Guide 6-1. Net Force and Acceleration for Systems of Objects
When 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 M1 that rests on a horizontal, frictionless surface. A block of mass M2 is in contact with block M1. If the goal is simply to find the acceleration of M1, then one can take the system to be that of both blocks and do a net force problem for that system. This will work, because both blocks will have the same acceleration. The system of two blocks is shown encircled by a yellow line in Figure 2.
The mass of this system is Msys = M1 + M2. The net, horizontal force on the system is just the push, which we'll denote as P. There are also two equal normal (or contact) forces that the blocks exert to the right and left on each other, but these are internal to our system so we can ignore them. We'll take the direction of +x to be to the right and solve for the acceleration of the system.
Fnet,sys = P
Msysasys = P
asys = P/Msys
= P/(M1 + M2)
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.
Figure 3 below shows a red block of mass M1 connected by a horizontal string passing over a pulley to a blue block of mass M2. The table is horizontal and frictionless, the pulley is massless and frictionless, and the string is massless and unstretchable. The textbook shows in example 6-6 how to solve this problem for the acceleration of either block as well as for the tension in the string. That method involves drawing a force diagram for each object, writing the net force equation for each object, and solving them simultaneously for acceleration and tension. If, however, one is only interested in the acceleration, then a simpler method is to treat the two blocks as a single system. This works because both blocks have the same acceleration. Figure 4 shows the system encircled by a yellow line.
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 and solve for the acceleration of the system.
Fnet,sys = M2g
Msysasys = M2g
asys = M2g/Msys
= M2g/(M1 + M2)
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 the weights of the blocks. (We assume M2 > M1 so that the difference M2 - M1 is positive.) This may be easier to understand if you first consider the situation where the blocks have equal mass so that (M2 - M1)g = 0. In that case, there is no acceleration, and the net force on the system is 0. Now considering the case of unequal mass and taking the direction of +y to be downward, we solve for the acceleration of the system.Fnet,sys = (M2 - M1)g
Msysasys = (M2 - M1)g
asys = (M2 - M1)g/Msys
= (M2 - M1)g/(M1 + M2)
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 M1 slides along a plane inclined at angle θ above the horizontal. When the system is selected as shown in Figure 8, the net, external force on the system is (M2 - M1sinθ)g. One can see that this expression reduces to the expressions for net, external force in Examples 2 and 3 when angles of 0° and 90° respectively are substituted for θ. When θ = 0°, the expression reduces to M2g, and when θ = 90°, the expression reduces to (M2 - M1)g.
Again taking the direction of +y to be downward, we solve for the acceleration of the system.
Fnet,sys = (M2 - M1sinθ)g
Msysasys = (M2 - M1sinθ)g
asys = (M2 - M1sinθ)g/Msys
= (M2 - M1sinθ)g/(M1 + M2)
A special case is that for which the system is in equilibrium so that asys = 0. In that case, we must have (M2 - M1sinθ) = 0, or the ratio of the masses M2/M1 = sinθ.
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