# 5.1 Forces  (Page 3/9)

 Page 3 / 9

Let’s analyze force more deeply. Suppose a physics student sits at a table, working diligently on his homework ( [link] ). What external forces act on him? Can we determine the origin of these forces? (a) The forces acting on the student are due to the chair, the table, the floor, and Earth’s gravitational attraction. (b) In solving a problem involving the student, we may want to consider the forces acting along the line running through his torso. A free-body diagram for this situation is shown.

In most situations, forces are grouped into two categories: contact forces and field forces . As you might guess, contact forces are due to direct physical contact between objects. For example, the student in [link] experiences the contact forces $\stackrel{\to }{C}$ , $\stackrel{\to }{F}$ , and $\stackrel{\to }{T}$ , which are exerted by the chair on his posterior, the floor on his feet, and the table on his forearms, respectively. Field forces, however, act without the necessity of physical contact between objects. They depend on the presence of a “field” in the region of space surrounding the body under consideration. Since the student is in Earth’s gravitational field, he feels a gravitational force $\stackrel{\to }{w}$ ; in other words, he has weight.

You can think of a field as a property of space that is detectable by the forces it exerts. Scientists think there are only four fundamental force fields in nature. These are the gravitational, electromagnetic, strong nuclear, and weak fields (we consider these four forces in nature later in this text). As noted for $\stackrel{\to }{w}$ in [link] , the gravitational field is responsible for the weight of a body. The forces of the electromagnetic field include those of static electricity and magnetism; they are also responsible for the attraction among atoms in bulk matter. Both the strong nuclear and the weak force fields are effective only over distances roughly equal to a length of scale no larger than an atomic nucleus ( ${10}^{-15}\phantom{\rule{0.2em}{0ex}}\text{m}$ ). Their range is so small that neither field has influence in the macroscopic world of Newtonian mechanics.

Contact forces are fundamentally electromagnetic. While the elbow of the student in [link] is in contact with the tabletop, the atomic charges in his skin interact electromagnetically with the charges in the surface of the table. The net (total) result is the force $\stackrel{\to }{T}$ . Similarly, when adhesive tape sticks to a piece of paper, the atoms of the tape are intermingled with those of the paper to cause a net electromagnetic force between the two objects. However, in the context of Newtonian mechanics, the electromagnetic origin of contact forces is not an important concern.

## Vector notation for force

As previously discussed, force is a vector; it has both magnitude and direction. The SI unit of force is called the newton    (abbreviated N), and 1 N is the force needed to accelerate an object with a mass of 1 kg at a rate of $1\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ : $1\phantom{\rule{0.2em}{0ex}}\text{N}=1\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m/s}}^{2}.$ An easy way to remember the size of a newton is to imagine holding a small apple; it has a weight of about 1 N.

We can thus describe a two-dimensional force in the form $\stackrel{\to }{F}=a\stackrel{^}{i}+b\stackrel{^}{j}$ (the unit vectors $\stackrel{^}{i}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{^}{j}$ indicate the direction of these forces along the x -axis and the y -axis, respectively) and a three-dimensional force in the form $\stackrel{\to }{F}=a\stackrel{^}{i}+b\stackrel{^}{j}+c\stackrel{^}{k}.$ In [link] , let’s suppose that ice skater 1, on the left side of the figure, pushes horizontally with a force of 30.0 N to the right; we represent this as ${\stackrel{\to }{F}}_{1}=30.0\stackrel{^}{i}\phantom{\rule{0.2em}{0ex}}\text{N}.$ Similarly, if ice skater 2 pushes with a force of 40.0 N in the positive vertical direction shown, we would write ${\stackrel{\to }{F}}_{2}=40.0\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}.$ The resultant of the two forces causes a mass to accelerate—in this case, the third ice skater. This resultant is called the net external force     ${\stackrel{\to }{F}}_{\text{net}}$ and is found by taking the vector sum of all external forces acting on an object or system (thus, we can also represent net external force as $\sum \stackrel{\to }{F}$ ):

${\stackrel{\to }{F}}_{\text{net}}=\sum \stackrel{\to }{F}={\stackrel{\to }{F}}_{1}+{\stackrel{\to }{F}}_{2}+\text{⋯}$

This equation can be extended to any number of forces.

In this example, we have ${\stackrel{\to }{F}}_{\text{net}}=\sum \stackrel{\to }{F}={\stackrel{\to }{F}}_{1}+{\stackrel{\to }{F}}_{2}=30.0\stackrel{^}{i}+40.0\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}$ . The hypotenuse of the triangle shown in [link] is the resultant force, or net force. It is a vector. To find its magnitude (the size of the vector, without regard to direction), we use the rule given in Vectors , taking the square root of the sum of the squares of the components:

${F}_{\text{net}}=\sqrt{{\left(30.0\phantom{\rule{0.2em}{0ex}}\text{N}\right)}^{2}+{\left(40.0\phantom{\rule{0.2em}{0ex}}\text{N}\right)}^{2}}=50.0\phantom{\rule{0.2em}{0ex}}\text{N}.$

The direction is given by

$\theta ={\text{tan}}^{-1}\left(\frac{{F}_{2}}{{F}_{1}}\right)={\text{tan}}^{-1}\left(\frac{40.0}{30.0}\right)=53.1\text{°},$

measured from the positive x -axis, as shown in the free-body diagram in [link] (b).

Let’s suppose the ice skaters now push the third ice skater with ${\stackrel{\to }{F}}_{1}=3.0\stackrel{^}{i}+8.0\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}$ and ${\stackrel{\to }{F}}_{2}=5.0\stackrel{^}{i}+4.0\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}$ . What is the resultant of these two forces? We must recognize that force is a vector; therefore, we must add using the rules for vector addition:

${\stackrel{\to }{F}}_{\text{net}}={\stackrel{\to }{F}}_{1}+{\stackrel{\to }{F}}_{2}=\left(3.0\stackrel{^}{i}+8.0\stackrel{^}{j}\right)+\left(5.0\stackrel{^}{i}+4.0\stackrel{^}{j}\right)=8.0\stackrel{^}{i}+12\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}$

Check Your Understanding Find the magnitude and direction of the net force in the ice skater example just given.

14 N, $56\text{°}$ measured from the positive x -axis

View this interactive simulation to learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats.

## Summary

• Dynamics is the study of how forces affect the motion of objects, whereas kinematics simply describes the way objects move.
• Force is a push or pull that can be defined in terms of various standards, and it is a vector that has both magnitude and direction.
• External forces are any outside forces that act on a body. A free-body diagram is a drawing of all external forces acting on a body.
• The SI unit of force is the newton (N).

## Conceptual questions

What properties do forces have that allow us to classify them as vectors?

Forces are directional and have magnitude.

## Problems

Two ropes are attached to a tree, and forces of ${\stackrel{\to }{F}}_{1}=2.0\stackrel{^}{i}+4.0\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}$ and ${\stackrel{\to }{F}}_{2}=3.0\stackrel{^}{i}+6.0\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}$ are applied. The forces are coplanar (in the same plane). (a) What is the resultant (net force) of these two force vectors? (b) Find the magnitude and direction of this net force.

a. ${\stackrel{\to }{F}}_{\text{net}}=5.0\stackrel{^}{i}+10.0\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}$ ; b. the magnitude is ${F}_{\text{net}}=11\phantom{\rule{0.2em}{0ex}}\text{N}$ , and the direction is $\theta =63\text{°}$

A telephone pole has three cables pulling as shown from above, with ${\stackrel{\to }{F}}_{1}=\left(300.0\stackrel{^}{i}+500.0\stackrel{^}{j}\right)$ , ${\stackrel{\to }{F}}_{2}=-200.0\stackrel{^}{i}$ , and ${\stackrel{\to }{F}}_{3}=-800.0\stackrel{^}{j}$ . (a) Find the net force on the telephone pole in component form. (b) Find the magnitude and direction of this net force. Two teenagers are pulling on ropes attached to a tree. The angle between the ropes is $30.0\text{°}$ . David pulls with a force of 400.0 N and Stephanie pulls with a force of 300.0 N. (a) Find the component form of the net force. (b) Find the magnitude of the resultant (net) force on the tree and the angle it makes with David’s rope.

a. ${\stackrel{\to }{F}}_{\text{net}}=660.0\stackrel{^}{i}+150.0\stackrel{^}{j}\phantom{\rule{0.2em}{0ex}}\text{N}$ ; b. ${F}_{\text{net}}=676.6\phantom{\rule{0.2em}{0ex}}\text{N}$ at $\theta =12.8\text{°}$ from David’s rope

a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
what is the centripetal force
Of?
John
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
Centripetal force is a force of attraction which keeps an object round the orbit towards the center of a circle. Mathematically Fc=mv²/r
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2.  By how much does the column shrink when a 5000- kg truck is on it?
what exactly is a transverse wave
does newton's first law mean that we don't need gravity to be attracted
no, it just means that a brick isn't gonna move unless something makes it move. if in the air, moves down because of gravity. if on floor, doesn't move unless something has it move, like a hand pushing the brick. first law is that an object will stay at rest or motion unless another force acts upon
Grant
yeah but once gravity has already been exerted .. i am saying that it need not be constantly exerted now according to newtons first law
Dharmee
gravity is constantly being exerted. gravity is the force of attractiveness between two objects. you and another person exert a force on each other but the reason you two don't come together is because earth's effect on both of you is much greater
Grant
maybe the reason we dont come together is our inertia only and not gravity
Dharmee
this is the definition of inertia: a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
Grant
the earth has a much higher affect on us force wise that me and you together on each other, that's why we don't attract, relatively speaking of course
Grant
quite clear explanation but i just want my mind to be open to any theory at all .. its possible that maybe gravity does not exist at all or even the opposite can be true .. i dont want a fixed state of mind thats all
Dharmee
why wouldn't gravity exist? gravity is just the attractive force between two objects, at least to my understanding.
Grant
earth moves in a circular motion so yes it does need a constant force for a circular motion but incase of objects on earth i feel maybe there is no force of attraction towards the centre and its our inertia forcing us to stay at a point as once gravity had acted on the object
Dharmee
why should it exist .. i mean its all an assumption and the evidences are empirical
Dharmee
We have equations to prove it and lies of evidence to support. we orbit because we have a velocity and the sun is pulling us. Gravity is a law, we know it exists.
Grant
yeah sure there are equations but they are based on observations and assumptions
Dharmee
g is obtained by a simple pendulum experiment ...
Dharmee
gravity is tested by dropping a rock...
Grant
and also there were so many newtonian laws proved wrong by einstein . jus saying that its a law doesnt mean it cant be wrong
Dharmee
pendulum is good for showing energy transfer, here is an article on the detection of gravitational waves: ***ligo.org/detections.php
Grant
yeah but g is calculated by pendulum oscillations ..
Dharmee
thats what .. einstein s fabric model explains that force of attraction by sun on earth but i am talking about force of attraction by earth on objects on earth
Dharmee
no... this is how gravity is calculated:F = G*((m sub 1*m sub 2)/r^2)
Grant
gravitational constant is obtained EXPERIMENTALLY
Dharmee
the G part
Dharmee
Calculate the time of one oscillation or the period (T) by dividing the total time by the number of oscillations you counted. Use your calculated (T) along with the exact length of the pendulum (L) in the above formula to find "g." This is your measured value for "g."
Dharmee
G is the universal gravitational constant. F is the gravity
Grant
search up the gravity equation
Grant
yeahh G is obtained experimentally
Dharmee
sure yes
Grant
thats what .. after all its EXPERIMENTALLY calculated so its empirical
Dharmee
yes... so where do we disagree?
Grant
its empirical whixh means it can be proved wrong
Dharmee
so cant just say why wouldnt gravity exists
Dharmee
the constant, sure but extremely unlikely it is wrong. gravity however exists, there are equations and loads of support surrounding the concept. unfortunately I don't have a high enough background in physics but have this discussion with a physicist
Grant
can u suggest a platform where i can?
Dharmee
stack overflow
Grant
stack exchange, physics section***
Grant
its an app?
Dharmee
there is! it is also a website as well
Grant
okayy
Dharmee
nice talking to you
Dharmee
***physics.stackexchange.com/
Grant
likewise :)
Grant
Gravity surely exist
muhammed
hi guys
Diwash
hi
muhammed
what is mathematics
What is the percentage by massof oxygen in Al2(so4)3
molecular mass of Al2(SO4)3 = (27×2)+3{(32×1)+(16×4)} =54+3(32+64) =54+3×96 =54+288 =342 g/mol molecular mass of Oxygen=12×16 =192 g/mol % of Oxygen= (molecular mass of Oxygen/ molecular mass of the compound)×100% =(192/342)×100% =19200/342% =56.14%
A spring with 50g mass suspended from it,has its length extended by 7.8cm 1.1 determine the spring constant? 1.2 it is observed that the length of the spring decreases by 4.7cm,from its original length, when a toy is place on top of it. what is the mass of the toy?
solution mass = 50g= 0.05kg force= 50 x 10= 500N extension= 7.8cm = 0.078m using the formula Force= Ke K = force/extension 500/.078 = 6410.25N/m
Sampson
1.2 Decrease in length= -4.7cm =-0.047m mass=? acceleration due to gravity= 10 force = K x e force= mass x acceleration m x a = K x e mass = K x e/acceleration = 6410.25 x 0.047/10 = 30.13kg
Sampson
1.1 6.28Nm-¹
Anita
1.2 0.03kg or 30g
Anita
I used g=9.8ms-²
Anita
you should explain how yoy got the answer Anita
Grant
ok
Anita
with the fomular F=mg I got the value for force because now the force acting on the spring is the weight of the object and also you have to convert from grams to kilograms and cm to meter
Anita
so the spring constant K=F/e where F is force and e is extension
Anita
mass=50g=50/1000 kg m=0.05kg extension=7.8 cm=7.8/100 e=0.078 m g=9.8 m/s² 1.1 F=ke k=F/e k=mg/e k=0.05×9.8/0.078 k=0.49/0.078 k=6.28 N/m 1.2 F=6.28e mg=6.28e m=6.28e/g e=4.7 cm =4.7/100 e=0.047 m=6.28×0.047/9.8 m=0.29516/9.8 m=0.0301 kg
In this first example why didn't we use P=P° + ¶hg where ¶ is density
Density = force applied x area p=fA =p = mga, then a=h therefore substitute =p =mgh
Hlehle
Hlehle
sorry I had a little typo in my question
Anita
Density = m/v (mass/volume) simple as that
Augustine
Hlehle vilakazi how density is equal to force * area and you also wrote p= mgh which is machenical potential energy ? how ?
Manorama
what is wave
who can state the third equation of motion
Alfred
wave is a distrubance that travelled in medium from one point to another with carry energy .
Manorama
wave is a periodic disturbance that carries energy from one medium to another..
Augustine
what exactly is a transverse wave then?
Dharmee
two particles rotate in a rigid body then acceleration will be ?
same acceleration for all particles because all prticles will be moving with same angular velocity.so at any time interval u find same acceleration of all the prticles
Zaheer
what is electromagnetism
It is the study of the electromagnetic force, one of the four fundamental forces of nature. ... It includes the electric force, which pushes all charged particles, and the magnetic force, which only pushes moving charges.
Energy
what is units?
units as in how
praise
units are measurements of quantities
What is th formular for force
F = m x a
Santos
State newton's second law of motion
can u tell me I cant remember
Indigo
force is equal to mass times acceleration
Santos
The acceleration of a system is directly proportional to the and in the same direction as the external force acting on the system and inversely proportional to its mass that is f=ma
David
The rate of change of momentum of a body is directly proportional to the force exerted on that body.
Rani By Stephen Voron By CB Biern By Janet Forrester By Darlene Paliswat By Laurence Bailen By Rhodes By Briana Knowlton By OpenStax By Stephen Voron By Cath Yu