Introduction
2.1 INTRODUCTION
2.2. FORCE ON A PARTICLE. RESULTANT OF TWO FORCES
2.3 VECTORS
2.4 ADDITION OF VECTORS
2.5 RESULTANT OF SEVERAL CONCURRENT FORCES
2.6 RESOLUTION OF A FORCE INTO COMPONENTS
2.7. RECTANGULAR COMPONENTS OF A FORCE. UNIT VECTORS
2.8. ADDITION OF FORCES BY SUMMING X AND Y COMPONENTS
2.9 EQUILIBRIUM OF A PARTICLE
2.10 NEWTON'S FIRST LAW OF MOTION
2.11 PROBLEMS INVOLVING THE EQUILIBRIUM OF A PARTICLE. FREE-BODY DIAGRAMS
2.12. RECTANGULAR COMPONENTS OF A FORCE IN SPACE
2.13 FORCE DEFINED BY ITS MAGNITUDE AND TWO POINTS ON ITS LINE OF ACTION
2.14 ADDITION OF CONCURRENT FORCES IN SPACE
2.15 EQUILIBRIUM OF A PARTICLE IN SPACE
Chapter 2 Ending
Page 21
2.6 RESOLUTION OF A FORCE INTO COMPONENTS

We have seen that two or more forces acting on a particle may be replaced by a single force which has the same effect on the particle. Conversely, a single force F acting on a particle may be replaced by two or more forces which, together, have the same effect on the particle. These forces are called the components of the original force F, and the process of substituting them for F is called resolving the force F into components.

Clearly, for each force F there exist an infinite number of possible sets of components. Sets of two components P and Q are the most important as far as practical applications are concerned. But, even then, the number of ways in which a given force F may be resolved into two components is unlimited (Fig.2.15). Two cases are of particular interest:

  1. One of the Two Components, P, Is Known. The second component, Q, is obtained by applying the triangle rule and joining the tip of P to the tip of F (Fig.2.16); the magnitude and direction of Q are determined graphically or by trigonometry. Once Q has been determined, both components P and Q should be applied at A.

  2. The Line of Action of Each Component Is Known. The magnitude and sense of the components are obtained by applying the parallelogram law and drawing lines, through the tip of F, parallel to the given lines of action (Fig.2.17). This process leads to two well-defined components, P and Q, which can be determined graphically or computed trigonometrically by applying the law of sines.

Fig.2.15
Fig.2.16
Fig.2.17

Many other cases can be encountered; for example, the direction of one component may be known, while the magnitude of the other component is to be as small as possible (see Sample Prob. 2.2). In all cases the appropriate triangle or parallelogram which satisfies the given conditions is drawn.

Page 22
SAMPLE PROBLEM 2.1

The two forces P and Q act on a bolt A. Determine their resultant.

SOLUTION

Graphical Solution.  A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant are measured and found to be

The triangle rule may also be used. Forces P and Q are drawn in tip-to-tail fashion. Again the magnitude and direction of the resultant are measured.

Trigonometric Solution.  The triangle rule is again used; two sides and the included angle are known. We apply the law of cosines.

Now, applying the law of sines, we write

(1)

Solving Eq. (1) for sin A, we have

Using a calculator, we first compute the quotient, then its arc sine, and obtain

We use 3 significant figures to record the answer (cf. Sec.1.6):

Alternative Trigonometric Solution.  We construct the right triangle BCD and compute

Then, using triangle ACD, we obtain

Again,

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SAMPLE PROBLEM 2.2

A barge is pulled by two tugboats. If the resultant of the forces exerted by the tugboats is a 5000-lb force directed along the axis of the barge, determine (a) the tension in each of the ropes knowing that α = 45°, (b) the value of α for which the tension in rope 2 is minimum.

SOLUTION

a. Tension for α = 45°. Graphical Solution.  The parallelogram law is used; the diagonal (resultant) is known to be equal to 5000 lb and to be directed to the right. The sides are drawn parallel to the ropes. If the drawing is done to scale, we measure

Trigonometric Solution.  The triangle rule can be used. We note that the triangle shown represents half of the parallelogram shown above. Using the law of sines, we write

With a calculator, we first compute and store the value of the last quotient. Multiplying this value successively by sin 45° and sin 30°, we obtain

b. Value of α for Minimum T2.  To determine the value of α for which the tension in rope 2 is minimum, the triangle rule is again used. In the sketch shown, line 1-1′ is the known direction of T1. Several possible directions of T2 are shown by the lines 2-2′. We note that the minimum value of T2 occurs when T1 and T2 are perpendicular. The minimum value of T2 is

Corresponding values of T1 and α are

Page 24
SOLVING PROBLEMS ON YOUR OWN
T

he preceding sections were devoted to the parallelogram law for the addition of vectors and to its applications.

Two sample problems were presented. In Sample Prob. 2.1, the parallelogram law was used to determine the resultant of two forces of known magnitude and direction. In Sample Prob. 2.2, it was used to resolve a given force into two components of known direction.

You will now be asked to solve problems on your own. Some may resemble one of the sample problems; others may not. What all problems and sample problems in this section have in common is that they can be solved by the direct application of the parallelogram law.

Your solution of a given problem should consist of the following steps:

1. Identify which of the forces are the applied forces and which is the resultant.  It is often helpful to write the vector equation which shows how the forces are related. For example, in Sample Prob. 2.1 we would have

You may want to keep that relation in mind as you formulate the next part of your solution.

2. Draw a parallelogram with the applied forces as two adjacent sides and the resultant as the included diagonal (Fig.2.2).  Alternatively, you can use the triangle rule, with the applied forces drawn in tip-to-tail fashion and the resultant extending from the tail of the first vector to the tip of the second (Fig.2.7).

3. Indicate all dimensions.  Using one of the triangles of the parallelogram, or the triangle constructed according to the triangle rule, indicate all dimensions—whether sides or angles—and determine the unknown dimensions either graphically or by trigonometry. If you use trigonometry, remember that the law of cosines should be applied first if two sides and the included angle are known [Sample Prob. 2.1], and the law of sines should be applied first if one side and all angles are known [Sample Prob. 2.2].

If you have had prior exposure to mechanics, you might be tempted to ignore the solution techniques of this lesson in favor of resolving the forces into rectangular components. While this latter method is important and will be considered in the next section, use of the parallelogram law simplifies the solution of many problems and should be mastered at this time.

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PROBLEMS
  • 2.1

    Two forces are applied at point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

    Fig.P2.1
  • 2.2

    The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.

    Fig.P2.2
  • 2.3

    Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

    Fig.P2.3 and P2.4
  • 2.4

    Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.

  • 2.5

    A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.

    Fig.P2.5
  • Page 26
  • 2.6

    A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant?

    Fig.P2.6 and P2.7
  • 2.7

    A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N.

  • 2.8

    A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

    Fig.P2.8 and P2.9
  • 2.9

    A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.

  • 2.10

    Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.

    Fig.P2.10
  • 2.11

    A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

    Fig.P2.11, P2.12, and P2.13
  • 2.12

    A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

  • 2.13

    A steel tank is to be positioned in an excavation. Determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R.

  • 2.14

    For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R.

  • Page 27
  • 2.15

    Solve Prob. 2.2 by trigonometry.

  • 2.16

    Solve Prob. 2.4 by trigonometry.

  • 2.17

    For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N.

  • 2.18

    For the hook support of Prob. 2.10, knowing that P = 75 N and α = 50°, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support.

  • 2.19

    Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.

    Fig.P2.19 and P2.20
  • 2.20

    Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 60 N and Q = 48 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.

Answers to all problems set in straight type (such as 2.1) are given at the end of the book. Answers to problems with a number set in italic type (such as 2.3) are not given.

Table of Contents
Chapter 11. Kinematics of Particles
Chapter 1. Introduction
Chapter 12. Kinetics of Particles: Newton's Second Law
Chapter 2. Statics of Particles
Chapter 13. Kinetics of Particles: Energy and Momentum Methods
Chapter 3. Rigid Bodies: Equivalent Systems of Forces
Chapter 14. Systems of Particles
Chapter 4. Equilibrium of Rigid Bodies
Chapter 15. Kinematics of Rigid Bodies
Chapter 5. Distributed Forces: Centroids and Centers of Gravity
Chapter 16. Plane Motion of Rigid Bodies: Forces and Accelerations
Chapter 6. Analysis of Structures
Chapter 17. Plane Motion of Rigid Bodies: Energy and Momentum Methods
Chapter 7. Forces in Beams and Cables
Chapter 18. Kinetics of Rigid Bodies in Three Dimensions
Chapter 8. Friction
Chapter 19. Mechanical Vibrations
Chapter 9. Distributed Forces: Moments of Inertia
Front Matter
Chapter 10. Method of Virtual Work
End Matter
Table of Contents
Index
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