## Atwoods Machine

Two weights connected by a string running over a pully is known as an

Atwoods Machine. An example is shown in the diagram below. Experimentally, an Atwoods Machine might be used to slow down an acceleration to make it easier to measure. Atwoods Machines, in various forms, allow us to isolate forces and understand (what else!?!?!) Newton's Second Law,F= ma, better. In solving Atwood Machine problems, we continue our well established pattern: identify all the forces, draw a clear free body diagram, apply Newton's Second Law,F= ma. As always, be careful to remember thatForce is a vector. Now, however, we havetwo massesto deal with so you must be sure that as you applyF= mayou are consideringallthe forces on a particular object andonlythose forces acting on that particular object.

Example: 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: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.

We might even have a "rough surface" -- just another way of saying "friction".

Example: Find the acceleration of an inclined Atwoods Machine with a hanging mass of m_{1}= 1 kg and a mass of m_{2}= 5 kg sitting on an inclined plane which is inclined at 30^{o}from the horizontal. The coefficient of kinetic friction between this mass and the plane is 0.25.The forces on the hanging mass, m

_{1}, are just as they were before:But the forces on the other mass, m

_{2}, which sits on the plane now have africtionforce to be included:Now we apply Newton's Second Law to these forces acting on mass m

_{2}.F _{y,net }= 0F

_{y,ne}= 0 because there is no motion -- and, certainly, no acceleration -- in the y-direction.F _{y,net }= n - m_{2}g cos 30^{o}= 0n = m

_{2}g cos 30^{o}n = (5 kg) (10 m/s

^{2}) (0.866)n = 43.3 N

Notice that the normal force is

notequal to the weight!This is important.Now that we know the normal force, we can immediately calculate the kinetic friction force,f _{k}= nf

_{k}= (0.25) (43.3 N)f

_{k}= 10.8 NNow we can apply

F= mato the x-component forces to findF _{x,net}= m_{2}g sin 30^{o}- T - 10.8 N = m_{2}a(5 kg) (10 m/s

^{2}) (0.5) - T - 10.8 N = (5 kg) a25 N - T - 10.8 N = (5 kg) a

14.2 N - T = (5 kg) a

We still have

one equationwithtwo unknowns. But from the forces on the hanging mass, m_{1}, we knowT - m _{1}g = m_{1}aT = m

_{1}g + m_{1}aT = (1 kg) (10 m/s

^{2}) + ( 1 kg) aT = 10 N + (1 kg) a

Now we substitute that to find

14.2 N - [10 N + (1 kg) a] = (5 kg) a 14.2 N - 10 N - (1 kg) a = (5 kg) a

4.2 N - (1 kg) a = (5 kg) a + (1 kg) a = (6 kg) a

a = 6 kg / 4.2 N

a = 1.43 m/s

^{2}

Return to ToC, Ch 6, Application of Newton's Laws(c) Doug Davis, 2005; all rights reserved