Guide 6-1. Net Force and Acceleration for Systems
of ObjectsWhen a system is composed of more than one object, and you're only interested in the external forces on the system and the acceleration of the system, you can treat all of the objects in the system as one. Basically, this reduces to a net force problem for one object. The limitation of this method is that you won't be able to determine any forces internal to the system.
Consider, for example, the situation
shown in Figure 1 below. A force (Push) is applied to the right to a block
of mass M
The mass of this system is M F M a
= P/(M Note what this method is unable to tell us. We can't determine the forces that the red and green blocks exert on each other. In order to do that, we would have to draw force diagrams for each of the blocks individually, write a net force equation for each block, and solve the two equations simultaneously. This procedure has been presented in the assessment E.05.02v2.
Figure 3 below shows a red block of mass M
The net, external force acting on the system is just the weight of the second block. That's because the weight of the first block is balanced by the normal force on it. There are also two equal tension forces that the strings exert on each block but these are internal to our system so we can ignore them. We'll take the direction of +y to be downward solve for the acceleration of the system. F M a
= M This is the same result obtained for the acceleration in textbook example 6-6.
In this example--depicted in Figure 5--our two blocks are connected by a massless, unstretchable string which passes over a massless, frictionless pulley. The pulley is supported from above so that it has no overall vertical motion. When the masses are released, one mass accelerates downward while the other accelerates upward. The magnitudes of the two accelerations are the same. The textbook shows in example 6-7 how to solve this problem for the acceleration of either block as well as for the tension in the string. If we need to know only the acceleration but not the tension, we may treat the two blocks as a system. Figure 6 shows the system encircled by a yellow line.
The net, external force acting on the system is just the
difference in weights of the blocks. (We assume M F M a
= (M This is the same result obtained for the acceleration in textbook example 6-7. For the final example we consider a situation which is
midway between those of Examples 3 and 4. In Figure 7, block M
Again taking the direction of +y to be downward, we solve for the acceleration of the system. F M a
= (M A special case is that for which the system is in
equilibrium so that a
When finding the acceleration of a system of objects each of which have the same acceleration, one may treat the system as a single object. The acceleration of the system is then the quotient of the net, external force on the system and the total mass of the system. If in addition to finding the acceleration, one must find the forces that the objects exert on each other, then a net force problem must be done for each object individually. This method will be described later. |