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.
The above method is illustrated with the following example. There are two additional examples in these these audio-video tutorials.
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)
The problem situation may look complicated. However, it's simplified by following a systematic procedure that starts with the force diagrams on the objects.
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
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
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