## Applications

Now that we

knowNewton's Laws of Motion, how do weapplythem? How can they let uspredictthe motion of an object if we know all the forces acting upon it? How can they let uspredictthe forces on an object if we know its motion?It is only a slight over statement to say that the rest of this course is just looking for devious ways to apply

F= ma

Example 1:Consider a crate being pulled along a frictionlessfloor (while such a floor isveryhard to find, this will still help us understand the conceptandwe can return to this situation later,afterconsidering friction, and solve it more realistically).Consider a crate being pulled along a horizontal, friction

lessfloor. A rope is tied around it and a man pulls on the rope with a force of T. T is thetensionin the rope. What happens to the crate?Before we can apply Newton's Second Law,

F= mawe must find the

netforce -- thevectorsum ofallthe forces -- acting on the object. In addition to the forceTexerted by the rope, what other forces acton the object?As discussed in class, in Mechanics, we can restrict our attention to "contact" forces and "gravity". That means gravity pulls

downon this crate with a force equal to its weight,w. But the floor supports the crate. The floor responds by pushingupon the crate with a force we call thenormal force. "Normal" means "perpendicular". We will call this forcen; you may also encounter it labeledNorF_{N}.These forces are shown on the "free body diagram" above. We have drawn in

allthe forces actingonthe object. The net force is thevectorsumof these forces.F_{net}=F=T+n+wwhere , the Greek upper case "sigma", means "the sum of". Remember, tho', vector notation is

alwayselegant shorthand notation. When we writeF_{net}=F=T+n+wwe have really written

F _{net, x}= F_{x}= T_{x}+ n_{x}+ w_{x}

andF _{net,y}= F_{y}= T_{y}+ n_{y }+ w_{y}What are these x- and y-components of the forces

T, n,andw? For this first, simple example, we can find -- by inspection -- that these components areT _{x}= TT

_{y}= 0n

_{x}= 0n

_{y}= nw

_{x}= 0w

_{y}= - wNow we are ready to apply

F= maBut that must first be written in terms of components,

F _{x}= F_{net,x}= F_{x}= m a_{x}F

_{x}= F_{net,x}= F_{x}= T_{x}+ n_{x}+ w_{x}= T = m a_{x}T = m a

_{x}a

_{x}= T / mThe crate has a

horizontal accelerationequal to the tension T divided by m, the mass of the crate. What about the forces in the vertical direction?F _{y}= F_{net,y}= F_{y}= m a_{y}F

_{y}= F_{net,y}= F_{y}= T_{y}+ n_{y}+ w_{y}= n - w = m a_{y}n - w = m a

_{y}Since we know the crate does not accelerate in the y-direction -- it does not jump up off the floor and it does not burrow down into the floor -- we know a

_{y}= 0, son = w The upward normal force exerted by the floor on the crate, in this situation, is equal to the weight, the downward force of gravity.

Example 2:What are the forces acting on book if you pushdownon it with a forceFwhile it sits on a smooth, horizontal table as shown in the sketch below?

Draw in all the forces.This is called a "free body diagram". We will restrict ourselves, in this Mechanics course, to "contact forces" and the force of gravity. Contact forces, for this case, will be the "normal" force -- the perpendicular force -- exerted by the table -- labelednin the diagram -- and the forceFexerted by the hand. Gravity exerts a force downward, called the weight and labeledw. Just as in the previous example, we can immediately writeF= mabut that is really elegant shorthand for

F _{x}= F_{net,x}= F_{x}= m a_{x}

andF _{y}= F_{net,y}= F_{y}= m a_{y}In this example, tho',

nothing happens in the horizontal direction.All of the forces haveonlyvertical components so all we really have isF _{y}= F_{net,y}= F_{y}= m a_{y}Taking up as positive, we have

F _{y}= F_{net,y}= F_{y}= n - w - FSince the book sits on a table, we know it does not accelerate so a

_{y}= 0. This meansn - w - F = 0 n = w + F

We can use Newton's Laws to determin the value of the normal force n.

This same idea and technique can be used is slightly different situation,

Example 3:Consider a lamp hanging from a chain. What is thetensionin the chain?As always, begin with a "free body diagram". Tension

Tactsupwardon the lamp while the force of gravity pullsdownwith forcew, theweightof the lamp. Thenet forceis thevector sumof these two forces. The lamp isnot acceleratingso the forceupmust equal the forcedown. In terms of magnitudes, this meansT = w

Tension:Tension is themagnitudeof the force exerted by a chain or a rope or a string. The direction of that force depends upon the rest of the situation and the object that we are concentrating on at the moment. If we concentrate on the chain shown below, the downward forceT'is the force exerted on the chain by the lamp while the upward forceT''is the force exerted on the chain by the ceiling. There is no substitute for good free-body diagrams.

Example 4:Consider a traffic light suspended by cords as shown in the sketch below. What is the tension in each of those cords?Tension T

_{3}is easy so we will look at the first. As we have seen in the previous two examples; this tension in a vertical cord supporting a weight is just equal to the weight. In the diagram below, we have drawn in the forces acting on the traffic light. The only forces acting on the traffic light arew, the weight acting downward, andT_{3}, the upward force due to the vertical cable. T_{3}is the tension in this cable. ClearlyT _{3}= wBut what of the tensions in the other two cables, T

_{2}and T_{1}? To find those, we must look at the junction where the three cables come together. That junction is in equilibrium soFnet = 0

F_{net}=F=T_{1}+T_{2}+T_{3}= 0However, we must remember that this single

vector equationis elegant, shorthand notation fortwo scalar equations,F _{net,x}= F_{x}= T_{1 x}+ T_{2 x}+ T_{3 x}= 0F

_{net,y}= F_{y}= T_{1 y}+ T_{2 y}+ T_{3 y}= 0So we must resolve all these forces into their x- and y-components,

T _{1x}= - T_{1}cos 37^{o}= - 0.8 T_{1}T

_{1y}= T_{1}sin 37^{o}= 0.6 T_{1}T

_{2x}= T_{2}cos 53^{o}= 0.6 T_{2}T

_{2y}= T_{2}sin 53^{o}= 0.8 T_{2}T

_{3x}= 0T

_{3y}= - T_{3}= - wThe

signsareimportant!Now we can go back to the component equations and solve for tension T_{1}and T_{2}.F _{net,x}= F_{x}= T_{1 x}+ T_{2 x}+ T_{3 x}= 0T

_{1 x}+ T_{2 x}+ T_{3 x}= 0- 0.8 T

_{1}+ 0.6 T_{2}+ 0 = 0T

_{1}= 0.75 T_{2}F

_{net,y}= F_{y}= T_{1 y}+ T_{2 y}+ T_{3 y}= 0T

_{1 y}+ T_{2 y}+ T_{3 y}= 00.6 T

_{1}+ 0.8 T_{2}2 - w = 00.6 T

_{1}+ 0.8 T_{2}= w0.6 (0.75 T

_{2}) + 0.8 T_{2}= w1.25 T

_{2}= wT

_{2}= 0.8 wT

_{1}= 0.75 (0.8 w)T

_{1}= 0.6 w

Now we expand our applications and look at anAtwoods Machine

Example 5: Consider the Atwoods Machine shown here with masses m_{1}and m_{2}. They are attached by a lightweight cord over a pulley as shown. What is the acceleration of the system?

We may say "acceleration of the system" for masses 1 and 2 will have the

sameacceleration since they are attached by a cord.If m

_{2}> m_{1}and the Atwoods machine is released from rest, mass m_{1}will accelerateupward while mass m_{2}acceleratesdownward. Actually, that will be their accelerations whether the system is released from rest or is moving. It is probably easier to visualize if you think of the system being released from rest.How can we apply

F= ma?We apply

F= mato the masses, one at a time.Look at the smaller mass, m

_{1}. What are the forces acting on this mass?The tension in the string exerts a force

upwhile gravity exerts a forcedown. We expect this mass to have an acceleration a that isup. There are no horizontal forces.We will take

upaspositive.F _{net}= F = T - w_{1}= m_{1}aF

_{net}= F = T - m_{1}g = m_{1}aT - m

_{1}g = m_{1}aThis

oneequation hastwo unknowns-- tension T and acceleration a. So we needmore inormation.We get that additional information by looking at the forces acting on the heavier mass, m

_{2}, and applying Newton's Second Law,F= ma, to that mass. There are no horizontal forces.The tension in the string exerts a force

upwhile gravity exerts a forcedown. We expect this mass to have an acceleration a that isdown. We can choose to calldown"positive" for this mass or we can callup"positive" and then we expect this mass to have an acceleration of - a. Either choice is fine.This time, let's choose

downas "positive".F _{net}= F = w_{2}- T = m_{2}aF

_{net}= F = m_{2}g - T = m_{2}am

_{2}g - T = m_{2}aOf course, this

oneequation also hastwo unknowns-- tension T and acceleration a.But now we have

twoequations withtwo unknownsand that is sufficient. We can solve for the tension T in the first equation,T - m _{1}g = m_{1}aT = m

_{1}g + m_{1}aand then stubstitute that into the second equation

m _{2}g - T = m_{2}am

_{2}g - ( m_{1}g + m_{1}a ) = m_{2}am

_{2}g - m_{1}g - m_{1}a = m_{2}am

_{2}g - m_{1}g = m_{1}a + m_{2}a( m

_{2}- m_{1}) g = ( m_{1}+ m_{2}) a( m

_{1}+ m_{2}) a = ( m_{2}- m_{1}) ga = ( m

_{2}- m_{1}) g / ( m_{1}+ m_{2})

Example 6:Now, let's consider an inclined Atwoods machine. Masses m_{1}and m_{2}are connected by a string which runs over a pulley and mass m_{2}sits on a smooth inclined plane. Remember, "smooth" is just a code word for "frictionless"; we'll get to friction shortly. This inclined Atwoods machine is sketched here:Now we want to apply Newton's Second Law,

F= ma. Newton's Second Law describes the effect of forces on one object. So we must isolate all the forces on mass m_{1}and apply it. Then we isolate all the forces on mass m_{2}and apply it again. This calls for good free-body diagrams.The hanging mass m1 has only two forces on it; the string pulls up with a force we label T while gravity pulls down with a force we label w:

We expect the acceleration to be upward and have that drawn beside the free-body diagram. As always, we are now ready to apply

F= mato these forces acting on this object.F _{net}= F = T - m_{1}g = m_{1}aT - m

_{1}g = m_{1}aAs we might expect by now, this

one equationhastwo unknowns-- tension T and acceleration a -- so we must look elsewhere for additional information. Of course, the place to look is at the other mass.Carefully construct a free-body diagram showing all the forces acting on mass m

_{2}. There arethreeforces acting on this mass -- the string exerts a forceT, the (frictionless) inclined plane exerts a "normal" forcen, and gravity pullsdownwith a force ofw_{1}= m_{1}g. To find thenet force, we must resolve these vectors into their components. Since the acceleration will be along the direction of the plane, we have chosen that direction as the x-axis.Notice that the angle in this diagram is measured from the

y-axis.That means the weight has components ofw _{x}= m_{2}g sinw

_{y}= - m_{2}g cosAnd we have

n _{x}= 0n

_{y}= nand

T

_{x}= - TT

_{y}= 0

Make sure you understand the signs and sines! Do not go on until all these components are clear to you!Now we can apply Newton's Second Law to this mass:

F= ma

F=F_{net}=T+n+w= maF

_{x}= F_{net,x}= F_{x}= m_{2}a_{x}F

_{x}= T_{x }+ n_{x }+ w_{x}= m_{2}a_{x}- T

_{ }+ 0_{ }+ m_{2}g sin = m_{2}a_{x}= m_{2 }awhere we have used

a _{x}= asince the acceleration is only in the positive x-direction.

- T _{ }+ m_{2}g sin = m_{2 }aThis provides all the information we need to solve for T and a. As before, we can solve one of these equations for T and substitute that into the other equation and solve for a.

T _{ }= m_{2}g sin - m_{2 }a[m

_{2}g sin - m_{2 }a] - m_{1}g = m_{1}am

_{1}a + m_{2 }a = m_{2}g sin - m_{1}g( m

_{1}+ m_{2 }) a = ( m_{2}sin - m_{1}) g

a = ( m_{2}sin - m_{1}) g / ( m_{1}+ m_{2 })What about the y-components of the forces on mass m

_{2}, on the inclined plane?F _{y}= F_{net,y}= F_{y}= m_{2 }a_{y}F

_{y}= T_{y }+ n_{y }+ w_{y}= m_{2}a_{y}= 0where we have used

a _{y}= osince the acceleration is only in the positive x-direction and there is no acceleration perpendicular to the plane.

T _{y }+ n_{y }+ w_{y}= 00 + n - m

_{2}g cos = 0n = m

_{2}g cosFrom the y-components of the forces on mass m

_{2}, we can solve for the normal force. This will be important when we takefrictioninto account.

Example7:Two blocks of masses m_{1}and m_{2}are placed in contact with each other on a smooth, horizontal plane as shown here. A constant horizontal forceFis applied to m_{1}. What is the acceleration of each mass?In one sense, we can (almost) solve this example intuitively -- in our head. A force F is applied to an object whose mass is m = m

_{1}+ m_{2}. So its acceleration must bea = F / m or

a = F / (m _{1}+ m_{2})That's the right answer! But is there nothing more to this question? Simple questions -- the intuitively obvious kind -- make wonderful templates or examples for more difficult problems.

Look at all the forces on m

_{1}. Make a good free-body diagram of the forces acting on m_{1}.Of course the external force

Fpushes to the right on the mass m_{1}. Gravity pulls down with forcew_{1}= m_{1}gand the plane responds with a normal forcen_{1}. But the other mass -- m_{2}-- exerts a force on mass m_{1}. This force is labeledP'and points to the left. We can apply Newton's Second Law to the y-component forces and find that n_{1}= w_{1}1. But now there is an additional and unknown force in the x-component of Newton's Second Law,F _{1,net}= F - P' = m_{1}aWe need

more informationso we turn to the other mass, m_{2}The first mass m1 exerts a force

Pon this mass, m2. Applying the y-component ofF= ma, we readily findn _{2}= w_{2}For the x-components, the

onlyforce acting on m2 is P soF _{2,net}= P = m_{2}aHowever, from Newton's Third Law,

F_{12}= -F_{21}, we know that P = P' soF - P' = m _{1}aF - m

_{2}a = m_{1}aF = m

_{1}a + m_{2}aF = ( m

_{1}+ m_{2}) aa = F / ( m

_{1}+ m_{2})That is the same answer we found so quickly earlier but this may provide a pattern to use for more complex situations.

Example 8:This particular example is stated in terms of weighing afishin an accelerating elevator. It is also fun to think of weighingyourselfin an accelerating elevator. When does an elevatoraccelerate upwards? When does an elevatoraccelerate downwards?The forces acting on the fish are shown in the free-body diagram. T is the tension supplied by the scale. This is the value the scale reads. We may call it the

apparent weightof the fish. The net force on the fish isF _{net}= T - wor

F _{net}= T - mgThe net force is (

always!) equal to the mass times the acceleration. This fish is moving along with the elevator. In this diagram we have taken the acceleration to beupso it is positive.F _{net}= T - m g = m aT = m g+ m a

T = m (g + a)

While the elevator accelerates

upward, theapparent weightof the fish isgreaterthan its true weight, mg.What happens as the elevator accelerates

downward?The forces on the fish are again shown in the free-body diagram,

F _{net}= T - wor

F _{net}= T - mgThe net force is (

always!) equal to the mass times the acceleration. This fish is moving along with the elevator. Now the acceleration to beupso it isnegative.F _{net}= T - m g = m ( - a )T = m g - m a

T = m (g - a)

While the elevator accelerates

downward, theapparent weightof the fish islessthan its true weight, mg.Try this

yourselfon an elevator -- not weighing a fish, put paying attention to your ownapparent weight!

Third LawFrictionReturn to ToC, Ch4, Newton's Laws of Motion(c) 2002, Doug Davis; all rights reserved