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Study Of Forces

Forces – Notes on forces and solving problems involving forces.

There are a number of forces that are considered when studying forces in physics.  They include gravitational, electromagnetic, and nuclear forces.

Gravitational force is a force of attraction that exists between all masses.  Gravitational force though weak compared to the other two forces acts over extremely long distances.

Electromagnetic forces are forces between charged particles.  Though considerably stronger than gravitational force, these forces act over much shorter distances. Electrical forces exist between stationary charged particles.  Magnetic forces are produced by the moving of charged particles.

Nuclear forces are much stronger than other forces, but only act over very short distances.  The strong nuclear force holds the nucleus of an atom together.  The weak force is a force in the nucleus that is observed during radioactive decay.

Four basic ideas about forces,

The Net force is the sum of all the forces acting on an object.  When the net force is equal to 0, all forces acting on an object are considered balanced.  When the net force is NOT equal to 0, one or more forces acting on an object are not considered to be balanced.

Inertia is the tendency for an object to remain in the same state of motion.  This means that an object’s natural state is either at rest or moving with constant velocity.  It is the resistance to change in motion of an object.

Newton’s First Law of Motion says that an object continues either in a state of rest, or in a state of constant velocity unless acted upon by an unbalanced (net force).  It says that unless one or more unbalanced force acts upon an object, the velocity of an object does not change. This law implies that there is no fundamental difference between an object at rest and one that is moving with constant velocity.  Zero velocity may be thought of as a special case of constant velocity.

Newton’s Second Law of Motion says that when an unbalanced force acts on an object the object will be accelerated. The acceleration will vary directly with the unbalanced force applied and will be in the same direction as the applied force.  It will vary inversely with the mass of the object.  Newton’s second law can be written as an equation.

The Equation for Newton’s Second Law is F = m a ; or F(net) = m a
 
 

Example Problem
An unbalanced force gives a 2.00 kg mass an acceleration of 5.00 m/s^2.  What is the force applied to this object?

G: mass = 2.00 kg ; acceleration = 5.00 m/s^2
F: the force applied
E: F = m a
S: F = 2.00 kg (5.00 m/s^2)
A: 10 N , because a kg m/s^2 is equal to a N (newton)

Mass is a measure of the amount or quantity of matter in an object.  Mass is expressed in kilograms in the metric system.  Mass is expressed in slugs in the British (English) system. The mass of a particular object remains the same at any location as long as no material is added or removed from the object.

Weight is a measure of the gravitational force acting on an object.  Weight is expressed in Newtons in the metric system.  Weight is expressed in pounds in the British (English) system. The weight of a particular object varies with the position in which the object is located. For example as an object moves away from the earth its weight (also called the force of gravity) decreases with the increase in vertical distance. Also objects in other gravitational fields such as the our moon have weights other than what we see on the earth. As an example consider that objects on the moon have a weight of 1/6 of that on earth.

The equation for weight is based on Newton’s Second Law.

The equation for weight is F(gravity) = W = m g , where g is the acceleration due to gravity at the particular location an object is being observed.

At the surface of the earth the average value for g is given at about 9.8 m/s^2 downward.  The value for g on earth decreases with increasing altitude.  The value for g is different for each star, planet, moon, asteroid, etc.

In problems where location is not indicated, you may assume that the location of an object is at the earth’s surface.
 
 

Example Problem - Calculating the weight of an object
What is the weight of a 7.2 kg mass on the earth’s surface?

G: mass = 7.2 kg ; acceleration = 9.8 m/s^2 , downwards
F: the weight of an object
E: W = m g
S: W = 7.2 kg (9.8 m/s^2) ; You could be literal and use –9.8 m/s^2, but since there are no other forces at work here and we commonly refer to weight as a positive number, knowing full well the direction of the force, we can express the answer without a negative sign.
A: W = 71 N  or  -71 N or 71 N downward

There are two ways to measure mass.

One method that you have been introduced to in this class is uses a balance beam.  The beam balance compares the weights of two objects.  In the process you are balancing an object whose mass is unknown to you against masses that are known.  When the balance is level, meaning that both sides contain equal mass, you know that the mass of the object is equal to the sum of all the known masses on the opposite side of the balance.  Mass determined by this method is usually called gravitational mass.
 
 

Example Problem - Finding Mass using a balance
An object is placed on a triple beam balance that has been previously “zeroed” (the process of balancing the balance when the pan(s) are empty.  The riders (weights of known mass) are slid along each of the three beams until the balance is level.  If the 100 g rider is at the 300 gram position, the 10 g rider is at the 70 gram position, and the 1 g rider is at the 6.5 gram position, what is the mass of the object placed on the pan? (Note: These riders are the "weights" you slide along the metal bars called beams.)

G: The 100 g rider is at the 300 gram position, the 10 g rider is at the 70 gram position, and the 1 g rider is at the 6.5 gram position
F: The (gravitational) mass of the object
E: Total Mass = mass(1) + mass(2) + …
S: Total Mass = 300 g + 70 g + 6.5 g
A: Total Mass = 376.5 g

The other method is to determine the mass of an object by using the property of inertia.  An object of unknown mass is placed on a frictionless surface.  Next, a known force is applied to the object and the acceleration experienced by the object is measured.  Finally, the mass of the object is calculated by using the equation F = m a.  Mass determined by this method is called inertial mass.
 
 

Example Problem - Calculating "Inertial" mass using F = m a
An object is placed upon a horizontal frictionless surface.  A force of 25 N is applied horizontally to the object.  The object goes from 0 m/s to 10 m/s in 20 s.  What is the mass of the object?

G: Applied force is 25 N, force of friction is zero newtons, intial velocity is 0 m/s, final velocity is 10 m/s, and the time interval during which the acceleration took place is 20 s.
F: The (inertial) mass of the object
E: a = (Vf – Vi) / t ;                      E:      m = F / a
S: a = (10 m/s – 0 m/s) / 20 s ;      S:      m = 25 N / 0.5 m/s^2
A: a = 0.5 m/s^2 ;                        A:      m = 50 kg

Gravitational mass and inertial mass are two essentially different concepts, but for any given object they are always the same value.

Information about the Force of Friction.

Friction is a force that opposes the motion of two objects that are in contact with each other.  For example the force a friction between a textbook and the surface of the table upon which it has been placed.

Friction is due to both the surface irregularities (hills and valleys) found in the surfaces of the two objects in contact and the electromagnetic forces of attraction between the two objects in contact.  Even a surface that appears to be smooth to the touch may be a rough surface at the microscopic level.

Static Friction is the force of friction encountered when a force is applied to a stationary object.  Most often this force is larger than the force of kinetic friction.

Kinetic Friction is the force of friction encountered when a force is applied to an object in order to keep it moving with a constant velocity.  Most often this force is smaller than the force of static friction.

The Normal force is a force that acts perpendicular to the surface that an object is either resting on or moving along on.  On a horizontal surface this force has the same magnitude as the force of gravity acting downwards on the object but is opposite in direction to the force of gravity.  As an example, consider the model of a textbook resting on a lab bench.  The book experiences the force of gravity pulling downward on it.  The gravitational force acting on the book is expressed as the book’s weight.  The lab bench,  obviously due to its design and the materials it is made up of, prevents the book from falling.  The upwards force that it produces that is equal in magnitude but opposite in direction to the force of gravity acting on the book is the normal force acting on the book.

The coefficient of friction is a number that is defined as the ratio of two forces.  The symbol for this coefficient is the Greek letter mu (?) which looks similar to the written lower case u from the English alphabet, but with a tail on the front extending downwards. On this page the abbreviation "Coef." will be used to represent the coefficient of friction. Unless otherwise indicated the "Coef." will refere to the coefficient of kinetic friction. Kinetic friction is also often called sliding friction. The value of the Coef. is expressed as a number between 0 and 1. It has no units because it represente the ratio between the Force of friction and the Normal Force. The values for the Coefficients of friction are found by observing and making measurements involving different surfaces being brought in contact with one another under both static friction and kinetic friction conditions.

The ratio (or equation) for the coefficient of friction is Coef. = F(friction) / F(normal)

This tells us that the equation for the force of friction is F(friction) = Coef. * F(normal)

Forces and Force Diagrams.

In the study of forces on horizontal surfaces the equation F(net) = F(Applied) + F(friction) comes into use. It links three basic forces in such a situation together. (If you are interested only in the scaler quantities you may drop the use of + and - when substituting values into the equation, if you change it to F(net) = F(Applied) - F(friction), which is a subtraction.)

In the study of forces involving vertical motion the equation F(net) = F(Applied) + F(gravity) comes into use. It links three basic forces in such a situation together. (If you are interested only in the scaler quantities you may drop the use of + and - when substituting values into the equation, if you change it to F(net) = F(Applied) - F(gravity), which is a subtraction.)

Horizontal Motion / Constant Velocity / and Forces

When studying forces it is always helpful to draw the force vectors associated with the situation being studied. Below is such a diagram. Diagrams of problems such as being described on this page are often referred to as Free body diagrams.

In a Constant Velocity situation all forces are balanced. In the diagram below we have a horizontal force diagram when velocity is constant.

Remember that the force of friction is opposite to the motion of the body.
 

Horizontal Force Diagram when velocity is constant

 
 
Example Problem - Forces under constant velocity conditions
A wooden block is placed on a horizontal wood surface.  It is discovered through observation and measurement that a force of 14.0 N is necessary to keep a 40.0 N block moving along at a constant velocity.
a. What is the normal force acting on the 40.0 N block?
b. What is the coefficient of sliding (kinetic) friction?
c. If another 20 N weight is set on top of the 40 N block, what force will be required to keep the block with the extra weight moving with constant velocity?

G: Weight of block is 40.0 N ; Applied force is 14.0 N ; extra weight is   20.0 N ; velocity is constant, which means that a = 0 m/s^2, and F(net) = 0 N

Part a:
E: | F(normal) | = | weight |
S: | F(normal) | = | - 40.0 N |
A: F(normal) = 40.0 N

Part b:
E:  Coef. = F(friction) / F(normal) ;
     And it is true that because velocity is constant
     F(net) = 0 N = F(applied) + F(friction)
     And therefore
     | F(friction) | = | F(applied) |
S: Coef = 14.0 N / 40.0 N
A: Coef = 0.350

Part c:
E: | F(applied) | = |F(friction)| ; And F(friction) = Coef * F(normal) ;
S: F(friction) = 0.35 (40.0 N + 20.0 N)
A: F(friction) = 21.0 N,  And so F(applied) = 21.0 N

Horizontal Motion / Uniform Acceleration / and Forces

When studying forces it is always helpful to draw the force vectors associated with the situation being studied. Below is such a diagram. Diagrams of problems such as being described on this page are often referred to as Free body diagrams.

In an acceleration situation forces are unbalanced. In the diagram below we have a horizontal force diagram when there is uniform acceleration occuring.

Remember that the force of friction is opposite to the motion of the body. Also, when there is acceleration, the applied force must contain both the magnitude of the force of friction (to cancel friction) and an additional amount equal to the net force (to produce acceleration).
 
 

Horizontal Force Diagram when there is uniform acceleration

 
 
Example Problem - Forces under uniform acceleration conditions
A 100 N object slides across a surface with an acceleration of 2.00 m/s^2.  The coefficient of kinetic friction is 0.200 for this situation.
a. What is the normal force?
b. What is the force of friction?
c. What is the net force?
d. What is the applied force?

Part a:
E: | F(normal) | = | weight |  ; note, if mass were given, then W = m g
S: | F(normal) | = | - 100 N |
A: F(normal) = 100 N

Part b:
E: F(friction) = Coef. * F(normal)
S: F(friction) = 0.200 (100 N)
A: F(friction) = 20 N

Part c:
E: F(net) = m a ; where m = W / g
S: F(net) = ( _____ ) (2.00 m/s^2) ;  m = 100 N / 9.8 m/s^2 = 10.20 kg
    F(net) = 10.20 kg (2.00 m/s^2)
A: F(net) = 20.40 N

Part d:
E: F(net) = F(applied) - F(friction)
    F(applied) = F(net) + F(friction)
S: F(applied) = 20.40 N + (20 N )
A: F(applied) = 40.40 N


 
 
2nd Example Problem - Forces under uniform acceleration conditions
A 200. kg object experiences an acceleration of 3.00 m/s^2. The force applied to the object is 1000 N.
a. What is the weight of the object?
b. What is the normal force?
c. What is the net force?
d. What is the force of friction?
e. What is the coefficient of friction?

Part a:
E: W = m g
S: W = 200. kg (9.8 m/s^2)
A: W = 1960 N

Part b:
E: | F(normal) | = | weight |
S: | F(normal) | = | -1960 N |
A: F(normal) = 1960 N

Part c:
E: F(net) = m a
S: F(net) = 200 kg (3.0 m/s^2)
A: F(net) = 600 N

Part d:
E: F(net) = F(applied) - F(friction), and so
    F(friction) = F(applied) - F(net)
S: F(friction) = 1000 N - 600 N
A: F(friction) = 400 N

Part e:
E: Coef. = F(friction) / F(normal)
S: Coef. = 400 N / 1960 N
A: Coef. = 0.204


 

Vertical Motion / Constant Velocity / and Forces

Remember that the weight of an object can be one of the forces acting on the object.  For vertical motion this is very important to remember.

When studying forces it is always helpful to draw the force vectors associated with the situation being studied. Below is such a diagram. Diagrams of problems such as being described on this page are often referred to as Free body diagrams.

In a Constant Velocity situation all forces are balanced. In the diagram below we have a vertical force diagram when velocity is constant.

Remember that the force of gravity opposes upward motion.
 
 

Vertical Force Diagram when velocity is constant

 
 
Example Problem - Forces under constant velocity conditions
A 10 kg object is suspended on a rope.  What is the tension (force) in the rope?

G: Mass is 10 kg ; velocity is 0 m/s, so a = 0 m/s^2, so F(net) = 0 N
F: F(tension) = F(applied force) in an upwards direction

E: F(net) = F(applied) - F(gravity) ; And F(gravity) = weight = m g
    And | F(applied) | = | F(gravity) | , when F(net) = 0 N
S: | F(applied) | = | 10 kg  * (-9.8 m/s^2) |
A: F(tension) = F(applied) = 98 N


 
 
2nd example problem - Forces under constant velocity conditions
A 25 kg mass, being pulled upwards on a cable, experiences an upwards constant velocity of 4 m/s.  What is the tension in the cable?

G: a 25 kg mass ; a constant velocity of 4 m/s, so a = 0 m/s^2, F(net) = 0 N
F: F(tension) = F(applied force) in an upwards direction

E: F(net) = F(applied) + F(gravity) ; And F(gravity) = weight = m g
    And | F(applied) | = | F(gravity) | ; when F(net) = 0 N
S: | F applied | = | 25 kg (-9.8 m/s^2) |
A: F(applied) = 245 N

Vertical Motion / Uniform Acceleration / and Forces

Remember that the weight of an object can be one of the forces acting on the object.  For vertical motion this is very important to remember.

When studying forces it is always helpful to draw the force vectors associated with the situation being studied. Below is such a diagram. Diagrams of problems such as being described on this page are often referred to as Free body diagrams.

In a uniform acceleration situation forces are unbalanced. In the diagram below we have a vertical force diagram when there is uniform acceleration.

Remember that the force of gravity opposes upward motion. Also, when there is acceleration, an applied force upwards must contain both the magnitude of the force of gravity (to cancel the gravity) and an additional amount equal to the net force (to produce an upwards acceleration).
 
 

Vertical Force Diagram when there is uniform acceleration

 
 
Example problem - Forces under uniform acceleration conditions
A 25 kg mass is pulled upwards on a cable.  The movement involves a uniform acceleration upwards of 0.50 m/s^2.  What is the tension in the cable?

G: a 25 kg mass ; a uniform acceleration of 0.50 m/s^2
F: F(tension) = F(applied) in an upwards direction

E: F(net) = F(applied) - F(gravity), and
    F(applied) = F(net) + F(gravity)

To solve this
1. calculate the F(net)
2. calculate the F(gravity)
3. calculate the F(applied)

1. E: F(net) = m a 
    S: F(net) = 25 kg (0.50 m/s^2) 
    A: F(net) = 12.5 N

2. E: F(gravity) = m g 
    S: F(gravity) = 25 kg (9.8 m/s^2)
    A: F(gravity) = 245 N

3. E: F(applied) = F(net) + F(gravity)
    S: F(applied) = 12.5 N + (245 N)
    A: F(applied) = 257.5 N

Inclined Planes / Trig Functions / Forces

Thus far forces on horizontal surfaces and forces in vertical motion scenarios have been the focus of these notes. Next is the study of Forces acting on objects that are on an inclined plane (a ramp, if you prefer) at an angle between 0 and 90 degrees. Objects on a horizontal surface are at 0 degrees (an extreme positon of an inclined plane, and object in a vertical motion scenario are at 90 degrees (the other extreme position of an inclined plane).

As you examine the diagram below, be sure to note labeling of the force vectors. Also note the two triangles in the diagram. They are congruent, which means that angle A is the same in each of the two different size, though congruent, triangles. Angle A is the degree of elevation of the ramp as measured from the horizontal.

Pay attention to the use of the sine and cosine functions. These two functions allow for the calculation of both the normal force and the parallel force. The normal force is defined, as stated above, as the contact force that is perpendicular to the surface upon which the object is placed. In this situation the object has been placed upon an inclined plane, so the normal force is perpendicular to the surface of the inclined plane. The parallel force is the force that is parallel to the surface upon which the object has been placed. It is a force which points down the ramp. Both normal and parallel force are functions of the force of gravity downwards, F(g).

The normal force opposes the force vector which is the product of F(g) and the Cosine of angle A, pointing upwards at 90 degrees to the surface of the ramp. The Parallel force points down the ramp. It is the product of F(g) and the Sine of the angle A. Forces down the ramp could be the parallel force only or could include an applied force, which would be an additional force, down the ramp. The force of friction would oppose these downward forces and would pont up the ramp.

However, a force could be applied pointing up the ramp. This could result in friction pointing down the ramp in the same direction as the parallel force. This parallel force will always point down the ramp.
 
 
 

Forces on an inclined plane

In the following example, the only forces present are the normal force and the parallel force. Friction is not present in this situation.
 
 

An inclined plane problem w/o friction
A 100 kg mass is placed 2 m up an inclined plane tilted 30 degrees from the horizontal in a frictionless environment.

a. Determine the normal force; b. Determine the parallel force; c. Find the applied force which will prevent the mass from sliding down the ramp.

Given: m = 100 kg; Angle = 30 degrees
Find: a. The normal force; b. The parallel force; c. The applied force which will prevent the mass from sliding down the ramp.

part a:
E: F(N) = mg Cos A
S: F(N) = 100 kg * 9.8 m/s^2 * Cos 30 deg.
A: F(N) = 849 N

part b:
E: F(p) = mg Sin A
S: F(p) = 100 kg * 9.8 m/s^2 * Sin 30 deg.
A: F(p) = 490 N

part c:
E: |F(A), upwards| = |F(p), downwards| + |F(f), downwards|
S: F(A) = 490 N + 0 N
A: F(A) = 490 N
 


 
 
 

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