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  5. Newton’s third law

5.5 Newton’s third law  (Page 3/6)

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Getting up to speed: choosing the correct system

A physics professor pushes a cart of demonstration equipment to a lecture hall ( [link] ). Her mass is 65.0 kg, the cart’s mass is 12.0 kg, and the equipment’s mass is 7.0 kg. Calculate the acceleration produced when the professor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance, total 24.0 N.

Figure shows a person pushing a cart from left to right. Near the feet of the person are the arrows labeled F subscript foot pointing left and F subscript floor pointing right. An arrow f pointing left is shown near the wheel of the cart. The arrows F subscript prof pointing right and F subscript cart pointing left are shown near her hands. The cart is circled and labeled system 2. The cart and person are circled together and this is labeled system 1. Two free body diagrams are shown. The first, of system 1 has F subscript floor pointing right, N pointing up, f pointing left and w pointing down. The second diagram, of system 2, has F subscript prof pointing right, N prime pointing up, f pointing left and w prime pointing down.
A professor pushes the cart with her demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces (except for f → , because it is too small to drawn to scale). System 1 is appropriate for this example, because it asks for the acceleration of the entire group of objects. Only F → floor and f → are external forces acting on System 1 along the line of motion. All other forces either cancel or act on the outside world. System 2 is chosen for the next example so that F → prof is an external force and enters into Newton’s second law. The free-body diagrams, which serve as the basis for Newton’s second law, vary with the system chosen.

Strategy

Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in [link] . The professor pushes backward with a force F foot of 150 N. According to Newton’s third law, the floor exerts a forward reaction force F floor of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force in the vertical direction. Therefore, the problem is one-dimensional along the horizontal direction. As noted, friction f opposes the motion and is thus in the opposite direction of F floor . We do not include the forces F prof or F cart because these are internal forces, and we do not include F foot because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the net external force can be found from all this information, we can use Newton’s second law to find the acceleration as requested. See the free-body diagram in the figure.

Solution

Newton’s second law is given by

a = F net m .

The net external force on System 1 is deduced from [link] and the preceding discussion to be

F net = F floor − f = 150 N − 24.0 N = 126 N .

The mass of System 1 is

m = ( 65.0 + 12.0 + 7.0 ) kg = 84 kg .

These values of F net and m produce an acceleration of

a = F net m = 126 N 84 kg = 1.5 m/s 2 .

Significance

None of the forces between components of System 1, such as between the professor’s hands and the cart, contribute to the net external force because they are internal to System 1. Another way to look at this is that forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the professor on the cart results in an equal and opposite force back on the professor. In this case, both forces act on the same system and therefore cancel. Thus, internal forces (between components of a system) cancel. Choosing System 1 was crucial to solving this problem.

Force on the cart: choosing a new system

Calculate the force the professor exerts on the cart in [link] , using data from the previous example if needed.

Strategy

If we define the system of interest as the cart plus the equipment (System 2 in [link] ), then the net external force on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart, F prof , is an external force acting on System 2. F prof was internal to System 1, but it is external to System 2 and thus enters Newton’s second law for this system.

Solution

Newton’s second law can be used to find F prof . We start with

a = F net m .

The magnitude of the net external force on System 2 is

F net = F prof − f .

We solve for F prof , the desired quantity:

F prof = F net + f .

The value of f is given, so we must calculate net F net . That can be done because both the acceleration and the mass of System 2 are known. Using Newton’s second law, we see that

F net = m a ,

where the mass of System 2 is 19.0 kg ( m = 12.0 kg + 7.0 kg ) and its acceleration was found to be a = 1.5 m/s 2 in the previous example. Thus,

F net = m a = ( 19.0 kg ) ( 1.5 m/s 2 ) = 29 N .

Now we can find the desired force:

F prof = F net + f = 29 N + 24.0 N = 53 N .

Significance

This force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that 150-N force is transmitted to the cart; some of it accelerates the professor. The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics of the situation (which are not necessarily the same things).

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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