Chapter 6: Applications of Newton's Laws Chapter Review 
63 Translational Equilibrium
A particularly important special case arises when the net force on an object is zero. Objects for which this is true are said to be in translational equilibrium. Since the net force equals zero, the acceleration is also zero. Why is this case important? Look around you; objects in translational equilibrium are everywhere. Notice that being in equilibrium (for short) does not mean motionless. Exercise 5.4 and example 5.5 were cases of translational equilibrium, in which the object moved with constant velocity, while example 6.3 is one in which the object of interest was motionless.
Physlet Illustration: Determining the Spring Constant  



A spring is suspended from the ceiling, as shown. The display grid is in meters. Vary the mass attached to the spring, and determine the spring constant. Start  
Hints
 
ReferenceSee Walker, Section 62 
Picture the Problem The left picture shows the box on the ramp being held in place by the cord. The right picture is the freebody diagram of the box.
Strategy The case of translational equilibrium has only one condition to apply = 0. So we apply this condition for each direction if necessary. Since the tension is just enough to balance the forces, static friction must have its maximum value.
Solution
1. Apply the equilibrium condition to the xdirection:  
2. Solve this equation for T:  
3. Apply the equilibrium condition to the ydirection:  
4. Substitute N into the equation for T:  
5. Solve for the tension: 
Insights Notice that for translational equilibrium in any direction the condition is always that the sum of the forces equal zero.
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64 Connected Objects
Physlet Illustration: Puck Traveling in Circular Path  



A puck travels in a circular path on a frictionless table, propelled by a string pulling from the center of the circle. How does the tension in the string depend upon the mass, the speed of the block, and the radius of the circle? You may adjust the mass (10 g < m < 500 g), the speed (1 m/s < v < 50 m/s), and/or the radius (.5 m < r < 3.5 m). The tension is displayed on the screen. Start  
Hints
 
ReferenceSee Walker, Section 65 
Physlet Illustration: Translational Equilibrium  



A 20kg block is being pulled by a rope across a rough floor, as shown. The rope makes an angle q with the horizontal. The horizontal velocity is given in m/s, and the block is initially moving at 5 m/s. Adjust the angle (q < 90°) and/or the magnitude of the pulling force (F < 200 N), and attempt to achieve translational equilibrium. Start  
Hints
 
ReferenceSee Walker, Section 63 
65 Circular Motion
In the previous chapter we noted that life isn't conveniently arranged to take place only on smooth, flat, horizontal surfaces which is why we needed to study inclined surfaces. Likewise, life doesn't always take place in straight line paths. Therefore, we need to study how to handle motion along curves. Specifically, this section focuses on circular motion, or at least motion along a circular section. If the object that moves along this circular path does so at constant speed then the acceleration must be perpendicular to the velocity and always points toward the center of the path. For this reason the acceleration is called centripetal acceleration; its magnitude is given by
,
where v is its speed and r is the radius of the circular path.
According to Newton's second law, where there is an acceleraion there must be a force that causes it. In the case of circular motion the force must also point toward the center of the circular path and is therefore called a centripetal force. Also is accordance with Newton's law, the magnitude of this centripetal force equals the product of the mass and the centripetal acceleration
It is important to recognize that centripetal force is not a new kind of force. The above expression is merely a condition that must be met by whatever force holds the object in this uniform circular motion as it is often called. In your studies, and in everyday life, you will come across circular motion caused by many different sources such as a normal force, a tension, static friction, gravity, and in later chapters, electric and magnetic forces as well.
Physlet Illustration: Atwood's Machine  




Two boxes of unequal mass are connected by a light string that passes over a massless, frictionless pulley, as shown. The boxes start from rest, and the velocity of the box on the left is given in m/s. Adjust the masses (1 kg < m_{1},m_{2} < 10 kg) and measure the acceleration of the system. Can you determine the tension in the string? Start  
Hints
 
ReferenceSee Walker, Section 64 
Picture the Problem The picture shows a side view for the freebody diagram of you and your bike. The center of the circular path is to the right.
Strategy The first thing to recognize is that static friction must supply the centripetal force to sustain the circular motion. So the problem is really asking for a comparison between the required centripetal force and f_{s,max}.
Solution
1. Convert the speed to m/s:  
2. Evaluate the centripetal force:  
3. Determine the maximum force of static friction: 
Insights The results show that the needed centripetal force is well within the range of static friction, so you're safe. Notice that in step 2 I set N = mg. I did this, without having to write out a separate Newton's second law equation for the ydirection, by inspecting the freebody diagram and recognizing that N and W must balance. As you do more and more problems you should begin to gain a similar feel for the freebody diagrams.
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