Engineering Mechanics Statics JL.Meriam Solution PDF

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Notes of Engineering Mechanics Statics by JL Mariam 6th Edition....

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Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587

By Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587

What is Mechanics? Mechanics is the physical science which deals with the effects of forces on objects. The subject of mechanics is logically divided into two parts: statics,which concerns the equilibrium of bodies under action of forces, and dynamics, which concerns the motion of bodies. BASIC CONCEPTS The following concepts and definitions are basic to the study of mechanics, and they should be understood at the outset. Space is the geometric region occupied by bodies whose positions are described by linear and angular measurements relative to a coordinate system. For three-dimensional problems, three independent coordinates are needed. For two-dimensional problems, only two coordinates are required. Time is the measure of the succession of events and is a basic quantity in dynamics. Time is not directly involved in the analysis of statics problems. Mass is a measure of the inertia of a body, which is its resistance to a change of velocity. Mass can also be thought of as the quantity of matter in a body. The mass of a body affects the gravitational attraction force between it and other bodies. This force appears in many applications in statics. Force is the action of one body on another. A force tends to move a body in the direction of its action. The action of a force is characterized by its magnitude, by the direction of its action, and by its point of application. Thus force is a vector quantity.. A particle is a body of negligible dimensions. In the mathematical sense, a particle is a body whose dimensions are considered to be near zero so that we may analyze it as a mass concentrated at a point. We often choose a particle as a differential element of a body. We may treat a body as a particle when its dimensions are irrelevant to the description of its position or the action of forces applied to it. Rigid body. A body is considered rigid when the change in distance between any two of its points is negligible for the purpose at hand. For instance, the calculation of the tension in the cable which supports the boom of a mobile crane under load is essentially unaffected by the small internal deformations in the structural members of the boom. For the purpose, then, of determining the external forces which act on the boom, we may treat it as a rigid body. Statics deals primarily with the calculation of external forces which act on rigid bodies in equilibrium. Determination of the internal deformations belongs to the study of the mechanics of deformable bodies, which normally follows statics in the curriculum.

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587

PROBLEMS Introductory Problems 2/1 The force F has a magnitude of 800 N. Express F as a vector in terms of the unit vectors i and j. Identify the x and y scalar components of F.

2/2 The magnitude of the force F is 600 N. Express F as a vector in terms of the unit vectors i and j. Identify both the scalar and vector components of F.

Soln. Soln. Step1: Free body Diagram

Step 1: Magnitude of force F=600 lb

Step 2: Free body diagram Step2:M agnitud e of force 800N x component of force, Fx=-Fsin35o F x=o 800sin35 F x=-459 N y component of force, F y=Fcos35o F y=800 cos35o F y=655 N Force vector, F=(-459i-655j)N

Step3: Force vector F=600cos30i - 600sin30j F=520i – 300j

Step 4:Scalar components of force Along x-axis Fx=520 lb Along y-axis Fy= -300 lb

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 =(-2.88i-3.84j)kN 2/4 The line of action of the 9.6kN force F runs through the points A and B as shown in the figure. Determine the x and y scalar components of F.

Step5:Vector components of force Along x-axis Fx=520i lb Along y-axis Fy= -300j lb 2/3 The slope of the 4.8-kN force F is specified as shown in the figure. Express F as a vector in terms of the unit vectors i and j.

Soln. Step1: Magnitude of force F=4800 lb Position of point A= -15i-20j Position of point B=30i+10j Step2: Free body diagram

Soln. Step1:Free body diagram

Step2: Magnitude of force, F=4.8 kN 3 Unit vector of force, n= - 5 4 5

i-

Stpe3: Position vector of AB, AB=OB – OA AB=45i+30j 2 2 Magnitude of AB, AB = √ 45 + 30 = 54.08 in

j

Force vector

F= F n F=4.8(-

3 5

i-

4 5

Unit vector of AB,

n=

AB AB

j) Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 =

nAB = 0.832i - 0.55j

45 i+30 j 54.08

TA= TA nAB

2/5 A cable stretched between the fixed supports A and B is under a tension T of 900 lb. Express the tension as a vector using the unit vectors i and j, first, as a forceTA acting on A and second, as a force TB acting on B.

TA=900(0.832i - 0.55j) TA=(749i499j)lb But TA= - TB =-(749i-499j)lb =(-749i+499j)lb 2/6 The 1800-N force F is applied to the end of the I beam. Express F as a vector using the unit vectors i and j.

Soln. Step1: Free body Diagram Soln. Stpe1: Magnitude of force F=1800 N 3 Unit vector of force, n= - 5 4 5 Step2: Magnitude of tension in cable AB, T = 900 lb 2 Unit vector of AB=nAB= 3

-

√ 2 +32 j 2

√ 22 + 3

j

Force vector

F= F n 3 F=1800(- 5

4 5

i 2

i-

i-

j)

=-1080iN-1440jN 2/7 The two structural members, one of which is in tension and the other in compression, exert the indicated forces on joint O. Determine the magnitude of the resultant R of the two

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 −3.23

forces and the angle θ which R makes with the positive x-axis.

=tan-1( −1.598 ) =26o Angle θ made by ‘R’ with positive x-axis θ=180o+26o =206o 2/8 Two forces are applied to the construction bracket as shown. Determine the angle θ which makes the resultant of the two forces vertical. Determine the magnitude R of the resultant.

Soln. Step1: Free body Diagram

x-components of resultant force R= ∑ F x

Soln. Step1: Free body Diagram

x

Rx= -3cos60o-2cos30o Rx= -3.23 kN y-components of resultant force R= ∑ F y

y

Ry= 3sin60o+2 sin30o Ry= -1.598 kN Magnitude of the resultant force R=

R=

√R + R 2 x

2 y

Step2:

√ (−3.23 ) +(−1.598 ) 2

R=3.6 kN

Step3: Angle θ made by ‘R’ Rx -1 θ=tan ( R y )

2

F1=800 lb F2=425 lb Given that the resultant force is normal to x-axis.Therefore the xcomponent of resultant force is zero. R= ∑ F =0 x

x

Rx= -425cosθ+800cos70o = 0 Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 800cos 70o Cosθ= 425

=-240iN-100jN Scalar component of P along x Px=-240 N Scalar component of P along y Py=-100 N 2/10 For the mechanism of Prob. 2/9, determine the scalar components Pt and Pn of P which are tangent and normal, respectively, to crank BC. Soln. Step1:Free body diagram

θ=49.90 y-components of resultant force R= ∑ F y

y

Ry= -425sin49.9-800 sin70o Ry= -1070 lb Magnitude of the resultant force √ R 2x + R2y R=

√ 0 +(−1077 ) 2

R=

2

R=1077 lb

Representative Problems 2/9 In the design of a control mechanism, it is determined that rod AB transmits a 260-N force P to the crank BC. Determine the x and y scalar components of P. Soln. Step1: Free body Diagram Step2:magnitude of force F=260 N 5 θ=¿ tan-1( 12 )

Step2:Magnitude of force P=260 N 12 Unit vector of force, n= - 13 i5 13

j

θ=22.6o Scalar components of force P along ‘n’ Pn=260 cos(30o-22.6o) Pn=258 N Scalar components of force P along ‘t’ Pt=260 sin(30o-22.6o) Pt=33. 5N 2/11 The t-component of the force F is known to be 75N. Determine the ncomponent and the magnitude of F.

Force vector P= P n P=260(5 13

12 13

i-

j)

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587

Soln. Step1: Free body Diagram

Step1: Free body diagram

Step2: The t-component of the force F,Ft=75 N.Let Fn be the n-component of the force F Ft=Fcos40o (1) Fn=-Fsinθ (2) Dividing equation (2) by (1) Fn F t =-tan40o o

Fn=-Ft tan40 Fn=-75 tan40o Fn=-62.9o By equation (1) Ft=Fcos40o 75=F cos40o F=97.9 N 2/12 A force F of magnitude 800 N is applied to point C of the bar AB as shown. Determine both the x-y and the n-t components of F.

Step2: Components of force Fx= -800 cos200 Fx= -752 lb Components of force Fy= 800 sin200 Fy= 274 lb Step3: Components of force Ft= -800 cos400 Ft= -613 lb Components of force Fn= -800 sin400 Fn= -514 lb

F along ‘x’ F along ‘y’

along ‘t’ along ‘n’

2/13 The two forces shown act at point A of the bent bar. Determine the resultant R of the two forces. Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 this purpose replace this force by its equivalent of two forces at A, Ft parallel and Fn perpendicular to the beam. Determine Ft and Fn.

Step1: Free body diagram

Step2: Components of force along ‘t’ Ft= 2 cos(200+300) Ft=1.286 kN Components of force along ‘n’ Fn=2 sin(200+300) Fn=1.532 kN

Step2: x-components of resultant force Rx= ∑ Fx Rx= 7cos45o-3cos30o Rx= 2.35 kips y-components of resultant force R= ∑ F y

Soln. Step1:Free body diagram

2/15 Determine the magnitude Fs of the tensile spring force in order that the resultant of Fs and F is a vertical force. Determine the magnitude R of this vertical resultant force.

y

Ry= -7sin60o+3 sin30o Ry= -3.45 kips The resultant of the two forces is R=Rxi+Ryj

R= (2.35i-3.45j)kips 2/14 To satisfy design limitations it is necessary to determine the effect of the 2-kN tension in the cable on the shear, tension, and bending of the fixed I-beam. For

Soln. Step1:Free body diagram

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Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587

Step 2:Magnitude of force F=120 lb

Tensile spring force be Fs Since the horizontal component of ‘R’ is zero F=Fs cos60o F=120 cos60o F=60 lb The resultant force R=Fsin60o R=120sin60o R=103.9 lb 2/16 The ratio of the lift force L to the drag force D for the simple airfoil is L/D =10. If the lift force on a short section of the airfoil is 200 N, compute the magnitude of the resultant force R and the angle θ which it makes with the horizontal.

Step2: Lift force L=50 lb Ratio of the lift force to drag force L D =10 50 D =10 D=5 lb Magnitude of resultant force R = √ L2+D2 R=

√ 52+ 502

R=50.2 lb Step3: Angle made by the resultant with ‘D’

θ=¿

L tan-1( D

θ=¿ tan-1(

)

50 5 )

θ=¿ 84.3o

Soln. Step1: Free body diagram

2/17 Determine the components of the 2-kN force along the oblique axes a and b. Determine the projections of F onto the a- and b-axes.

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587

Soln. Step1:Free body diagram

Let Pa and Pb be the projections of force P along a and b. Pa=2 cos45o Pa=1.414 kN Pb=2 cos15o Pb=1.932 kN 2/18 Determine the scalar components Ra and Rb of the force R along the non rectan -gular axes a and b. Also determine the orthogonal projection Pa of R onto axis a.

Step2: By the law of sine’s Fa 2 o = sin 120 o sin15 sin 15 o Fa= sin 120 o

×2

Fa=0.598 kN Fb

2

o

sin 45

= sin 120 o

sin 45o Fb= sin 120 o

Soln. Step1: Free body diagram

×2

Fb=1.633 kN Step3:

Step2: By the law of sine Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 800 Rb = sin 30 sin 40 Rb= 622 N 800 Ra sin 110 = sin 40 Ra= 1170 N Step3:Let Pa be the orthogonal projection of P onto a-axis

Step4: Pa=R cos300 =800 cos300 =693 N 2/19 Determine the resultant R of the two forces shown by (a) applying the parallelogram rule for vector addition and (b) summing scalar components.

Step2:(a) From the law of cosine’s R2=4002+6002 – 2(400)(600)cos1202 R2=760000 R2=872 N Let θ be the angle made with the vertical,then by the law of sine’s 872 600 sinθ = sin 120 o θ =36.6o

Step3:(b) x-components of resultant force Rx= ∑ Fx Rx=600cos30o Rx=520 N

y-components of resultant force R= ∑ F y

R y=

y

-400-600sin30o

Ry=-700 N Magnitude of the resultant force √ R 2x + R y2 R=

Soln. Step1: Free body diagram

R=

520 ¿ ¿ ¿ √¿

R=872 N Angle θ made by ‘R’ Rx -1 θ=tan ( R y )

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Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 700 =tan ( 520 ) -1

=53.4o So the angle made by resultant with the vertical θ=90o-53.4o θ=36.6o 2/20 It is desired to remove the spike from the timber by applying force along its horizontal axis. An obstruction A prevents direct access, so that two forces, one 1.6 kN and the other P, are applied by cables as shown. Compute the magnitude of P necessary to ensure a resultant T directed along the spike. Also find T.

θ2=36.87o α=180o – (θ1+θ2) α=180o – (26.57o+36.87o) α=116.56o By the law of sine 400 P sinθ 2 = sinθ 1 P= 537 lb P T = sinθ sinα 2 T= 800.541 lb 2/21 At what angle θ must the 800-N force be applied in order that the resultant R of the two forces has a magnitude of 2000 N? For this condition, determine the angle θ between R and the vertical.

Soln. Step1: Free body diagram Soln. Step1: Free body diagram

Step2: From figure 4 θ1=tan-1( 8 ) θ1=26.57o 6 θ2=tan-1( 8 )

Step2: Magnitude of the resultant force R=2000 lb From the law of cosine’s 20002=14002+8002 – 2(1400) (800)cos(180o-θ) But cos(180o-θ)=-cosθ Therefore

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Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 20002=14002+8002 + 2(1400)(800)cosθ θ=51.3o From the law of sine’s 800 R sinβ = sin(180−θ)

sinβ =R×

sin(180−θ) 800

β=18.19o 2/22 The unstretched length of the spring of modulus k =1.2 kN/m is 100 mm.When pin P is in the position θ=30o determine the x- and ycomponents of the force which the spring exerts on the pin. (The force in a spring is given by F=kx, where x is the deflection from the unstretched length.)

Position of point A=0.06i+0.04j Position of point P=0.08sin30i0.08cos30j P=0.04i0.0693j Position vector PA=OA-OP PA=(0.06-0.04)i-[(0.04)-(0.0693)]j PA=0.02i+0.1093j 2 2 Magnitude of PA= √ 0.02 + 0.1093

Unit vector

n

PA=0.1111m PA nPA= PA

0.02 ¿ ¿ 0.1093 ¿ ¿ ¿2 ¿ √¿ 0.02i+0.1093 j ¿

=

PA

n

Soln. Step1: Free body diagram

PA

=0.18i+0.984j

Magnitude of spring force F= Kx Where x is change in length. x=PA- l x =0.1111-0.1 x=0.0111m F=1.2 kN×0.0111 F=13.32 N Step3: Spring force vector F=F

n

PA

=13.32(0.18i+0.984j) =(2.4i+13.1j) N

Step2: Givent hat spring modulus k=1.2 kN/m Unstretched length of spring l =0.1m

x and y components of force are Fx=2.4 N Fy=13.1 N 2/23 Refer to the statement and figure of Prob. 2/22. When pin P is in the position θ=20o,determine the n- and tcomponents of the force F which the spring of modulus k =1.2 kN/m exerts on the pin. Step1:Given that

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore.fb:[email protected]

Solution Manual by Khalid Yousaf BS(Continue)Civil Engineering The University of Lahore. Cell#03338189587 Spring constant of the spring k=1.2 kN/m Unstrectched length of the spring,

F=0.0237 kN F=23.7 kN

l=0.1m

2/24 The cable AB prevents bar OA from rotating clockwise about the pivot O. If the cable tension is 750N.Determine the n- and tcomponents of this force acting on point A of the bar.

The angle made by the pin with the vertical θ=20o Let the centre of the coordinate system be positioned at 0. Step2:The arrangement of the spring system at the present instant is as shown by the free body diagram

Soln. Step1: Free body diagram Step3: The coordinates of points A and P are A=(0.06i,0.04j) P=(0.08 sin20oi-0.08cos20oj) P=(0.0274i-0.0752j) Hence the position vector PA is given by PA=OA-OP PA=(0.06-0.0274)i + [0.04-(-0.0752]j PA=0.0326i+0.1152j The magnitude of the vector is given by 0.0326 ¿ ¿ PA= ¿ √¿ PA=0.1197m The magnitude of the spring force is given by F=kx Where x=PA - 0.01 x=0.1197-0.01 x=0.0197m F=1.2×0.0197

Step2: Tension in the cable T=750 N Let Tn and Tt be the components of force ‘T’ along ‘n’ and ‘t’axes respectively By the cosine law AB= √ OA 2+ OB2−2( OA )(OB )cos 120o AB= √ (1.5 )2+(1.2 )2−2(1.5 )(1.2)(−0.866 ) AB=2.34 m

Solution manual by Khalid Yousaf BS(Continue)Civil Engineering The Univ...