Chapter 6: Applications of Newton's Laws
Chapter Review


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6-3 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.


Determining the Spring Constant

Physlet Illustration: Determining the Spring Constant

 

 

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m kg 

 
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

  1. How much mass must you add to stretch the spring by one meter?
  2. How much force does a hanging mass exert on the spring?

Reference

See Walker, Section 6-2


Example 6.4 Held in Place: A 5.32 kg box is held stationary, on a ramp inclined at 40.0 degrees to the horizontal, by a cord that is attached to a vertical wall. The length of the cord is parallel to the ramp and it provides just enough force to hold the box in place. If the coefficient of static friction between the box and the ramp is 0.101, what is the tension in the cord?

Picture the Problem The left picture shows the box on the ramp being held in place by the cord. The right picture is the free-body 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 x-direction:
2. Solve this equation for T:
3. Apply the equilibrium condition to the y-direction:
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.

Practice Quiz

 
If an object is in translational equilibrium...
it must be at rest
there are no forces acting on it
it is not rotating
it moves with constant acceleration
[none of the above]


 
Can an object be in translational equilibrium if only one force acts on it?
Yes, if itis a gravitational force.
No, because even one force can have more than one component.
Yes, trying to lift an object that is too heavy for you to lift is an example.
No, an object with only one force on it must accelerate.
[none of the above]

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your answer: [none of the above]

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your answer: No, an object with only one force on it must accelerate.

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6-4 Connected Objects

A Puck Traveling in Circular Path

Physlet Illustration: Puck Traveling in Circular Path

 

mgrams  vm/s  rm    
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

  1. If you only vary the mass, how does the tension change?
  2. If you only vary the velocity, how does the tension change?
  3. If you only vary the radius, how does the tension change?
  4. What is the centripetal acceleration of the puck?

Reference

See Walker, Section 6-5



Translational Equilibrium

Physlet Illustration: Translational Equilibrium

 

 

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FNewtons  q ° 
A 20-kg 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

  1. How should the box's velocity change if it is in translational equilibrium?.
  2. What would be the box's acceleration in this case?
  3. What net force on the box would result in such an acceleration?

Reference

See Walker, Section 6-3


6-5 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.


Atwood's Machine

Physlet Illustration: Atwood's Machine

m1 = kg
m2 = kg

 

 

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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 < m1,m2 < 10 kg) and measure the acceleration of the system. Can you determine the tension in the string? Start

Hints

  1. How do the accelerations of the two boxes compare?
  2. Can you draw a free-body diagram for each box?
  3. What force results in a box's acceleration?
  4. Apply Newton's 2nd Law to each box's motion, equating that force to the product of mass times acceleration.

Reference

See Walker, Section 6-4


Example 6.5 Making a Turn: Suppose that the combined mass of you and your bike is 75 kg. You are riding down the street and have to make a turn in a circular section of road whose radius is approximately 6.5 m. The speedometer on your bike reads 20 km/h. If the coefficient of static friction between your wheels and the road is 0.79, are you going too fast to make the turn safely?

Picture the Problem The picture shows a side view for the free-body 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 fs,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 y-direction, by inspecting the free-body 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 free-body diagrams.

Practice Quiz

 
For an object undergoing circular motion at constant speed...
the velocity is constant
the acceleration is constant
the direction of the velocity is constant
the direction of the acceleration is constant
[none of the above]
 
Which of the following correctly identifies the relationship between the directions of the velocity and acceleration for objects in circular motion at constant speed?
They are in the same direction.
They are in opposite directions.
They are perpendicular to each other.
Their directions may differ by any angle depending on the curve of the arc.
[none of the above]
 
With what constant speed must an object traverse in a circular path of radius 1.34 m in order to have a centripetal acceleration equal to the acceleration of gravity?
3.13 m/s
13.1 m/s
1.34 m/s
9.81 m/s
3.63 m/s

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your answer: [none of the above]

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your answer: They are perpendicular to each other.

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your answer: 3.63 m/s


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