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Example Problem 1:
situationA 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. Ignore friction forces and determine the acceleration of the boxes and the force acting between the boxes.
The first approach to this problem involves the dual combination of a system analysis and an individual object analysis. As mentioned, the system analysis is used to determine the acceleration and the individual object analysis is used to determine the forces acting between the objects. In the system analysis, the two objects are considered FBD of Systemto be a single object. The dividing line that separates the objects is ignored. The mass of the system of two objects is 15.0 kg. The free-body diagram for the system is shown at the right. There are three forces acting upon the system - the gravity force (the Earth pulls down on the 15.0 kg of mass), the normal force (the floor pushes up on the system to support its weight), and the applied force (the hand is pushing on the back part of the system). The force acting between the 5.0-kg box and the 10.0-kg box is not considered in the system analysis since it is an internal force. Just as the forces holding atoms together within an object are not included in a free-body diagram, so the forces holding together the parts of a system are ignored. These are considered internal forces; only external forces are considered when drawing free-body diagrams. The magnitude of the force of gravity is m•g or 147 N. The magnitude of the normal force is also 147 N since it must support the weight (147 N) of the system. The applied force is stated to be 45.0 N. Newton's second law (a = Fnet/m) can be used to determine the acceleration. Using 45.0 N for Fnet and 15.0 kg for m, the acceleration is 3.0 m/s2.
Now that the acceleration has been determined, an individual object analysis can be performed on either object in order to determine the force acting between them. It does FBD front boxnot matter which object is chosen; the result will be the same in either case. Here the individual object analysis is conducted on the 10.0 kg object (only because there is one less force acting on it). The free-body diagram for the 10.0-kg object is shown at the right. There are only three forces acting upon it - the force of gravity on the 10.0-kg, the support force (from the floor pushing upward) and the rightward contact force (Fcontact). As the 5.0-kg object accelerates to the right, it will be pushing rightward upon the 10.0-kg object; this is known as a contact force (or a normal force or an applied force or …). The vertical forces balance each other since there is no vertical acceleration. The only unbalanced force on the 10.0-kg object is the Fcontact. This force is the net force and is equal to m•a where m is equal to 10.0 kg (since this analysis is for the 10.0-kg object) and a was already determined to be 3.0 m/s2. The net force is equal to 30.0 N. This net force is the force of the 5.0-kg object pushing the 10.0-kg object to the right; it has a magnitude of 30.0 N. So the answers to the two unknowns for this problem are 3.0 m/s2 and 30.0 N.
Now we will consider the solution to this same problem using the second approach - the use of two individual object analyses. In the process of this second approach, we will ignore the fact that we know what the answers are and presume that we are solving the problem for the first time. In this approach, two separate free-body diagram analyses are performed. The diagrams below show the free-body diagrams for the two objects.
FBDs for Both objects
HOPE IT IS HELPFUL
