Dynamics 14th edition by r c hibbeler chapter 05

NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the beam’s FBD shown in Fig... NA and By can be d

Determine the components of the support reactions at the

fixed support A on the cantilevered beam.

1.5 m

1.5 m

3030

Equations of Equilibrium: From the free-body diagram of the cantilever beam, Fig a,

A x , A y , and M Acan be obtained by writing the moment equation of equilibrium about

Equations of Equilibrium NA and By can be determined directly by writing the

moment equations of equilibrium about points B and A, respectively, by referring to

the beam’s FBD shown in Fig a.

3 m

A

B

Determine the horizontal and vertical components of

reaction at the pin A and the reaction of the rocker B on

the beam

SOLUTION

Equations of Equilibrium: From the free-body diagram of the beam, Fig a, N Bcan

be obtained by writing the moment equation of equilibrium about point A.

Equations of Equilibrium NA and By can be determined directly by writing the

moment equations of equilibrium about points B and A, respectively, by referring to

the FBD of the beam shown in Fig a.

a+ ΣM B = 0; 600(6)(3) + 12 (300)(3)(5) - N A(6) = 0

a+ ΣM A = 0; B y(6) - 12 (300)(3)(1) - 600(6)(3) = 0

Also, Bx can be determined directly by writing the force equation of equilibrium

along the x axis.

Equations of Equilibrium NA can be determined directly by writing the moment

equation of equilibrium about point B by referring to the FBD of the beam shown

3 m

800 N/m

3 m

1 m

Equations of Equilibrium Ay and NB can be determined by writing the moment

equations of equilibrium about points B and A, respectively, by referring to the FBD

of the truss shown in Fig a.

Determine the tension in the cable and the horizontal and

vertical components of reaction of the pin A The pulley at

Dis frictionless and the cylinder weighs 80 lb

SOLUTION

Equations of Equilibrium: The tension force developed in the cable is the same

throughout the whole cable The force in the cable can be obtained directly by

summing moments about point A.

Ax = 33.4 lb

Ax - 74.583¢ 1

25≤ = 0:+ ©Fx = 0;

T = 74.583 lb = 74.6 lb

T152 + T¢ 2

25≤1102 - 801132 = 0+ ©MA = 0;

B A

D

C

2 1

The man attempts to support the load of boards having a

weight W and a center of gravity at G If he is standing on a

smooth floor, determine the smallest angle at which he can

hold them up in the position shown Neglect his weight

u

Ans:

u = 41.4°

Equations of Equilibrium: Since the roller at offers no resistance to vertical

movement, the vertical component of reaction at support is equal to zero From

the free-body diagram, , , and can be obtained by writing the force

equations of equilibrium along the and axes and the moment equation of

equilibrium about point , respectively

MA

By

Ax

AA

Ans:

A x = 0

B y = P

M A = PL2

The man has a weight W and stands at the center of the

plank If the planes at A and B are smooth, determine the

tension in the cord in terms of W and u.

A uniform glass rod having a length L is placed in the smooth

hemispherical bowl having a radius r Determine the angle of

inclination for equilibrium.u

+Q ©Fx = 0; NAcos u- W sin u = 0 NA= W tanu

+ ©MA= 0; NB(2r cos u) - WaL2 cos ub = 0 NB = WL4r

f = u

B r

A

u

Ans:

u = cos-1aL + 2L 16r2 + 12r2b

T BC = 113 N

Solution

Equations of Equilibrium TBC can be determined by writing the moment equation

of equilibrium about point O by referring to the FBD of the rod shown in Fig a.

a+ ΣM O = 0; 40(9.81)(1.5 cos 600°) - T BC(3 sin 60°) = 0

The uniform rod AB has a mass of 40 kg Determine the

force in the cable when the rod is in the position shown

There is a smooth collar at A.

Equations of Equilibrium NA can be determined directly by writing the moment

equation of equilibrium about point B by referring to the FBD of the beam shown

If the intensity of the distributed load acting on the beam

is w = 3 kN>m, determine the reactions at the roller A and

pin B.

A

B w

3 m30

4 m

w = 2.67 kN>m

Solution

Equations of Equilibrium NA can be determined directly by writing the moment

equation of equilibrium about point B by referring to the FBD of the beam shown

If the roller at A and the pin at B can support a load up

to 4 kN and 8 kN, respectively, determine the maximum

intensity of the distributed load w, measured in kN>m, so

that failure of the supports does not occur

A

B w

3 m30

4 m

The relay regulates voltage and current Determine the force

in the spring CD, which has a stiffness of k 120 N m, so

that it will allow the armature to make contact at A in figure

(a) with a vertical force of 0.4 N Also, determine the force

in the spring when the coil is energized and attracts the

armature to E, figure (b), thereby breaking contact at A.

50 mm 50 mm 30 mm10°

)b()

a(

D D

k k

Ans:

F s = 1.33 N

F s = 1.96 N

Determine the reactions on the bent rod which is supported

by a smooth surface at B and by a collar at A, which is fixed

to the rod and is free to slide over the fixed inclined rod 3 ft 3 ft

3 4 5

The mobile crane is symmetrically supported by two

outriggers at A and two at B in order to relieve the

suspension of the truck upon which it rests and to provide

greater stability If the crane and truck have a mass of

18 Mg and center of mass at , and the boom has a mass

of 1.8 Mg and a center of mass at , determine the vertical

reactions at each of the four outriggers as a function of the

boom angle when the boom is supporting a load having a

mass of 1.2 Mg Plot the results measured from to

the critical angle where tipping starts to occur

u = 70.3°

NA = 0

NA = 58 860 - 62 539 sinu+ 1.2A103B(9.81) (2 - 12.25 sinu) = 0+ ©MB = 0; -NA(4) + 18A103B(9.81)(1) + 1.8A103B(9.81) (2 - 6 sinu)

Equations of Equilibrium NB can be determined directly by writing the moment

equation of equilibrium about point A by referring to the FBD of the bar shown in

Determine the reactions acting on the smooth uniform bar,

which has a mass of 20 kg

A linear torsional spring deforms such that an applied couple

moment M is related to the spring’s rotation u in radians by

the equation M = (20 u) N#m If such a spring is attached to

the end of a pin-connected uniform 10-kg rod, determine

the angle u for equilibrium The spring is undeformed

P = 272 N

Solution

Equations of Equilibrium P can be determined directly by writing the moment equation of

Equilibrium about point B, by referring to the FBD of the roller shown in Fig a.

a+ ΣM B = 0; P cos 30°(0.25) + P sin 30° (20.32 - 0.2522 - 50(9.81)20.32- 0.252 = 0

Determine the force P needed to pull the 50-kg roller over

the smooth step Take u = 30°

A B

P

300 mm

50 mm u

Pmin = 271 N

Solution

Equations of Equilibrium P will be minimum if its orientation produces the greatest

moment about point B This happens when it acts perpendicular to AB as shown in

Fig a Thus

u = f = cos-1a0.250.3 b = 33.56° = 33.6° Ans.

Pmin can be determined by writing the moment equation of equilibrium about point

B by referring to the FBD of the roller shown in Fig b.

a+ ΣM B = 0; Pmin (0.3) - 50(9.81)(0.3 sin 33.56°) = 0

5–30.

Determine the magnitude and direction u of the minimum

force P needed to pull the 50-kg roller over the smooth step.

A B

P

300 mm

50 mm u

The operation of the fuel pump for an automobile depends

on the reciprocating action of the rocker arm ABC, which

is pinned at B and is spring loaded at A and D When the

smooth cam C is in the position shown, determine the

horizontal and vertical components of force at the pin and

the force along the spring DF for equilibrium The vertical

force acting on the rocker arm at A is , and at C

B A

F E

Bx = 43.3 N:+ ©Fx = 0; -Bx + 86.6025 sin 30° = 0

FB = 86.6025 = 86.6 N+ ©MB = 0; - 60(50) - FBcos 30°(10) + 125(30) = 0

Ans:

F B = 86.6 N

B x = 43.3 N

B y = 110 N

Determine the magnitude of force at the pin and in the

cable needed to support the 500-lb load Neglect the

weight of the boom

SOLUTION

Equations of Equilibrium: The force in cable can be obtained directly by

summing moments about point

By = 22.8 kN+ c ©Fy = 0; By - 800 (9.81) - 15 000 = 0

Ax = 25.4 kN+ ©MB= 0; Ax(2) - 800 (9.81) (0.75) - 15 000(3) = 0

The dimensions of a jib crane, which is manufactured by the

Basick Co., are given in the figure If the crane has a mass of

800 kg and a center of mass at G, and the maximum rated

force at its end is F 15 kN, determine the reactions at its

bearings The bearing at A is a journal bearing and supports

only a horizontal force, whereas the bearing at B is a thrust

bearing that supports both horizontal and vertical components

F

G A

The dimensions of a jib crane, which is manufactured by the

Basick Co., are given in the figure The crane has a mass of

800 kg and a center of mass at G.The bearing at A is a journal

bearing and can support a horizontal force, whereas the

bearing at B is a thrust bearing that supports both horizontal

and vertical components Determine the maximum load F that

can be suspended from its end if the selected bearings at A

and B can sustain a maximum resultant load of 24 kN and

Equations of Equilibrium NA can be determined directly by writing the force equation of

equilibrium along the x axis by referring to the FBD of the pipe shown in Fig a.

S+ ΣF x = 0; N A cos 30° - 300 sin 30° = 0 N A = 173.21 N = 173 N Ans.

Using this result to write the moment equations of equilibrium about points B and C,

a+ ΣM B = 0; 300 cos 30°(1) - 173.21 cos 30°(0.26) - 173.21 sin 30°(0.15) - N C(0.5) = 0

a+ ΣM C = 0; 300 cos 30°(0.5) - 173.21 cos 30°(0.26) - 173.21 sin 30°(0.65) - N B(0.5) = 0

The smooth pipe rests against the opening at the points of

contact A, B, and C Determine the reactions at these points

needed to support the force of 300 N Neglect the pipe’s

thickness in the calculation

0.26 m

0.15 m

u = 3.82°

Solution

Equations of Equilibrium FA and FB can be determined directly by writing the

moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the beam shown in Fig a.

Assuming that the angle of tilt is small,

of tilt of the beam when the load is applied

The cantilevered jib crane is used to support the load of

780 lb If , determine the reactions at the supports

Note that the supports are collars that allow the crane to

rotate freely about the vertical axis.The collar at B supports a

force in the vertical direction, whereas the one at A does not.

A

SOLUTION

Equations of Equilibrium: Referring to the of the jib crane shown in Fig a, we

notice that and can be obtained directly by writing the moment equation of

equilibrium about point B and force equation of equilibrium along the y axis,

: ©Fx = 0;

NA

By = 780

By - 780 = 0+ c ©Fy = 0;

NA = 975 lb

NA(4) - 780(5) = 0a

The cantilevered jib crane is used to support the load of

780 lb If the trolley T can be placed anywhere between

determine the maximum magnitude of

reaction at the supports A and B Note that the supports

are collars that allow the crane to rotate freely about the

vertical axis The collar at B supports a force in the vertical direction, whereas the one at A does not.

1.5 ft … x … 7.5 ft,

SOLUTION

Require a

Ax = 1462.5 = 1462 lb:+ ©Fx = 0; Ax - 1462.5 = 0

Bx = 1462.5 lb+ ©MA = 0; -780(7.5) + Bx (4) = 0

Equations of Equilibrium FA and FB can be determined directly by writing the

moment equation of equilibrium about points B and A respectively by referring to

the FBD of the bar shown in Fig a.

Note: The moment equations are set up assuming small u, but even with non-small

u the reactions come out with the same FA , F B, and then the rest of the solution goes

through as before

The bar of negligible weight is supported by two springs,

each having a stiffness k = 100 N>m If the springs are

originally unstretched, and the force is vertical as shown,

determine the angle u the bar makes with the horizontal,

A

B C

k

Ans:

u = 17.5°

k = 116 N>m

Solution

Equations of Equilibrium FA and FB can be determined directly by writing the

moment equation of equilibrium about points B and A respectively by referring to

the FBD of the bar shown in Fig a.

Note: The moment equations are set up assuming small u, but even with non-small

u the reactions come out with the same FA , F B, and then the rest of the solution goes

through as before

*5–40.

Determine the stiffness k of each spring so that the  30-N

force causes the bar to tip u = 15° when the force is applied

Originally the bar is horizontal and the springs are

unstretched Neglect the weight of the bar

2 m

1 m

A

B C

k

The bulk head AD is subjected to both water and

soil-backfill pressures Assuming AD is “pinned” to the

ground at A, determine the horizontal and vertical

reactions there and also the required tension in the

ground anchor BC necessary for equilibrium The bulk

head has a mass of 800 kg

SOLUTION

Equations of Equilibrium: The force in ground anchor BC can be obtained directly

by summing moments about point A.

Ax = 460 kN

Ax + 311.375 + 236 - 1007.5 = 0:+ ©Fx = 0;

F = 311.375 kN = 311 kN

1007.512.1672 - 23611.3332 - F162 = 0+ ©MA = 0;

Ans:

F = 311 kN

A x= 460 kN

A y= 7.85 kN

Ax = 625 N:+ ©Fx = 0; Ax - 45(781.6) = 0

FCB= 781.6 = 782 N+ 45FCB (2.5 sin 30°) + 35FCB(2.5cos 30°) = 0+ ©MA= 0; -800(1.5 cos 30°) - 350(2.5 cos 30°)

The boom supports the two vertical loads Neglect the size

of the collars at D and B and the thickness of the boom,

and compute the horizontal and vertical components of

force at the pin A and the force in cable CB Set

The boom is intended to support two vertical loads, and

If the cable CB can sustain a maximum load of 1500 N before

it fails, determine the critical loads if Also, what is

the magnitude of the maximum reaction at pin A?

Ax = 1200 N:+ ©Fx = 0; Ax - 45(1500) = 0

F1 = 1.45 kN

F1 = 2F2 = 1448 N

F2 = 724 N+ 45(1500)(2.5 sin 30°) + 35(1500)(2.5 cos 30°) = 0

u = 24.6°

*5–44.

The 10-kg uniform rod is pinned at end A If it is also

subjected to a couple moment of 50 N#m, determine the

smallest angle u for equilibrium The spring is unstretched

when u = 0, and has a stiffness of k = 60 N>m.

Equations of Equilibrium Here the spring stretches x = 2 sin u The force in the

spring is F sp = kx = 60 (2 sin u) = 120 sin u Write the moment equation of

equilibrium about point A by referring to the FBD of the rod shown in Fig a,

a+ ΣM A = 0; 120 sin u cos u (2) - 10(9.81) sin u (1) - 50 = 0

240 sin u cos u - 98.1 sin u - 50 = 0Solve numerically

Equations of Equilibriums Py can be determined directly by writing the force

equation of equilibrium along y axis by referring to the FBD of the hand truck

shown in Fig a.

+cΣF y = 0; P y - 50(9.81) = 0 P y = 490.5 N

Using this result to write the moment equation of equilibrium about point A,

a+ ΣM A = 0; P x sin 60°(1.3) - P x cos 60°(0.1) - 490.5 cos 30°(0.1)

-490.5 sin 30°(1.3) - 50(9.81) sin 60°(0.5)+50(9.81) cos 60°(0.4) = 0

The man uses the hand truck to move material up the step

If the truck and its contents have a mass of 50 kg with center

of gravity at G, determine the normal reaction on both

wheels and the magnitude and direction of the minimum

force required at the grip B needed to lift the load.

A

B

60

0.4 m0.5 m0.4 m

0.4 m0.1 m

Three uniform books, each having a weight W and length a,

are stacked as shown Determine the maximum distance d

that the top book can extend out from the bottom one so

the stack does not topple over

SOLUTION

Equilibrium: For top two books, the upper book will topple when the center of

gravity of this book is to the right of point A.Therefore, the maximum distance from

the right edge of this book to point A is a/2.

Equation of Equilibrium: For the entire three books, the top two books will topple

about point B.

a

Ans.

d = 3a4+ ©MB = 0; W(a-d)-Wad-a2 b = 0

Ans:

d = 3a4

Determine the reactions at the pin A and the tension in cord

BC Set F = 40 kN Neglect the thickness of the beam.

5 34

If rope BC will fail when the tension becomes  50  kN,

determine the greatest vertical load F that can be applied to

the beam at B What is the magnitude of the reaction at A

for this loading? Neglect the thickness of the beam

5 34

The rigid metal strip of negligible weight is used as part of an

electromagnetic switch If the stiffness of the springs at A

and B is and the strip is originally horizontal

when the springs are unstretched, determine the smallest

force needed to close the contact gap at C.

k

Ans:

F C = 10 mN