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Guide 6-2. Solving Connected Object Problems

In the previous guide, Net Force and Acceleration for Systems of Objects, you learned how you could treat a collection of objects as a system and solve for the acceleration of the system. We also mentioned that there's a limitation to this method when one wants to know the forces that the objects of the system exert on each other. We give an example.

Consider this situation: A helicopter is lifting a Hum-V by a cable as shown in Figure 1 below. A second cable attaches the Hum-V to a large duck. The mass of the Hum-V is mh, and the mass of the duck is md. The tension in the upper cable is TU. Assume the cables have negligible mass and do not stretch.

Goal: Determine the acceleration of the duck.

Solution: We can use the method learned previously and take the Hum-V and the duck as a system. By solving for the acceleration of the system, we'll also have the acceleration of the duck. The force diagram for the system is shown in Figure 2. The solution for the acceleration, a, is shown below the force diagram. Note that a represents the acceleration of both the duck and the Hum-V as long as the cable joining the two objects doesn't stretch. Note also that the mass of the system, msys = md + mh.

Goal: Determine the tension in the cable between the Hum-V and the duck.

Limitation: As long as we treat the duck and Hum-V as a system, we can't determine the tension in that cable. Newton's 2nd Law doesn't give us a way to determine forces internal to a system.This is an example of the limitation that we referred to previously.

Solution: The answer is to treat the duck alone as the system. Then we can examine the forces acting on the duck. The tension in the cable is one of those forces. The force diagram is shown in Figure 3. We use the symbol TL to represent the tension in the lower cable. The downward force is just the weight of the duck, as the Hum-V is no longer part of the system. The solution for the tension in the lower cable is shown in Figure 3. Note that the earlier result for the acceleration, a, is substituted to give an expression for TL in terms of the givens.

 Figure 1. Picture of the situation Figure 2. Finding the acceleration of the system Figure 3. Finding the tension in the lower cable
 When you need to find the forces that connected objects exert on each other, you must set up and solve individual net force problems for each of the objects. Here is an encapsulation of the method to use. For each object, draw a force diagram. Distinguish forces of the same type on different objects with subscripts to denote the objects. Write the net force equations for each object. Connect the net force equations as follows: Use Newton's 3rd Law to say that the forces that the two objects exert on each other have equal magnitudes. Use a characteristic of the system (such as strings that don't stretch and pulleys that exert no friction) to say that the objects have the same acceleration. With these connections, you can solve the net force equations simultaneously for the magnitude of the connecting force as well as the acceleration of the system.

The above method is illustrated with the following example. There are two additional examples in these these audio-video tutorials.

Problem statement

Two blocks are connected with a string passing over a pulley. Block 1 is on a horizontal plane and is being pushed by a force F to the right. Block 2 is on a plane inclined at 30° with respect to the horizontal. Both planes are frictionless, and the pulley is frictionless and massless. The string likewise is massless, and it cannot stretch. Find the acceleration of the system and the tension in the connecting string. (Additional given information is provided on the picture.)

Picture of the situation (with givens)

 Assumptions: The pulley is frictionless so that it won't exert a force along the string and introduce unequal tension forces on either side of the pulley. The string is unstretchable so that the accelerations of the two objects will be the same.

Goals

1. Find the acceleration of the system.
2. Find the tension in the string.

Force diagrams

The problem situation may look complicated. However, it's simplified by following a systematic procedure that starts with the force diagrams on the objects.

 Note that the normal and tension forces and the masses must be distinguished with subscripts. Also, the force diagram for each object has its own set of axes. As always, one of the axes is in the direction of the acceleration.

Special conditions

As long as the cord is taut but doesn’t stretch and the pulley is massless and frictionless, we can say, by the 3rd law, that T1 = T2. We can therefore remove the subscript and just call the tension, T.  In that case, the two masses will always remain the same distance apart and will have the same acceleration, which we will denote as ax.

Net force equations

 Mass 1 Mass 2 Fnet,1x= F + T Fnet,1y = N1 – m1g Fnet,2x = m2gsinθ - T Fnet,2y = N2 – m2gcosθ

The Fnet,y equations won’t be needed. There’s no friction, so forces along one axis don’t affect those along the other.

Applying the 2nd Law and solving for the acceleration

Applying the 2nd law to the Fnet,x equations:

m1ax = F + T

m2ax = m2gsinθ - T

Solving the first equation for T,

T = m1ax - F

and substituting into the second,

m2ax = m2gsinθ - (m1ax - F)

Solving for ax,

.

Units and sign check

All quantities to the right of the equal sign are positive, so acceleration is positive as expected. The units on the right are N/kg, also expected.

Checks of special cases

It makes sense that F is added to m2gsinθ, the component of m2g along the inclined plane. These two forces act in the same direction and both act to increase the acceleration of the system. For the following checks, let’s simplify things by setting F = 0. Then, we have:

.

Now let’s look at the angle. If θ = 0°, we expect no acceleration, since everything is horizontal in this special case. The equation above does indeed give ax = 0 when sinθ = 0. If θ = 90°, we get:

.  This is a result we've seen before.

Note that if the roles of the masses are reversed, that is, m1 becomes m2 and vice versa, the equation still holds. This is called a symmetry check.

Let’s examine the equation using special values for the masses.

Case 1. m2 = 0. The formula gives ax = 0. This makes sense, as there would be no accelerating force acting on m1.

Case 2. m1 = 0. The formula gives ax = g. With no m1, there would be no tension in the cord, so this result makes sense.

Solving for the tension

We’re now ready to solve for the tension, T. We can start with Fnet,2x and solve for T.

Substitute the expression for ax and simplify.

Check: Note that if F = 0 and θ = 90°, we get . This is the expected result.  (See textbook Example Problem 6-6.)

Substituting numerical values and reducing units