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**Chapter 3: Force System Resultants**

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**Cross Product The Cross product of two vectors Magnitude: C=AB sin**

Direction: C is perpendicular to both A and B

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**Laws of Operation for Cross Product**

Commutative law is not valid Scalar Multiplication Distributive Law

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**Cartesian Vector Formulation (sec 3.1)**

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**Cross Product of Two Vectors (sec 3.1)**

Let

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Moment Systems (sec 3.2) The moment of a force about an axis (sometimes represented as a point in a body) is the measure of the force’s tendency to rotate the body about the axis (or point). The magnitude of the moment is:

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**Moment Systems of System of Forces (sec 3.2)**

It is customary to assume CCW as the positive direction.

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**Moment of a Force-Vector Formulation (sec 3.3)**

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**Moment of a Force-Vector Formulation (sec 3.3)**

The axis of the moment is perpendicular to the plane that contains both F and r The axis passes through point O

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**Moment of Force Systems-Vector Formulation (sec 3.3)**

Let a system of forces act upon a body. We like to compute the net moment of all the forces about the point O.

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**Problem 3-10 (page 84, Section 3.1-3.3)**

3.10 Determine the resultant moment about point B on the three forces acting on the beam. Solution:

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**Problem 3-20 (page 86, Section 3.1-3.3)**

3.20 The cable exerts a 140-N force on the telephone pole as shown. Determine the moment of this force at the base A of the pole. Solve the problem two ways, i.e., by using a position vector from A to C, then A to B. Solution:

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**3. 20 The cable exerts a 140-N force on the telephone pole as shown**

3.20 The cable exerts a 140-N force on the telephone pole as shown. Determine the moment of this force at the base A of the pole. Solve the problem two ways, i.e., by using a position vector from A to C, then A to B. Solution-Con’t

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**3. 20 The cable exerts a 140-N force on the telephone pole as shown**

3.20 The cable exerts a 140-N force on the telephone pole as shown. Determine the moment of this force at the base A of the pole. Solve the problem two ways, i.e., by using a position vector from A to C, then A to B. Solution-Con’t

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**Section 3.1-3.3 (In-class Exercise)**

Solution: Net moment should be zero Problem 4-15, page 130 of Statics book. Picture pasted from 04-PPT-06 (25 October 02)

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**Section 3.1-3.3 (In-class Challenge Exercise)**

Solution: This is Problem 4-21 of Statics book, page no Figure is taken from 04PPT-06 (26 October 2002)

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**3.4 Principle of Moments (sec 3.4)**

The moment of a force is equal to the sum of the moment of the force’s component about a point. (Varginon’s theorem )

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**3.5 Moment of a Force about a specified axis**

If we need the moment about other axis still through O, we can use either scalar or vector analysis. Here we have F=20 N applied. Though the typical equation gives moment with respect to b-axis, we require it through y-axis.

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**3.5 Moment of a Force about a specified axis-2**

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**3.5 Moment of a Force about a specified axis-3**

The two steps in the previous analysis can be combined with the definition of a scalar triple product. Since dot product is commutative

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3.6 Moment of a Couple-1 A couple is defined as two parallel forces with same magnitude and opposite direction. Net force is zero, but rotates in specified direction.

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3.6 Equivalent Couples -2 Two couples are equivalent if they produce the same moment. The forces should be in the same or parallel planes for two couple to be equivalent. Couple moments are free vectors. They can be added at any point P in the body. On the bottom figure (page 149, statics), to obtain a moment of 12 N.m, we can have 30 N applied at 0.4 m apart, or 40 N at 0.3 m apart. The final vector is a 12 N.m free moment vector with axis along the shaft.

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**Problem 3-39 (page 95, Section 3.4-3.6)**

3.39The bracket is acted upon by a 600-N force at A. Determine the moment of this force about the y axis. Solution:

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**3. 39The bracket is acted upon by a 600-N force at A**

3.39The bracket is acted upon by a 600-N force at A. Determine the moment of this force about the y axis. Solution-Con’t

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**Problem 3-54 (page 103, Section 3.4-3.6)**

3.54 Two couples act on the frame. If d = 6 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 3-14) and (b) summing the moments of all the force components about point A Solution:

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**3. 54 Two couples act on the frame**

3.54 Two couples act on the frame. If d = 6 ft, determine the resultant couple moment. Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq. 3-14) and (b) summing the moments of all the force components about point A Solution-Con’t

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**Section 3.4-3.6 (In-class Exercise)**

Two couples act on the frame. If d = 4ft, find the resultant couple moment by (a) direct method, and (b) resolving the x and y components (take moment about A). Solution: a. Find the normal distance for each case first. Problem 4-83, page 130 of Statics book. Picture pasted from 04-PPT-06 (25 October 02)

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**Section 3.4-3.6 (In-class Challenge Exercise)**

The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles. Solution: This is Problem 4-94 of Statics book, page no Figure is taken from 04PPT-08 (27 October 2002)

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**Section 3.4-3.6 (In-class Challenge Exercise-2)**

Solution—contd. This is Problem 4-94 of Statics book, page no Figure is taken from 04PPT-08 (27 October 2002)

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**3.7 Movement of a Force on a Rigid Body-1**

A single force on a body can cause it to rotate (moment) and translate (force). In the first example, the ruler causes a force F and in addition a moment M=Fd. In the example, the ruler causes a force F and NO ADDITIONAL moment.

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**3.7 Movement of a Force on a Rigid Body-2**

Extend this idea to a general 3-D case. Now, the force can be moved Force now causes the force at any point 0 and then a couple.

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**3.8 Resultant of a Force and Couple System-1**

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**3.9 Further Reduction on Force/Couples-1**

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**3.9 Further Reduction on Force/Couples-2**

Concurrent Force Systems Only equivalent force Coplanar Force Systems A single force at d from point 0

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**3.9 Further Reduction on Force/Couples-2**

Parallel Force Systems

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**Problem 3-103 (page 124, Section 3.7-3.9)**

3.103 The weights of the various components of the truck shown. Replace this system of forces by an equivalent resultant force and couple moment acting at point A. Solution:

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**Problem 3-93 (page 122, Section 3.7-3.9)**

3.93 The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x,y) on the slab. Solution:

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**Problem 3-69 (page 119, Section 3.7-3.9)**

3.69 The gear is subjected to the two forces shown. Replace these forces by an equivalent resultant force and couple moment acting at point O. Solution:

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**3. 69 The gear is subjected to the two forces shown**

3.69 The gear is subjected to the two forces shown. Replace these forces by an equivalent resultant force and couple moment acting at point O. Solution-Con’t

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**Section 3.7-3.9 (In-class Exercise)**

Solution: Problem 4-130, page 178 of Statics book. Picture pasted from 04-PPT-09 (27 October 02)

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**Section 3.7-3.9 (In-class Exercise..2)**

Solution (contd). Problem 4-130, page 178 of Statics book. Picture pasted from 04-PPT-09 (27 October 02)

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**Section 3.7-3.9 (In-class Challenge Exercise)**

The weights of the various components of the truck are shown. Replace this system by an equivalent resultant force and specify its location from point A. Solution: Equivalent force This is Problem 4-21 of Statics book, page no Figure is taken from 04PPT-06 (26 October 2002) Location of force

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**Chapter 3: Force System Resultants…concludes**

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