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**Equilibrium of Rigid Bodies**

CE 102 Statics Chapter 4 Equilibrium of Rigid Bodies

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**Contents Introduction Free-Body Diagram**

Reactions at Supports and Connections for a Two-Dimensional Structure Equilibrium of a Rigid Body in Two Dimensions Statically Indeterminate Reactions Sample Problem 4.1 Sample Problem 4.2 Sample Problem 4.3 Equilibrium of a Two-Force Body Equilibrium of a Three-Force Body Sample Problem 4.4 Equilibrium of a Rigid Body in Three Dimensions Reactions at Supports and Connections for a Three-Dimensional Structure Sample Problem 4.5

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Introduction For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body. The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces form a system equivalent to zero, Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium,

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Free-Body Diagram First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram. Select the extent of the free-body and detach it from the ground and all other bodies. Indicate point of application, magnitude, and direction of external forces, including the rigid body weight. Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body. Include the dimensions necessary to compute the moments of the forces.

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**Reactions at Supports and Connections for a Two-Dimensional Structure**

Reactions equivalent to a force with known line of action.

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**Reactions at Supports and Connections for a Two-Dimensional Structure**

Reactions equivalent to a force of unknown direction and magnitude. Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude

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**Equilibrium of a Rigid Body in Two Dimensions**

For all forces and moments acting on a two-dimensional structure, Equations of equilibrium become where A is any point in the plane of the structure. The 3 equations can be solved for no more than 3 unknowns. The 3 equations can not be augmented with additional equations, but they can be replaced

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**Statically Indeterminate Reactions**

More unknowns than equations Fewer unknowns than equations, partially constrained Equal number unknowns and equations but improperly constrained

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**Sample Problem 4.1 SOLUTION: Create a free-body diagram for the crane.**

Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A. Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components. A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero.

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Sample Problem 4.1 Determine B by solving the equation for the sum of the moments of all forces about A. Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces. Create the free-body diagram. Check the values obtained.

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**Sample Problem 4.2 SOLUTION:**

Create a free-body diagram for the car with the coordinate system aligned with the track. Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle. Determine the cable tension by solving the equation for the sum of force components parallel to the track. A loading car is at rest on an inclined track. The gross weight of the car and its load is 5500 lb, and it is applied at at G. The cart is held in position by the cable. Determine the tension in the cable and the reaction at each pair of wheels. Check the values obtained by verifying that the sum of force components perpendicular to the track are zero.

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**Sample Problem 4.2 Determine the reactions at the wheels.**

Create a free-body diagram Determine the cable tension.

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**Sample Problem 4.3 SOLUTION:**

Create a free-body diagram for the frame and cable. Solve 3 equilibrium equations for the reaction force components and couple at E. The frame supports part of the roof of a small building. The tension in the cable is 150 kN. Determine the reaction at the fixed end E.

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Sample Problem 4.3 Solve 3 equilibrium equations for the reaction force components and couple. Create a free-body diagram for the frame and cable.

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**Equilibrium of a Two-Force Body**

Consider a plate subjected to two forces F1 and F2 For static equilibrium, the sum of moments about A must be zero. The moment of F2 must be zero. It follows that the line of action of F2 must pass through A. Similarly, the line of action of F1 must pass through B for the sum of moments about B to be zero. Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense.

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**Equilibrium of a Three-Force Body**

Consider a rigid body subjected to forces acting at only 3 points. Assuming that their lines of action intersect, the moment of F1 and F2 about the point of intersection represented by D is zero. Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about any axis must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D. The lines of action of the three forces must be concurrent or parallel.

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**Sample Problem 4.4 SOLUTION:**

Create a free-body diagram of the joist. Note that the joist is a 3 force body acted upon by the rope, its weight, and the reaction at A. The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R. A man raises a 10 kg joist, of length 4 m, by pulling on a rope. Find the tension in the rope and the reaction at A. Utilize a force triangle to determine the magnitude of the reaction force R.

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**Sample Problem 4.4 ( ) Create a free-body diagram of the joist.**

Determine the direction of the reaction force R. ( ) 636 . 1 414 313 2 tan m 2.313 515 828 20 25 45 cot( cos 4 = - + AE CE BD BF CD AF AB a

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Sample Problem 4.4 Determine the magnitude of the reaction force R.

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**Equilibrium of a Rigid Body in Three Dimensions**

Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three dimensional case. These equations can be solved for no more than 6 unknowns which generally represent reactions at supports or connections. The scalar equations are conveniently obtained by applying the vector forms of the conditions for equilibrium,

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**Reactions at Supports and Connections for a Three-Dimensional Structure**

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**Reactions at Supports and Connections for a Three-Dimensional Structure**

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**Sample Problem 4.5 SOLUTION: Create a free-body diagram for the sign.**

Apply the conditions for static equilibrium to develop equations for the unknown reactions. A sign of uniform density weighs 270 lb and is supported by a ball-and-socket joint at A and by two cables. Determine the tension in each cable and the reaction at A.

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**Sample Problem 4.5 Create a free-body diagram for the sign.**

Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading.

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Sample Problem 4.5 Apply the conditions for static equilibrium to develop equations for the unknown reactions. Solve the 5 equations for the 5 unknowns,

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**Problem 4.6 P The semicircular rod ABCD is a**

maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that a = 45o, determine the reactions at B , C , and D.

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**Solving Problems on Your Own**

The semicircular rod ABCD is maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that a = 45o, determine the reactions at B , C , and D. Problem 4.6 P a A O D 45o 45o B C 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

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**Solving Problems on Your Own**

The semicircular rod ABCD is maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that a = 45o, determine the reactions at B , C , and D. Problem 4.6 P 45o O D C A B a 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = SFy = SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis or SMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line.

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**Draw a free-body diagram of the body.**

Problem 4.6 Solution P 45o O D C A B a Draw a free-body diagram of the body. 45o O D C A B P sina P cosa R C/ 2 B/ 2 P

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**+ S Fx = 0: P cosa + B/ 2 _ C / 2 = 0 (2)**

Problem 4.6 Solution 45o O D C A B P sina P cosa R C/ 2 B/ 2 P Write three equilibrium equations and solve for the unknowns. + S MO = 0: (P sina) R _ D (R) = D = P sina (1) + S Fx = 0: P cosa + B/ 2 _ C / 2 = (2) + S Fy = 0: _P sina + B/ C / 2 _ P sina = 0 _2P sina + B/ C / 2 = (3)

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**(2) + (3) P(cosa _ 2sina) + 2 B/ 2 = 0 B = (2sina _ cosa) P (4) **

D C A B P sina P cosa R C/ 2 B/ 2 Problem 4.6 Solution P (2) + (3) P(cosa _ 2sina) + 2 B/ 2 = 0 B = (2sina _ cosa) P (4) (2) _ (3) P(cosa + 2sina) _ 2 C/ 2 = 0 C = (2sina + cosa) P (5) 2 2

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45o O D C A B P sina P cosa R C/ 2 B/ 2 Problem 4.6 Solution P For a = 45o sina = cosa = 1/ 2 2 2 2 1 1 2 1 2 EQ. (4) : B = ( _ ) P = P ; B = P o EQ. (5) : C = ( _ ) P = P ; C = P o EQ. (1) : D = P/ D = P/ 2 2 2 2 1 3 2 3 2

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**Problem 4.7 4 in 4 in 20 lb 40 lb The T-shaped bracket shown is**

supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. A B 2 in C D 3 in 3 in E q

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4 in 4 in Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. Problem 4.7 20 lb 40 lb A B 2 in C D 3 in 3 in E q 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

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4 in 4 in Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. Problem 4.7 20 lb 40 lb A B 2 in C D 3 in 3 in E q 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = SFy = SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis or SMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line.

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**Draw a free-body diagram of the body. 2 in C D 3 in 4 in 4 in 3 in E **

Problem 4.7 Solution 20 lb 40 lb A B Draw a free-body diagram of the body. 2 in C D 3 in 4 in 4 in 3 in E 20 lb 40 lb q A B C 2 in C D 3 in D 3 in E E 30o

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**+ S Fy = 0: E cos 30o _ 20 _ 40 = 0 E = = 69.28 lb 60 lb**

4 in 4 in Problem 4.7 Solution 20 lb 40 lb A B C 2 in Write equilibrium equations and solve for the unknowns. C D 3 in D 3 in E E 30o + S Fy = 0: E cos 30o _ 20 _ 40 = 0 E = = lb E = 69.3 lb o 60 lb cos 30o

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**( 20 lb)( 4 in) _ ( 40 lb)( 4 in) _ C ( 3 in) + E sin 30o ( 3 in) = 0 **

Problem 4.7 Solution 20 lb 40 lb A B C 2 in C D 3 in D 3 in E E + S MD = 0: ( 20 lb)( 4 in) _ ( 40 lb)( 4 in) _ C ( 3 in) + E sin 30o ( 3 in) = 0 _ 80 _ 3C ( 0.5 )( 3 ) = 0 C = lb C = 7.97 lb 30o

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4 in 4 in Problem 4.7 Solution 20 lb 40 lb A B C 2 in C + S Fx= 0: E sin 30o + C _ D = 0 ( lb )( 0.5 ) lb _ D = 0 D = 42.6 lb D 3 in D 3 in E E 30o

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**Problem 4.8 y A 3-m pole is supported by a E**

z x 5 kN A B C E D f 1 m 2 m A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle f=30o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A.

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**Solving Problems on Your Own**

z x 5 kN A B C E D f 1 m 2 m Solving Problems on Your Own A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle f=30o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

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**Solving Problems on Your Own A 3-m pole is supported by a ball- **

z x 5 kN A B C E D f 1 m 2 m Problem 4.8 Solving Problems on Your Own A 3-m pole is supported by a ball- and-socket joint at A and by the cables CD and CE. Knowing that the line of actionof the 5-kN force forms an angle f=30o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A. 2. Write equilibrium equations and solve for the unknowns. For three-dimensional body the six scalar equations SFx = SFy = SFz = 0 SMx = SMy = 0 SMz = 0 should be used and solved for six unknowns. These equations can also be written as SF = SMO = S (r x F ) = 0 where F are the forces and r are position vectors.

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**Draw a free-body diagram of the body.**

Problem 4.8 Solution 1.2 m y 1.5 m z x 5 kN A B C E D f 1 m 2 m Draw a free-body diagram of the body. 1.2 m y 1.5 m z x A B C E D 1 m 2 m 30o 5 kN TCE TCD Ay j Ax i Az k

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**rB/A = 2 i rC/A = 3 i CD = _ 3 i+ 1.5 j + 1.2 k CD = 3.562 m**

y 1.5 m z x A B C E D 1 m 2 m 30o 5 kN TCE TCD Ay j Ax i Az k Problem 4.8 Solution Write equilibrium equations and solve for the unknowns. 5 unknowns and 6 equations of equilibrium, but equilibrium is maintained, S MAC = 0 . rB/A = 2 i rC/A = 3 i Load at B, FB = _ ( 5 cos 30o ) j + ( 5 sin 30o ) k = _ 4.33 j k CD = _ 3 i+ 1.5 j k CD = m TCD = TCD = (_ 3 i j k) TCE = TCE = (_ 3 i j _ 1.2 k) CD TCD 3.562 CE TCD 3.562

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**SMA = 0: rC/A x TCD + rC/A x TCE + rB/A x FB = 0**

y 1.5 m z x A B C E D 1 m 2 m 30o 5 kN TCE TCD Ay j Ax i Az k Problem 4.8 Solution SMA = 0: rC/A x TCD + rC/A x TCE + rB/A x FB = 0 i j k _ i j k _3 1.5 _1.2 i j k 0 _ TCD 3.562 TCE 3.562 + + = 0

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**Eq. (1): _3.6TCD + 3.6 (5.902) _ 17.81 = 0 TCD = 0.954 kN**

1.2 m y 1.5 m z x A B C E D 1 m 2 m 30o 5 kN TCE TCD Ay j Ax i Az k Problem 4.8 Solution Equate coefficients of unit vectors to zero. j: _ _ 5 = 0 _3.6 TCD+3.6 TCE_17.81 = 0 (1) TCD 3.562 TCE k: _ 8.66 = 0 4.5 TCD+4.5 TCE = (2) TCD 3.562 TCE (2) (1): TCE _ = 0 ; TCE = 5.90 kN Eq. (1): _3.6TCD (5.902) _ = TCD = kN

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**SF = 0: A + TCD + TCE + FB = 0 i: Ax + (_3) + (_3) = 0 Ax = 5.77 kN**

1.2 m y 1.5 m z x A B C E D 1 m 2 m 30o 5 kN TCE TCD Ay j Ax i Az k Problem 4.8 Solution SF = 0: A + TCD + TCE + FB = 0 i: Ax (_3) (_3) = 0 Ax = 5.77 kN 3.562 0.954 5.902 j: Ay (1.5) (1.5) _ 4.33 = Ay = kN 3.562 0.954 5.902 3.562 0.954 3.562 5.902 k: Az (1.2) (_1.2) = Az = _ kN A = ( 5.77 kN) i + ( kN ) j - ( kN ) k

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**Problem 4.9 Rod AC is supported by a pin and**

bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B. A B C a 20 in 10 in 60 lb

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Problem 4.9 A B C a 20 in 10 in 60 lb Solving Problems on Your Own Rod AC is supported by a pin and bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

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Problem 4.9 A B C a 20 in 10 in 60 lb Solving Problems on Your Own Rod AC is supported by a pin and bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B. 2. For a three-force body, solution can be obtained by constructing a force triangle. The resultants of the three forces must be concurrent or parallel. To solve a problem involving a three-force body with concurrent forces, draw the free-body diagram showing that the three forces pass through the same point. Complete the solution by using a force triangle.

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**tan a = a = 26.57o Problem 4.9 Solution A (a) a = 8 in a**

C a 20 in 10 in 60 lb (a) a = 8 in Draw a free-body diagram of the body. C 10 in 60 lb A g a 8 in 12 in B 2 1 G D F E tan a = a = 26.57o 1 2

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**AE = EB = (8) = 4 in. EF = BG = 10 _ 4 = 6 in DG = BG = (6) = 3 in.**

C 10 in 60 lb A g a 8 in 12 in B 2 1 G D F E Problem 4.9 Solution Construct a force triangle. 3 - FORCE BODY Reaction at A passes through D where B and 60-lb load intersect AE = EB = (8) = 4 in. EF = BG = 10 _ 4 = 6 in DG = BG = (6) = 3 in. FD = FG _ DG = 8 _ 3 = 5 in. Tan g = = ; g = 26.57o 1 2 FD AF 5 10

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10 in Problem 4.9 Solution A FORCE TRIANGLE g E F A a = 26.57o 8 in D 30 lb A G 60 lb a B B 12 in B 2 30 lb a = 26.57o 1 C 60 lb 10 in A = B = = lb A = 67.1 lb o B = 67.1 lb o sin 26.57o 30 lb

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**a Problem 4.9 Solution A (b) For A horizontal a**

C a 20 in 10 in 60 lb (b) For A horizontal Draw a free-body diagram of the body. 10 in A F a A a a G a B B 2 a = 26.57o 1 C 60 lb

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**a D ABF : BF = AF cos a D BFG : FG = BF sin a a = FG = AF cos a sin a**

Problem 4.9 Solution A Construct a force triangle. F a A a a G a B D ABF : BF = AF cos a D BFG : FG = BF sin a a = FG = AF cos a sin a a = (10 in.) cos 26.57o sin 26.57o a = 4.00 in. B 2 a = 26.57o 1 C 60 lb

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10 in Problem 4.9 Solution A FORCE TRIANGLE F a A a a G a B 60 lb A B a = 26.57o B 2 a = 26.57o 1 C 60 lb tan a 60 lb A = = 120 lb A = lb B = = lb B = lb o sin a 60 lb

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**Problem 4.10 Rod AD supports a vertical load**

P and is attached to collars B and C, which may slide freely on the rods shown. Knowing that the wire attached at D forms an angle a = 30o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C. D C B A a 30o P

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**Solving Problems on Your Own**

Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing that the wire attached at D forms an angle a = 30o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C. Problem 4.10 D C B A a 30o P 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it.

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**Solving Problems on Your Own**

D C B A a 30o P Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing that the wire attached at D forms an angle a = 30o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C. 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = SFy = SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis or SMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line.

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**Draw a free-body diagram of the body.**

Problem Solution D C B A a 30o P Draw a free-body diagram of the body. D C B A 30o P a T

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**+ S MB = 0: P a _ (C sin 30o) a + T cos 30o (2a) = 0 **

Problem Solution D C B A 30o P a T Write equilibrium equations and solve for the unknowns. 30o S F = 0: _ P cos 30o + T cos 60o = 0 T = P = P T = P cos 30o cos 60o 3 / 2 1 / 2 + S MB = 0: P a _ (C sin 30o) a + T cos 30o (2a) = 0 P a _ ( C ) a P ( ) 2a = 0 _ C + (1 + 3) P = 0; C = 8 P C = 8 P o 1 2 3 2 1 2

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**+ S F = 0: _ B cos 30o + C cos 30o _ T sin 30o = 0 **

Problem Solution D C B A 30o P a T + S F = 0: _ B cos 30o + C cos 30o _ T sin 30o = 0 _ B P _ P ( ) = 0; B = 7 P B = 7 P o 3 2 3 2 1 2

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