balance of Beth’s forces changed,
too, when she let the rope go. Where could that force that was suddenly removed have
come from? The only thing that changed is that Ashley and Beth are no longer connected
by the rope. Ashley must have been causing a force that acted on Beth.
Newton recognized this fact—if Ashley causes a force that acts on Beth, then
reciprocally Beth causes a force that acts on Ashley. Furthermore, another simple
experiment shows that the size of the forces must have been the same and that the forces
had to act in opposite directions. Candace replaces the rope. Both Ashley and Beth pull.
If Beth lets go, Candace and Ashley briefly move away from her (their motion had
changed). They hold hands and pull again, then Ashley lets go. Now Candace and Beth
briefly move away from her, in the direction opposite to that Candace and Ashley had
moved). When they both pull on Candace, Candace doesn’t move (she’s not changing her
motion).
According to the First Law, the two forces on her (by Ashley and by Beth) are balanced.
They must be the same size, but in opposite directions.
So, if Ashley pulls on Beth, Beth must pull equally on Ashley, but in the opposite
direction. Sometimes, these are referred to as “action” and “reaction” forces, but that is an invitation to think that some forces are more fundamental, or precede other forces. The
two forces (the force on Ashley caused by Beth and the force on Beth caused by Ashley
in the rope example) act pairwise, with essentially no time lag. The two forces are acting
on different objects, in distinction to the First and Second Laws, which refer to all the
forces acting on ONE object.
Steps for solving problems involving Newton's equations. |
- Read the problem carefully. Make sure that you understand the given data and the asked information.
- Select the order in which you are going to work the problem.
- Start by solving from the object to which less unknowns are associated with.
- Also, start solving for objects for which there is less number of forces applied.
- Draw the free body diagram for the object.
- Break every force and acceleration into their x and y components.
- The axis should be selected such that the minimal number of vectors need to be break down into components.
- If possible, select the x-axis along the horizontal component and the y-axis along the vertical component.
- From the diagram, obtain the net force acting on the selected object for each vector component, one for x-axis and another for y-axis components.
- As a direct application of Newton's second law, the previous net force is also related to the acceleration of the object for each individual component.
- Notice that results from step 5 and step 6 are obtained from two independent analysis. However, these results must be the same. Therefore, equating these two independent results for the net force an equation can be obtained.
- Repeat steps 3 to 7 as need it until all the unknown can be solve from the different equations
Free-Body Diagrams
Free-body diagrams are used to represent all the forces on an object to determine the net force on it. They are termed free-body diagrams because each diagram considers only the forces acting on the particular object considered.
Lifting an object
Dragging an object (on ice)
Forces on inclined planes
Atwood's Machine
 | For this system (Atwood's Machine) need to consider free-body diagrams for two objects. is the same in magnitude for each weight, the sign relative to gravity must be opposite on one side from another. The weights are both connected by the same rope and thus the force due to tension is the same for each object. The concept behind this device is used in elevators and funiculars |
More pulleys
More pulleys solved
Pendulum style accelerometer
example.
1. a box with a mass of 10 kg sliding on an inclined plane with μ k = 0.3 and θ = 45. barapa determine the force experienced by these boxes if the boxes in a constant speed ?
solution :
we know that m= 10 kg
μ k = 0.3
θ = 45
g = 9.8 m/s²
we use the formula
ΣF = ma
F - FS = 0
F - μ kN = 0
F = μ kN = μ k.mg sin θ
= 0.3.10.9.8 sin 45
= 20.58 N
2. a lightweight rope whose mass is negligible suspended two boxes. See the picture above.
determine T1 and T2?
Solution :
This example is sufficiently simple to be solved without calculations. Thus, it is a good example to start understanding the method for solving problems involving Newton's laws of motion.
Following step two described above, it is simpler to solve for the bottom weight first, w2. The corresponding FBD is (Notice that the effect of the rope is replaced by the tension. This tension pulls up the object)
Free Body Diagram Object 2
Free Body Diagram Object 2
Vertical Direction
Using step 5, the net force is the result of the vector addition between the tension of the rope (vector pointing up; then, positive) and the weight of the object (vector pointing down; then, negative).
Following step 6, the net force is related to the acceleration of the block. But the acceleration of the block is zero as indicated in the diagram. Therefore, .
Equating these two relations for the net force acting on object 2, the following equation is obtained
This equation can be immediately solved for T2
Vertical Direction
In this case, the net force is the result of the vector addition between the tension of the rope (vector pointing up; then, positive) and the vectors pointing down (vectors pointing down; then, negatives); the weight of the object and the tension of the bottom rope. Notice that T2 pulls on object one downward, for a review on tension go to the Tension section of these notes. Therefore, the net force is
SOURCES ;
http://www.enc.edu
http://tycphysics.org
and physic book
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