# 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 = m a, 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 = m a. As always, be careful to remember that Force is a vector. Now, however, we have two masses to deal with so you must be sure that as you apply F = m a you are considering all the forces on a particular object and only those forces acting on that particular object.

Example: Consider the Atwoods Machine shown here with masses m1 and m2. 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 same acceleration since they are attached by a cord. If m2 > m1 and the Atwoods machine is released from rest, mass m1 will accelerate upward while mass m2 accelerates downward. 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 = m a ? We apply F = m a to the masses, one at a time. Look at the smaller mass, m1. What are the forces acting on this mass? The tension in the string exerts a force up while gravity exerts a force down. We expect this mass to have an acceleration a that is up. There are no horizontal forces. We will take up as positive. Fnet = F = T - w1 = m1 a Fnet = F = T - m1 g = m1 a T - m1 g = m1 a This one equation has two unknowns -- tension T and acceleration a. So we need more inormation. We get that additional information by looking at the forces acting on the heavier mass, m2, and applying Newton's Second Law, F = m a, to that mass. There are no horizontal forces. The tension in the string exerts a force up while gravity exerts a force down. We expect this mass to have an acceleration a that is down. We can choose to call down "positive" for this mass or we can call up "positive" and then we expect this mass to have an acceleration of - a. Either choice is fine. This time, let's choose down as "positive". Fnet = F = w2 - T = m2 a Fnet = F = m2 g - T = m2 a m2 g - T = m2 a Of course, this one equation also has two unknowns -- tension T and acceleration a. But now we have two equations with two unknowns and that is sufficient. We can solve for the tension T in the first equation,

T - m1 g = m1 a

T = m1 g + m1 a

and then stubstitute that into the second equation

m2 g - T = m2 a

m2 g - ( m1 g + m1 a ) = m2 a

m2 g - m1 g - m1 a = m2 a

m2 g - m1 g = m1 a + m2 a

( m2 - m1 ) g = ( m1 + m2 ) a

( m1 + m2 ) a = ( m2 - m1 ) g

a = ( m2 - m1 ) g / ( m1 + m2 )

Example: Now, let's consider an inclined Atwoods machine. Masses m1 and m2 are connected by a string which runs over a pulley and mass m2 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 = m a. Newton's Second Law describes the effect of forces on one object. So we must isolate all the forces on mass m1 and apply it. Then we isolate all the forces on mass m2 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 = m a to these forces acting on this object.

Fnet = F = T - m1 g = m 1 a

T - m1 g = m 1 a

As we might expect by now, this one equation has two 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 m2. There are three forces acting on this mass -- the string exerts a force T, the (frictionless) inclined plane exerts a "normal" force n, and gravity pulls down with a force of w1 = m1 g. To find the net 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 of

wx = m2 g sin wy = - m2 g cos And we have

nx = 0

ny = n

and

Tx = - T

Ty = 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 = m a

F = Fnet = T + n + w = m a

Fx = Fnet,x = Fx = m 2 ax Fx = T x + n x + w x = m 2 ax

- T + 0 + m2 g sin = m 2 ax = m 2 a

where we have used

ax = a

since the acceleration is only in the positive x-direction.

- T + m2 g sin = m2 a

This 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 = m2 g sin - m2 a

[m2 g sin - m2 a] - m1 g = m 1 a

m 1 a + m2 a = m2 g sin - m1 g

( m 1 + m2 ) a = ( m2 sin - m1 ) g

a = ( m2 sin - m1 ) g / ( m 1 + m2 )

What about the y-components of the forces on mass m2, on the inclined plane?

Fy = Fnet,y = Fy = m 2 ay Fy = T y + n y + w y = m 2 ay = 0

where we have used

ay = o

since the acceleration is only in the positive x-direction and there is no acceleration perpendicular to the plane.

T y + n y + w y = 0

0 + n - m2 g cos = 0

n = m2 g cos From the y-components of the forces on mass m2, we can solve for the normal force. This will be important when we take friction into 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 m1 = 1 kg and a mass of m2 = 5 kg sitting on an inclined plane which is inclined at 30o from the horizontal. The coefficient of kinetic friction between this mass and the plane is 0.25. The forces on the hanging mass, m1, are just as they were before: But the forces on the other mass, m2, which sits on the plane now have a friction force to be included: Now we apply Newton's Second Law to these forces acting on mass m2.

Fy,net = 0

Fy,ne = 0 because there is no motion -- and, certainly, no acceleration -- in the y-direction.

Fy,net = n - m2 g cos 30o = 0

n = m2 g cos 30o

n = (5 kg) (10 m/s2) (0.866)

n = 43.3 N

Notice that the normal force is not equal to the weight! This is important. Now that we know the normal force, we can immediately calculate the kinetic friction force,

fk = n

fk = (0.25) (43.3 N)

fk = 10.8 N

Now we can apply F = m a to the x-component forces to find

Fx,net = m2 g sin 30o - T - 10.8 N = m2 a

(5 kg) (10 m/s2) (0.5) - T - 10.8 N = (5 kg) a

25 N - T - 10.8 N = (5 kg) a

14.2 N - T = (5 kg) a

We still have one equation with two unknowns. But from the forces on the hanging mass, m1, we know

T - m1 g = m1 a

T = m1 g + m1 a

T = (1 kg) (10 m/s2) + ( 1 kg) a

T = 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/s2  Vertical Circles Summary Return to ToC, Ch 6, Application of Newton's Laws

(c) Doug Davis, 2005; all rights reserved