Chapter 4
The Giancoli onLine Tutor

Motion and Force: Dynamics

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Problems in this chapter divide themselves into two categories, the least important is...

I. Problems with No Acceleration

Problems with no acceleration are called "statics" problems and are covered in more detail in a later chapter. For the purpose of this chapter it is sufficient to note that if an object is at rest the net force on it must be zero. If the object rests on a level surface the supporting force must equal the weight.

II. Problems with Acceleration

Problems with One Force


Problems with only one active force (or multiple forces in the same direction) are one dimensional problems. A typical problem is a box on a level table subject to a horizontal force. There are other forces in the problem, the weight and supporting force, but they cancel out and can usually be ignored. Problems in this category are actually Chapter 2 problems in disguise. Recall that Chapter 2 dealt with problems with 6 unknowns:
t is the time at the end of the problem assuming that t=0 at the start,
xo is the location at the start of the problem (i.e. when t=0),
x is the location at the end of the problem,
vo is the speed at the start,
v is the speed at the end,
a is the acceleration.
To solve these problems, you had the use of two of the three equations:

  1. v = vo + at,
  2. x = xo + vot + 0.5t2,
  3. v2 = vo2 + 2a(x - xo).
(Review Chapter 2 hints.).
One dimensional force problems add two more unknown variables: and one more equation:
F = ma.

A solution involves identifying five variable values. Typically a problem statement gives the mass and enough information (four variable values just as in Chapter 2) to calculate the acceleration. The mass and acceleration allow you to calculate the force using F = ma.

To summarize when working a problem of this type:
  1. Find the acceleration.
  2. Use F = ma to find the answer, the force (F) if mass (m) is given or the mass if force is given.

Other problems give values for force (F) and mass (m) and expect you to find the acceleration (a) and use it to find one of the other 5 kinematic variables, vo, v, xo, x, and t. This is basically a Chapter 2 problem with acceleration as one of the given variable values.

Problems with Many Forces

Problems with many forces require the use of the component method of vector addition to reach a solution. The following exercise will help you review that method.

Interactive Exercise

In the figure on the left, three forces are shown acting on an object. The force F2 makes a 45o angle with F3. Remember that your calculator assumes all forces are measured counter-clockwise from East.

What is the calculator angle for F2?

Complete the following table (see Chapter 3 hints if you dont understand what to do). Enter the value in each box then click outside the box. If you make a mistake you will receive coaching comments in the feedback field at the bottom of the form.
Force X-Component Y-Component
F1
F2
F3
TOTALS (F)
FEEDBACK:
When you have successfully completed the table above, you have found the x and y components, Fx and Fy, of the total force on the yellow charge. Scroll down to see the results!


The Fx and Fy components are shown in the figure along with the resulting net force vector (F).
To find the magnitude of F use the Pythagorean Theorem.

|F|=√(Fx2+Fy2)=√(1.3542+.6462)= 1.5

Use the inverse tangent function to find the angle, as follows:

q = tan-1(Fy/Fx)= tan-1(.646/-1.354)= tan-1(-0.477)
Checking this on your calculator gives an answer of -25.5o. The angle defined by the blue line in the figure (clockwise angles are considered negative), which is not the angle of the force (F).
This occurs because the inverse tangent function is double valued. It cannot distinguish between

tan-1(.646/.-1.354) [second quadrant vector]

and

tan-1(-.645/1.355) [fourth quadrant vector]
By convention the calculator returns the smallest angle. All second quadrant vectors will produce results in the fourth quadrant. It is up to you to recognize this and add 180o to the result to obtain the correct answer. Similarly, all third quadrant results will be mirrored in the first quadrant, again you must add 180o. If you get a result that is actually in the fourth quadrant, convert it to the correct positive angle by adding 360o.

Types of Forces

There are four basic forces that are used in this chapter:
The "movie" at the left shows a box at rest on a rough (high static coefficient of friction) surface and shows how three of these forces behave as the surface is tilted. Use the buttons on the right side of the controller to step back and forth through the frames.
w
(=mg)		This is the weight, the force of gravity; it is always directed downward. (See the movie to the left.)
FNThis is the Normal (ie, perpendicular, or supporting) force. It is always perpendicular to the supporting surface and becomes smaller as the surface tilts.
fs or kThis is the force of friction. It is equal to mFN, and it is always parallel to the surface and opposed to the motion. m is the coefficient of friction and is usually a given. The coefficient of friction depends on the nature of the two surfaces and on whether or not the object is moving. Static friction, as is the case here, is greater than kinetic (moving) friction.
FThis is usually a driving force like that from the driving wheels of a car and is usually in the direction of motion.
Notice that as the incline gets steeper the friction force gets greater to counter the tendency of the box to slide down the incline under the increasing influence of gravity.

If you don't have the Quicktime plugin there is probably a broken image at the left. You can view a GIF animation version of the movie here.


Problem Solving Strategies

Types of Problems

There are three basic variations on problems with multiple forces:
  1. Objects on plains, inclined or not,
  2. Objects hung over pulleys,
  3. A combination of the above.


The general approach to solving any of these problems is outlined in the following Interactive Activity.

Interactive Activity

Consider the problem on the right below. To solve this problem and others like it follow the steps:
  1. DRAW A PICTURE, like the one on the right below.
  2. Pick out an object or objects in the picture to concentrate upon (that is your "system") and draw all forces acting on the object from outside the system.
     
  3. Shrink the object to a point leaving the forces in place.
     
  4. Choose an x and y axis (choose the x axis parallel to the surface and the y axis perpendicular to the x axis.)
     
  5. Replace any force not directed along the x or y axis by its x and y component.
     
  6. Set the sum of the forces in the y direction equal to zero.
  7. Set the sum of the forces in the x direction equal to the mass times the acceleration and solve for the unknown force.
     
  1. The 'Box' on the inclined plain problem


    Use the buttons on the right side of the movie control panel to step back and forth to see the effect of the tilt on the problem. The only difference in this problem from the one above is that in this problem the weight has to be resolved into components instead of the "pulling" force. Otherwise the problems are worked in exactly the same way.
    If you don't have the Quicktime plugin there is probably a broken image at the left. You can view a GIF animation version of the movie here.



    Interactive Example

    If the mass (m) in the above illustration is 10 kg and the angle (q) is 30o, what is the acceleration (a)? The acceleration of gravity (g) is 9.80 m/s2.
    Enter your answer then click outside the box.

    To the nearest newton, what is the normal force (FN)?
    Enter your answer then click outside the box.
  2. Problem involves the use of pulleys.




    The figure on the left above shows a simple problem, in which a mass of 22 kg is being pulled upward by a force of 120 N and downwards by a force of 100 N.

    The acceleration will be
    a = (120 - 100)/22
    In the middle figure we have cut the mass into two parts and attached them via a string. Everything else is unchanged and the answer is the same, because the string adds nothing to the problem.

    The figure on the right shows the same two masses hung over a pulley and hanging there under the influence of their weight (w=mg; g=10.0 m/s2, for illustration). The only effect of the pulley is to change up into down. The 12 kg mass cannot tell whether it is in the middle figure or the one on the right; and, consequently the answer is the same!
    Put another way:
    The two problems are the SAME. You can get rid of the pulley by simply straightening the line out, leaving everything else the same.

Interactive Activity

Find the acceleration of the two blocks in the figure on the right below:
  1. DRAW A PICTURE, like the one on the right.
  2. Pick out the objects in the picture to concentrate upon (that is your "system") and draw all forces acting on the objects from outside the system. In this case the objects of interest are the two blocks and the cord that connects them.
     
  3. Replace the pulley by straightening the cord.
     
  4. Choose an x and y axis (choose the x axis parallel to the surface and the y axis perpendicular to the x axis.)
     
  5. Shrink the objects to a point leaving the forces in place. Note: The mass of the system is 20 kg.
     
  6. Replace any force not directed along the x or y axis by its x and y component.
     
  7. Set the sum of the forces in the y direction equal to zero.
  8. Set the sum of the forces in the x direction equal to the mass times the acceleration and solve for the acceleration.
     
© 1997-2000 Carl Adler