Rigid Bodies: Equivalent Systems of Forces Introduction Treatment
Rigid Bodies: Equivalent Systems of Forces Introduction Treatment of a body as a single particle is not always possible. In general, the size of the body and the specific points of application of the forces must be considered. Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the conditions of equilibrium or motion of the body. Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with a simpler equivalent system. moment of a force about a point moment of a force about an axis moment due to a couple Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple. External and Internal Forces Forces acting on rigid bodies are divided into two groups: - External forces - Internal forces External forces are shown in a free-body diagram.
If unopposed, each external force can impart a motion of translation or rotation, or both. Principle of Transmissibility: Equivalent Forces Principle of Transmissibility Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F are equivalent forces. Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck. Principle of transmissibility may not always apply in determining internal forces and deformations. 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 M O r F The moment vector MO is perpendicular to the plane containing O and the force F. Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along
M O rF sin Fd MO. The sense of the moment may be determined by the rule. right-hand 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. 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. Varignons 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 Varignon'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. Rectangular Components of the Moment of a Force The moment of F about O, M O r F , r xi yj zk F Fx i Fy j Fz k M O M x i M y j M z k i x Fx
j y Fy k z Fz yFz zF y i zFx xFz j xFy yFx k The moment of F about B, M B rA / B F
rA / B rA rB x A x B i y A y B j z A z B k F Fx i Fy j Fz k i M B x A xB Fx j y A yB Fy k z A zB Fz
Rectangular Components of the Moment of a Force For two-dimensional structures, M O xFy yFx k M O M Z xFy yFx M B x A xB Fy y A y B Fx k M B M Z x A xB Fy y A y B Fx Sample Problem 1 A 100-lb 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-lb vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original force. 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. M O Fd d 24 in. cos 60 12 in. M O 100 lb 12 in. M 1200 lb in O b) Horizontal force at A that produces the same moment, d 24 in. sin 60 20.8 in. M O Fd 1200 lb in. F 20.8 in. 1200 lb in. F 20.8 in. F 57.7 lb
c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA. M O Fd 1200 lb in. F 24 in. 1200 lb in. F 24 in. F 50 lb d) To determine the point of application of a 240 lb force to produce the same moment, M O Fd 1200 lb in. 240 lb d 1200 lb in. d 5 in. 240 lb OB 10 in. OB cos60 5 in. e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb 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 lb force. 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 M O 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 M B rA B F rA B rA rB The result is independent of the point B along the given axis. Sample Problem 2
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. Moment of P about A, M A rF A P rF A a i a j a i j P P 1 / 2 j 1 / 2 k P / 2 j k M A a i j P / 2 j k
MA 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 3 - 18 Moment of P about the diagonal AG, M AG M A
rA G ai aj ak 1 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 M AG aP 6 3 - 19
Perpendicular distance between AG and FC, P 1 P P j k i j k 0 1 1 2 3 6 0 Therefore, P is perpendicular to AG. M AG aP Pd 6 d a 6 3 - 20 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 sin 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. 3 - 21 Moment of a Couple Two couples will have equal moments if F1d1 F2 d 2 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. 3 - 22 Addition of Couples
Consider two intersecting planes P1 and P2 with each containing a couple M 1 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 Varigons theorem M r F1 r F2 M 1 M 2 Sum of two couples is also a couple that is equal to the vector sum of the two couples 3 - 23 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. 3 - 24 Resolution of a Force Into a Force at O and a Couple 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. 3 - 25 Moving F from A to a different point O requires the addition of a different couple vector MO M O ' r 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. 3 - 26 Sample Problem 3 Attach equal and opposite 20 lb forces in the +x direction at A, thereby producing 3 couples for which the moment components are easily computed. Alternatively, compute the sum of the moments of the four forces about an arbitrary single point. The point D is a good choice as only two of the forces will produce non-zero moment contributions.. Determine the components of the single couple equivalent to the couples shown. 3 - 27 SOLUTION: Attach equal and opposite 20 lb forces in the +x direction at A The three couples may be represented by three couple vectors,
M x 30 lb18 in. 540 lb in. M y 20 lb12 in. 240lb in. M z 20 lb 9 in. 180 lb in. M 540 lb in. i 240lb in. j 180 lb in. k 3 - 28 Alternatively, compute the sum of the moments of the four forces about D. Only the forces at C and E contribute to the moment about D. M M D 18 in. j 30 lb k 9 in. j 12 in. k 20 lb i
M 540 lb in. i 240lb in. j 180 lb in. k 3 - 29 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. 3 - 30 System of coplanar forces isreduced to a force-couple system R and M OR that is mutually perpendicular. System can be reduced to a single
force by moving the line of action of R until its moment about O becomes M OR In terms of rectangular coordinates, xR y yRx M OR 3 - 31 Sample Problem 4 a) Compute the resultant force for the forces shown and the resultant couple for the moments of the forces about A. For the beam, reduce the system of forces shown to (a) an equivalent force-couple system at A, (b) an equivalent force couple system at B, and (c) a single force or resultant. Note: Since the support reactions are not included, the given system will not maintain the beam in equilibrium. b) Find an equivalent force-couple system at B based on the forcecouple system at A. c) Determine the point of application for the resultant force such that its moment about A is equal to the resultant couple at A.
3 - 32 SOLUTION: a) Compute the resultant force and the resultant couple at A. R F 150 N j 600 N j 100 N j 250 N j R 600 N j R M A r F 1.6 i 600 j 2.8 i 100 j 4.8 i 250 j R
M A 1880 N m k 3 - 33 b) Find an equivalent force-couple system at B based on the force-couple system at A. The force is unchanged by the movement of the force-couple system from A to B. R 600 N j The couple at B is equal to the moment about B of the force-couple system found at A. R R M B M A rB A R 1880 N m k 4.8 m i 600 N j 1880 N m k 2880 N m k R M B 1000 N m k 3 - 34
Sample Problem 5 Determine the relative position vectors with respect to A. rB A 0.075 i 0.050k m rC A 0.075 i 0.050k m rD A 0.100 i 0.100 j m Resolve the forces into rectangular components. FB 700 N rE B 75 i 150 j 50k
rE B 175 0.429 i 0.857 j 0.289k FB 300 i 600 j 200k N FC 1000 N cos 45 i cos 45 j 707 i 707 j N FD 1200 N cos 60 i cos 30 j 600 i 1039 j N 3 - 35 Compute the equivalent force,
R F 300 707 600 i 600 1039 j 200 707 k R 1607i 439 j 507k N Compute the equivalent couple, R M A r F i j k rB A F B 0.075
0 0.050 30i 45k 300 i rC A F c 0.075 707 i rD A F D 0.100 600 600 200 j k 0 0.050 17.68 j 0 707 j k
0.100 0 163.9k 1039 0 R M A 30 i 17.68 j 118 .9k 3 - 36 Problem 6 y C 10 ft The 15-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 570 lb, determine the moment about A of the force exerted by the cable at B. 6 ft A z
15 ft B x 37 y C 10 ft Problem 7 6 ft Solving Problems on Your Own A z 15 ft B x The 15-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the
vertical wall. If the tension in the cable is 570 lb, determine the moment about A of the force exerted by the cable at B. 1. Determine the rectangular components of a force defined by its magnitude and direction. If the direction of the force is defined by two points located on its line of action, the force can be expressed by: F F = F = (dx i + dy j + dz k) d 38 y C 10 ft Problem 7 6 ft Solving Problems on Your Own A
z 15 ft B x The 15-ft boom AB has a fixed end A. A steel cable is stretched from the free end B of the boom to a point C located on the vertical wall. If the tension in the cable is 570 lb, determine the moment about A of the force exerted by the cable at B. 2. Compute the moment of a force in three dimensions. If r is a position vector and F is the force the moment M is given by: M=rxF 39 y C 10 ft A Problem 7 Solution
6 ft 570 N Determine the rectangular components of a force defined by its magnitude and direction. First note: z 15 ft B x dBC = (_15)2 + (6) 2 + (_10) 2 dBC = 19 ft Then: TBC = 570 lb(_15 i + 6 j _ 10 k) = _ (450 lb) i + (180 lb) j _ (300 lb)k 19 40 y
Problem 7 Solution C 10 ft 6 ft A Compute the moment of a force in three dimensions. 570 N z 15 ft Have: MA = rB/A x TBC B x Where: rB/A = (15 ft) i
Then: MA = 15 i x (_ 450 i + 180 j _ 300 k) MA = (4500 lb.ft) j + (2700 lb.ft) k 41 Problem 8 y 0.35 m 0.875 m G H O 0.925 m 0.75 m A z 0.5 B
m 0.5 D x 0.75 m C m The frame ACD is hinged at A and D and is supported by a cable which passes through a ring at B and is attached to hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment about the diagonal AD of the force exerted on the frame by portion BH of the cable. P 42 y 0.35 m
0.875 m G O 0.925 m A z 0.5 B m 0.5 m C The frame ACD is hinged at A H and D and is supported by a cable which passes through a ring at B and is attached to 0.75 m hooks at G and H. Knowing that the tension in the cable is 450 N, determine the moment
D x about the diagonal AD of the 0.75 m force exerted on the frame by portion BH of the cable. P 1. Determine the moment MOL of a force about a given axis OL. MOL is defined as MOL = . MO = . ( r x F ) where is the unit vector along OL and r is a position vector from 43 any point on the line OL to any point on the line of action of F. y 0.35 m Determine the moment MAD of a force about line AD. 0.875 m G H O
0.925 m TBH A AD z B rB/A 0.5 m 0.5 MAD 0.75 m D x 0.75 m C m MAD = AD . ( rB/A x TBH ) Where
1 AD = 5 (4 i _ 3 k) rB/A = (0.5 m ) i P dBH = ( 0.375 )2 + ( 0.75 )2 + ( _ 0.75 )2 = 1.125 m 450 N TBH = ( 0.375 i + 0.75 j _ 0.75 k ) 1.125 = ( 150 N) i + ( 300 N) j _ ( 300 N) k 44 y 0.35 m 0.875 m G
Finally: H MAD = AD . ( rB/A x TBH ) O 0.925 m TBH A AD z rB/A 0.5 m B 0.5 MAD 0.75 m D x 0.75 m
C m P MAD = 1 = 5 4 0 _3 1 5 0.5 0 _ 0 150 300 300 [(_3)(0.5)(300)] MAD = _ 90 N.m 45 Problem 9 C A B
300 mm P 200 mm The force P has a magnitude of 250 N and is applied at the end C of a 500 mm rod AC attached to a bracket at A and B. Assuming = 30o and = 60o, replace P with (a) an equivalent force-couple system at B , (b) an equivalent system formed by two parallel forces applied at A and B. 46 Solving Problems on Your Own C A B
300 mm P 200 mm The force P has a magnitude of 250 N and is applied at the end C of a 500 mm rod AC attached to a bracket at A and B. Assuming = 30o and = 60o, replace P with (a) an equivalent force-couple system at B , (b) an equivalent system formed by two parallel forces applied at A and B. 1. Replace a force with an equivalent force-couple system at a specified point. The force of the force-couple system is equal to the original force, while the required couple vector is equal to the moment of the original force about the given point. 47 C A Problem 9 Solution B
300 mm P 200 mm Replace a force with an equivalent force-couple system at a specified point. (a) Equivalence requires: F: F=P or F = 250 N 60o MB : M = _ (0.3 m)(250 N) = _75 N The equivalent force couple system at B is: F = 250 N
60o, M = 75 N . m 48 C A B 300 mm P 200 mm (b) Require: The two force systems are equivalent. C 60o B A y
30 o 250 N C x B = A FB FA 49 C y 60o B
o A 0 = FA cos + FB cos FA = _ F B Fy : If FB FA Equivalence then requires: Fx : x B = 250 N 30
A C _ or cos = 0 250 = _ FA sin _ FB sin FA = _ F B Consequently then _ 250 = 0 cos = 0 or reject 50 = 90o and FA + FB = 250
C y 60o B o 30 A x B = 250 N C A FB
FA Also: + MB : _ (0.3 m)( 250 N) = (0.2 m) FA or and FA = 375 N FA = _ 375 N FB = + 675 N 60o, FB = 625 N 60o 51 Problem 10 10 lb 12 in M A
B C 30 lb A couple of magnitude 60o M = 54 lb.in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. 8 in 45 lb 52 Solving Problems on Your Own 10 lb 12 in M
A B C 30 lb A couple of magnitude 60o M = 54 lb.in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. 8 in 45 lb 1. Determine the resultant of two or more forces. Determine the rectangular components of each force. Adding these components will yield the components of the resultant. 53
10 lb 12 in 30 lb 60o Solving Problems on Your Own A couple of magnitude M = 54 lb.in. and the three forces shown are applied to A B an angle bracket. (a) Find the M 8 in resultant of this system of forces. (b) Locate the points C 45 lb where the line of action of the resultant intersects line AB and line BC. 2. Reduce a force system to a force and a couple at a given point. The force is the resultant R of the system obtained by adding the various forces. The couple is the moment resultant of the system M, obtained by adding the moments about the point of the various forces. R=F
M=(rxF) 54 10 lb 12 in 30 lb 60o Solving Problems on Your Own A couple of magnitude M = 54 lb.in. and the three forces shown are applied to A B an angle bracket. (a) Find the M 8 in resultant of this system of forces. (b) Locate the points C 45 lb where the line of action of the resultant intersects line AB and line BC. 3. Reduce a force and a couple at a given point to a single force.
The single force is obtained by moving the force until its moment about the point (A) is equal to the couple vector MAR. A position vector r from the point, to any point on the line of action of the single force R must satisfy the equation r x R = MAR 55 10 lb 30 lb 12 in M A 60o B C (a) Problem 10 Solution 8 in
Determine the resultant of two or more forces. 45 lb Adding the components of the forces: F : R = ( _ 10 j ) + (_ 45 i ) + ( 30 cos 60o i + 30 sin 60o j ) = _ ( 30 lb ) i + ( 15.98 lb ) j R = 34.0 lb 28.0o 56 10 lb 30 lb 12 in M A 60o B C 8 in
Reduce a force system to a force and a couple at a given point. 45 lb (b) First reduce the given forces and couple to an equivalent force-couple system (R, MB) at B. + M : M = ( 54 lb.in ) + ( 12 in )(10 lb) _ ( 8 in ) ( 45 lb) B B = _ 186 lb.in 57 a A D B R E C
c Reduce a force and a couple at a given point to a single force. With R at D: + M : B _ 186 lb.in = _ a ( 15.9808 lb) or a = 11.64 in and with R at E + MB : _ 186 lb.in = c ( 30 lb) or c = 6.20 in
The line of action of R intersects line AB 11.64 in. to the left of B and intersects line BC 6.20 in below B. 58
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