Force Resultants, Torques, Scalar Products, Equivalent Force Systems

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Force resultants, Torques, Scalar Products, Equivalent Force systems

• Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force.

• The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs.

• Force is a vector quantity.

• Concurrent forces: set of forces which all pass through the same point.

A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.

• Vector force components: two or more force vectors which, together, have the same effect as a single force vector.

© Tata McGraw-Hill Companies, 2008 Rectangular Components of a Force: Unit Vectors • May resolve a force vector into perpendicular components so that the resulting parallelogram is a ! ! rectangle. are referred to as rectangular Fx and Fy vector components and ! ! ! F = Fx + Fy ! ! • Define perpendicular unit vectors i a n d j which are parallel to the x and y axes.

• Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components. ! ! ! F = Fxi + Fy j ! Fx and Fy are referred to as the scalar components of F

• Resolve each force into rectangular components ! ! ! ! ! ! ! ! R i + R j = P i + P j + Q i + Q j + S i + S j x y x y x! y x y! = (Px + Qx + Sx )i + (Py + Qy + S y )j • The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.

Rx = Px + Qx + Sx Ry = Py + Qy + S y = ∑ Fx = ∑ Fy • To find the resultant magnitude and direction,

2 2 −1 Ry R = Rx + Ry θ = tan Rx © Tata McGraw-Hill Companies, 2008 Sample Problem 2.3 SOLUTION: • Resolve each force into rectangular components.

• Determine the components of the resultant by adding the corresponding force components.

• Calculate the magnitude and direction Four forces act on bolt A as shown. of the resultant. Determine the resultant of the force on the bolt.

© Tata McGraw-Hill Companies, 2008 Sample Problem 2.3 SOLUTION: • Resolve each force into rectangular components. force mag x − comp y − comp ! F 150 +129.9 + 75.0 !1 F 80 − 27.4 + 75.2 !2 F 110 0 −110.0 !3 F4 100 + 96.6 − 25.9 Rx = +199.1 Ry = +14.3 • Determine the components of the resultant by adding the corresponding force components. • Calculate the magnitude and direction.

R = 199.12 +14.32 R = 199.6N 14.3N tanα = α = 4.1° 199.1N © Tata McGraw-Hill Companies, 2008 Vector Product of Two Vectors • Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product. • Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions: 1.Line of action of V is perpendicular to plane containing P and Q. 2.Magnitude of V is, V = PQsinθ 3.Direction of V is obtained from the right- hand rule. • Vector products: 1.are not commutative, Q × P = −(P ×Q) 2.are distributive, P ×(Q1 + Q2 )= P ×Q1 + P ×Q2 3.are not associative, P ×Q × S ≠ P × Q × S ( ) ©( Tata McGraw-Hill) Companies, 2008 Vector Products: Rectangular Components • Vector products of Cartesian unit vectors, ! ! ! ! ! ! ! ! i ×i = 0 j ×i = −k k ×i = j ! ! ! ! ! ! ! ! i × j = k j × j = 0 k × j = −i ! ! ! " ! ! ! ! i × k = − j j × k = i k × k = 0

• Vector products in terms of rectangular coordinates ! ! ! ! ! ! ! V = (Pxi + Py j + Pzk )× (Qxi + Qy j + Qzk ) ! ! = (PyQz − PzQy )i + (PzQx − PxQz )j ! + (PxQy − PyQx )k ! ! ! i j k

= Px Py Pz Qx Qy Qz © Tata McGraw-Hill Companies, 2008 Moment of a Force About a Point • A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application. • The moment of F about O is defined as

MO = r × F R • The moment vector MO is perpendicular to the plane containing O and the force F. RxF?

• Magnitude of MO measures the tendency of the force to cause rotation of the body about an

axis along MO. M O = rF sinθ = Fd The sense of the moment may be determined by the right-hand rule. • Any force F’ that has the same magnitude and direction as F, is equivalent if it also has the same line of action

and therefore, produces the same moment.© Tata McGraw-Hill Companies, 2008 Moment of a Force About a Point • Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure.

• The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane.

• If the force tends to rotate the structure clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive.

• If the force tends to rotate the structure counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative.

© Tata McGraw-Hill Companies, 2008 Varignon’s Theorem • The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same point O. " " " " " " " r × (F1 + F2 +!)= r × F1 + r × F2 +!

• Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F.

The moment of F about B, ! ! ! M B = rA/ B × F ! ! ! rA/ B = rA − rB ! ! ! = (xA − xB )i + (yA − yB )j + (z A − zB )k ! ! ! ! F = Fxi + Fy j + Fz k

! ! ! i j k ! M B = (xA − xB ) (y A − yB ) (z A − zB )

Fx Fy Fz

M O = M Z = xFy − yFz

! ! M B = [(xA − xB )Fy − (yA − yB )Fz ]k

M B = M Z

= (xA − xB )Fy − (yA − yB )Fz

© Tata McGraw-Hill Companies, 2008 Sample Problem 3.1 A 100-N vertical force is applied to the end of a lever which is attached to a shaft at O. Determine: a) moment about O, b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, d) location for a 240-N vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original force.

© Tata McGraw-Hill Companies, 2008 Sample Problem 3.1 a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper.

MO = Fd d = (24cm)cos60° =12 cm

MO = (100 N)(12 cm)

MO =1200 N ⋅ cm

MO = Fd 1200 N ⋅ cm = F(20.8 cm) 1200 N ⋅ cm F = F = 57.7 N 20.8 cm

MO = Fd 1200 N ⋅ cm = F(24 cm) 1200 N ⋅ cm F = F = 50 N 24 cm

MO = Fd 1200 N ⋅ cm = (240 N)d 1200 N ⋅ cm d = = 5 cm 240 N OBcos60° = 5 cm OB =10 cm

© Tata McGraw-Hill Companies, 2008 Sample Problem 3.1 e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 N force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 N force.

SOLUTION:

The moment MA of the force F exerted by the wire is obtained by evaluating the vector product, ! ! ! M A = rC A × F

The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C. © Tata McGraw-Hill Companies, 2008 Sample Problem 3.4 SOLUTION: ! ! ! M A = rC A × F ! ! ! ! ! rC A = rC − rA = (0.3 m)i + (0.08 m)j ! ! ! r F = Fλ = (200 N) C D rC D ! ! ! − (0.3 m)i + (0.24 m)j − (0.32 m)k = (200 N) 0.5 m ! ! ! = −(120 N)i + (96 N)j − (128 N)k ! ! ! i j k ! M A = 0.3 0 0.08 −120 96 −128 " ! ! ! M A = −(7.68 N ⋅ m)i + (28.8 N ⋅ m)j + (28.8 N ⋅ m)k © Tata McGraw-Hill Companies, 2008 Scalar Product of Two Vectors • The scalar product or dot product between two vectors P and Q is defined as ! ! P •Q = PQ cosθ (scalar result) • Scalar products: ! ! ! ! - are commutative, P •Q = Q • P ! ! ! ! ! ! ! - are distributive, P • (Q1 + Q2 )= P •Q1 + P •Q2 ! ! ! - are not associative, (P •Q)• S = undefined • Scalar products with Cartesian unit components, ! ! ! ! ! ! ! ! P •Q = (Pxi + Py j + Pz k )• (Qxi + Qy j + Qz k ) ! ! ! ! ! ! ! ! " ! ! ! i • i = 1 j • j = 1 k • k = 1 i • j = 0 j • k = 0 k • i = 0 ! ! P •Q = PxQx + PyQy + PzQz ! ! 2 2 2 2 P • P = Px + Py + Pz = P © Tata McGraw-Hill Companies, 2008 Scalar Product of Two Vectors: Applications

• Angle between two vectors: ! ! P •Q = PQ cosθ = PxQx + PyQy + PzQz P Q + P Q + P Q cosθ = x x y y z z PQ

• Projection of a vector on a given axis:

POL = P cosθ = projection of P along OL ! ! P •Q = PQ cosθ ! ! P •Q = P cosθ = P Q OL • For an axis defined by a unit vector: ! ! POL = P • λ

= Px cosθ x + Py cosθ y + Pz cosθ z

• The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign, ! ! ! ! ! ! ! ! ! S • (P×Q)= P • (Q× S )= Q • (S × P) ! ! ! ! ! ! ! ! = −S • (Q× P)= −P • (S ×Q)= −Q • (P× S )

• Evaluating the mixed triple product, ! ! ! S • (P×Q)= Sx (PyQz − PzQy )+ S y (PzQx − PxQz ) + Sz (PxQy − PyQx )

Sx S y Sz = Px Py Pz Qx Qy Qz © Tata McGraw-Hill Companies, 2008 Some basic facts in Euclidean geometry

• Two parallel lines form a unique plane • Three non-colinear points form a unique plane • A line and a non-colinear point form a unique plane • Two non-parallel but intersecting lines form unique plane Moment of a Force About a Given Axis

• Moment MO of a force F applied at the point A about a point O, ! ! ! M O = r × F

• Scalar moment MOL about an axis OL is the

projection of the moment vector MO onto the axis, ! ! ! ! ! M OL = λ • M O = λ • (r × F ) • Moments of F about the coordinate axes,

M x = yFz − zFy

M y = zFx − xFz

M z = xFy − yFx

• Moment of a force about an arbitrary axis, ! ! M BL = λ • M B ! ! ! = λ • (rA B × F ) ! ! ! rA B = rA − rB

• The result is independent of the point B along the given axis.

A cube is acted on by a force P as shown. Determine the moment of P a) about A b) about the edge AB and c) about the diagonal AG of the cube. d) Determine the perpendicular distance between AG and FC.

© Tata McGraw-Hill Companies, 2008 Sample Problem 3.5 • Moment of P about A, ! ! ! M A = rF A × P ! ! ! ! ! rF A = ai − a j = a(i − j ) ! ! ! ! ! P = P / 2 (i + j )= P / 2 (i + j ) ! ! ! ! ! M A = a(i − j )× P / 2(i + j ) ! ! ! ! M A = (aP / 2)(i + j + k ) • Moment of P about AB, ! ! M AB = i • M A ! ! ! ! = i • (aP / 2)(i + j + k )

M AB = aP / 2

© Tata McGraw-Hill Companies, 2008 Sample Problem 3.5 • Moment of P about the diagonal AG, ! ! M AG = λ • M A ! ! ! ! ! rA G ai − aj − ak 1 ! ! ! λ = = = (i − j − k ) rA G a 3 3 ! aP ! ! ! M A = (i + j + k ) 2 1 ! ! ! aP ! ! ! M AG = (i − j − k )• (i + j + k ) 3 2 aP = (1−1−1) 6

aP M = − AG 6

© Tata McGraw-Hill Companies, 2008 Sample Problem 3.5 • Perpendicular distance between AG and FC, ! ! P ! ! 1 ! ! ! P P • λ = (j − k )• (i − j − k )= (0 −1+1) 2 3 6 = 0 Therefore, P is perpendicular to AG. aP M = = Pd AG 6 a d = 6

This is straightforward currently since

© Tata McGraw-Hill Companies, 2008 General rule to get distance between lines Moment of a Couple • Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple.

• Moment of the couple, ! ! ! ! ! M = rA × F + rB × (− F ) ! ! ! = (rA − rB )× F ! ! = r × F M = rF sinθ = Fd

• The moment vector of the couple is independent of the choice of the origin of the coordinate axes, i.e., it is a free vector that can be applied at any point with the same effect.

Two couples will have equal moments if

• F1d1 = F2d2 • the two couples lie in parallel planes, and

• the two couples have the same sense or the tendency to cause rotation in the same direction.

• Consider two intersecting planes P1 and P2 with each containing a couple ! ! ! M1 = r × F1 in plane P1 ! ! ! M 2 = r × F2 in plane P2

• Resultants of the vectors also form a couple ! ! ! ! ! ! M = r × R = r × (F1 + F2 )

• By Varignon’s theorem ! ! ! ! ! M = r × F1 + r × F2 ! ! = M1 + M 2

• Sum of two couples is also a couple that is equal to the vector sum of the two couples © Tata McGraw-Hill Companies, 2008 Couples Can Be Represented by Vectors

• A couple can be represented by a vector with magnitude and direction equal to the moment of the couple.

• Couple vectors obey the law of addition of vectors.

• Couple vectors are free vectors, i.e., the point of application is not significant.

• Couple vectors may be resolved into component vectors.

• Force vector F can not be simply moved to O without modifying its action on the body.

• Attaching equal and opposite force vectors at O produces no net effect on the body.

• The three forces may be replaced by an equivalent force vector and couple vector, i.e, a force-couple system.

• Moving F from A to a different point O’ requires the

addition of a different couple vector MO’ ! ! ! M O' = r ʹ× F • The moments of F about O and O’ are related, ! ! ! ! ! ! ! ! ! ! M O' = r '×F = (r + s )× F = r × F + s × F ! ! ! = M O + s × F • Moving the force-couple system from O to O’ requires the addition of the moment of the force at O about O’. © Tata McGraw-Hill Companies, 2008 System of Forces: Reduction to a Force and Couple

• A system of forces may be replaced by a collection of force-couple systems acting a given point O • The force and couple vectors may be combined into a resultant force vector and a resultant couple vector, ! ! ! R ! ! R = ∑ F M O = ∑(r × F ) • The force-couple system at O may be moved to O’ with the addition of the moment of R about O’ , ! R ! R ! ! M O' = M O + s × R • Two systems of forces are equivalent if they can be reduced to the same force-couple system. © Tata McGraw-Hill Companies, 2008 Further Reduction of a System of Forces • If the resultant force and couple at O are mutually perpendicular, they can be replaced by a single force acting along a new line of action.

• The resultant force-couple system for a system of forces will be mutually perpendicular if: 1) the forces are concurrent, 2) the forces are coplanar, or 3) the forces are parallel.

• System of coplanar forces is reduced to a ! ! R force-couple system R a n d M O that is mutually perpendicular.

• System can be reduced to a single force ! by moving the line of action of until !R its moment about O becomes R M O • In terms of rectangular coordinates, R xRy − yRx = M O

Simplest Resultant