Solution manual engineering mechanics statics 13th edition by r c hibbeler13e chap 08

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Equations of Equilibrium: The normal reactions acting on the wheels at (A and B)

are independent as to whether the wheels are locked or not Hence, the normal

reactions acting on the wheels are the same for both cases

a

the wheels do not slip Thus, the mine car does

+ FB max = 23.544 kN 7 10 kN,1FB2max = msNB = 0.4142.3162 = 16.9264 kN. 1FA2max

NB = 42.316 kN = 42.3 kN

NB + 16.544 - 58.86 = 0+ c ©Fy = 0;

NA = 16.544 kN = 16.5 kN

NA11.52 + 1011.052 - 58.8610.62 = 0+ ©MB = 0;

The mine car and its contents have a total mass of 6 Mg and

a center of gravity at G If the coefficient of static friction

between the wheels and the tracks is when the

wheels are locked, find the normal force acting on the front

wheels at B and the rear wheels at A when the brakes at

both A and B are locked Does the car move?

ms = 0.4

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Determine the maximum force P the connection can

support so that no slipping occurs between the plates There

are four bolts used for the connection and each is tightened

so that it is subjected to a tension of 4 kN The coefficient of

static friction between the plates is

SOLUTION

Free-Body Diagram: The normal reaction acting on the contacting surface is equal

to the sum total tension of the bolts Thus, When the plate is

on the verge of slipping, the magnitude of the friction force acting on each contact

surface can be computed using the friction formula As

indicated on the free-body diagram of the upper plate, F acts to the right since the

plate has a tendency to move to the left

Equations of Equilibrium:

Ans.

p = 12.8 kN0.4(16) - P

2

P

2

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The winch on the truck is used to hoist the garbage bin onto

the bed of the truck If the loaded bin has a weight of 8500 lb

and center of gravity at G, determine the force in the cable

needed to begin the lift The coefficients of static friction at

A and B are and respectively Neglect

the height of the support at A.

T(0.86603) - 0.67321 NB = 1390.91

- 0.2NBcos 30° - NBsin 30° - 0.3(4636.364) = 0:+ ©Fx = 0; T cos 30°

NA = 4636.364 lb+ ©MB = 0; 8500(12) - NA(22) = 0

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The tractor has a weight of 4500 lb with center of gravity at

G The driving traction is developed at the rear wheels B,

while the front wheels at A are free to roll If the coefficient

of static friction between the wheels at B and the ground is

determine if it is possible to pull at

without causing the wheels at B to slip or the front wheels at

A to lift off the ground.

P = 1200 lb

ms = 0.5,

4 ft2.5 ft

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The 15-ft ladder has a uniform weight of 80 lb and rests

against the smooth wall at B If the coefficient of static

friction at A is determine if the ladder will slip

FA = 23.094 lb:+ ©Fx = 0;

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The ladder has a uniform weight of 80 lb and rests against

the wall at B If the coefficient of static friction at A and B is

, determine the smallest angle at which the ladder

will not slip

SOLUTION

Free-Body Diagram: Since the ladder is required to be on the verge to slide down,

the frictional force at A and B must act to the right and upward respectively and

their magnitude can be computed using friction formula as indicated on the FBD,

Fig a.

Equations of Equlibrium: Referring to Fig a.

(1) (2)

Solving Eqs (1) and (2) yields

Using these results,

= 1.05413.79sinu - 434.48cosu = 0

0.4(27.59)(15cosu) + 27.59(15sinu) - 80cosu(7.5) = 0+ ©MA = 0;

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The block brake consists of a pin-connected lever and

friction block at B The coefficient of static friction between

the wheel and the lever is and a torque of

is applied to the wheel Determine if the brake can hold

the wheel stationary when the force applied to the lever is

5 N m

50 mm

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The block brake consists of a pin-connected lever and

friction block at B The coefficient of static friction between

the wheel and the lever is , and a torque of

is applied to the wheel Determine if the brake can hold

the wheel stationary when the force applied to the lever is

5 N m

50 mm

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The block brake is used to stop the wheel from rotating

when the wheel is subjected to a couple moment If the

coefficient of static friction between the wheel and the

block is determine the smallest force P that should be

r

C

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The block brake is used to stop the wheel from rotating

when the wheel is subjected to a couple moment If the

coefficient of static friction between the wheel and the block

is , show that the brake is self locking, i.e., the required

r

C

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The block brake is used to stop the wheel from rotating

when the wheel is subjected to a couple moment If the

coefficient of static friction between the wheel and the

block is , determine the smallest force P that should be

r

C

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If a torque of is applied to the flywheel,

determine the force that must be developed in the hydraulic

cylinder CD to prevent the flywheel from rotating The

coefficient of static friction between the friction pad at B

and the flywheel is

SOLUTION

Free-BodyDiagram: First we will consider the equilibrium of the flywheel using the

free-body diagram shown in Fig a Here, the frictional force must act to the left to

produce the counterclockwise moment opposing the impending clockwise rotational

motion caused by the couple moment Since the wheel is required to be on

the verge of slipping, then Subsequently, the free-body

diagram of member ABC shown in Fig b will be used to determine F CD

Equations of Equilibrium: We have

NB = 2500 N0.4 NB(0.3) - 300 = 0

1 m

O

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The cam is subjected to a couple moment of

Determine the minimum force P that should be applied to

the follower in order to hold the cam in the position shown

The coefficient of static friction between the cam and the

follower is ms = 0.4.The guide at A is smooth.

NB = 147.06 N+ ©MO= 0; 5 - 0.4 NB(0.06) - 0.01 (NB) = 0

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N = 200 lb+ c ©Fy = 0;

W = 318 lb

-W3 cos 45° + 0.6 N = 0:+ ©Fx = 0;

W

3 sin 45° + N - 200 = 0+ c ©Fy = 0;

Determine the maximum weight W the man can lift with

constant velocity using the pulley system, without and then

with the “leading block” or pulley at A The man has a

weight of 200 lb and the coefficient of static friction

between his feet and the ground is ms = 0.6

(b)

w A w

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The car has a mass of 1.6 Mg and center of mass at G If the

coefficient of static friction between the shoulder of the road

and the tires is determine the greatest slope the

shoulder can have without causing the car to slip or tip over

if the car travels along the shoulder at constant velocity

5 ft

2.5 ft

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The uniform dresser has a weight of 90 lb and rests on a tile

floor for which If the man pushes on it in the

horizontal direction determine the smallest

magnitude of force F needed to move the dresser Also, if

the man has a weight of 150 lb, determine the smallest

coefficient of static friction between his shoes and the floor

so that he does not slip

Nm = 150 lb+ c ©Fy = 0; Nm - 150 = 0

F = 22.5 lb:+ ©Fx = 0; F - 0.25(90) = 0

ND = 90 lb+ c ©Fy = 0; ND - 90 = 0

F

u

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Fm = 26.295 lb

Nm = 134.82 lb:+ ©Fx = 0; Fm - 30.363 cos 30° = 0

+ c ©Fy = 0; Nm - 150 + 30.363 sin 30° = 0

F = 30.363 lb = 30.4 lb

N = 105.1 lb:+ ©Fx = 0; F cos 30° - 0.25 N = 0

+ c ©Fy = 0; N - 90 - F sin 30° = 0

The uniform dresser has a weight of 90 lb and rests on a tile

floor for which If the man pushes on it in the

direction determine the smallest magnitude of force F

needed to move the dresser Also, if the man has a weight

of 150 lb, determine the smallest coefficient of static friction

between his shoes and the floor so that he does not slip

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The 5-kg cylinder is suspended from two equal-length cords

The end of each cord is attached to a ring of negligible mass

that passes along a horizontal shaft If the rings can be

separated by the greatest distance and still

support the cylinder, determine the coefficient of static

friction between each ring and the shaft

SOLUTION

Equilibrium of the Cylinder: Referring to the FBD shown in Fig a,

Equilibrium of the Ring: Since the ring is required to be on the verge to slide, the

frictional force can be computed using friction formula as indicated in the

FBD of the ring shown in Fig b Using the result of I,

Ff = mN

T = 5.2025 m

2BT¢232

6 ≤ R - m(9.81) = 0+ c ©Fy = 0;

d = 400 mm

d

600 mm

600 mm

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The 5-kg cylinder is suspended from two equal-length

cords The end of each cord is attached to a ring of

negligible mass, which passes along a horizontal shaft If the

coefficient of static friction between each ring and the shaft

is determine the greatest distance d by which the

rings can be separated and still support the cylinder

ms = 0.5,

SOLUTION

Friction: When the ring is on the verge to sliding along the rod, slipping will have to

occur Hence, From the force diagram (T is the tension developed

F = mN = 0.5N

d

600 mm

600 mm

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The board can be adjusted vertically by tilting it up and

sliding the smooth pin A along the vertical guide G When

placed horizontally, the bottom C then bears along the edge

of the guide, where Determine the largest

dimension d which will support any applied force F without

causing the board to slip downward

ms = 0.4

G A

G C

0.75 in

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The uniform pole has a weight W and length L Its end B is

tied to a supporting cord, and end A is placed against the

wall, for which the coefficient of static friction

is Determine the largest angle at which the pole can be

placed without slipping

:+ ©Fx = 0; NA - T sin u2 = 0

+ ©MB = 0; -NA(L cos u) - msNA(L sin u) + W aL2sin ub = 0 A

C

B L

L

u

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If the clamping force is and each board has a

mass of 2 kg, determine the maximum number of boards the

clamp can support The coefficient of static friction between

the boards is , and the coefficient of static friction

between the boards and the clamp is

SOLUTION

Free-Body Diagram: The boards could be on the verge of slipping between

the two boards at the ends or between the clamp Let n be the number of

boards between the clamp Thus, the number of boards between the two boards

at the ends is If the boards slip between the two end boards, then

Equations of Equilibrium: Referring to the free-body diagram shown in Fig a,

we have

referring to the free-body diagram shown in Fig b, we have

a

Thus, the maximum number of boards that can be supported by the clamp will be

the smallest value of n obtained above, which gives

n = 8

n = 9.172(90) - n(2)(9.81) = 0

+ c ©Fy = 0;

F¿ = ms¿N = 0.45(200) = 90 N

n = 8.122(60) - (n - 2)(2)(9.81) = 0

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A 35-kg disk rests on an inclined surface for which

Determine the maximum vertical force P that may be

applied to link AB without causing the disk to slip at C.

Friction: If the disk is on the verge of moving, slipping would have to occur at

point C Hence, Substituting this value into Eqs (1) and (2)

and solving, we have

NCsin 60° - FCsin 30° - 0.6667P - 343.35 = 0+ c ©Fy = 0

P16002 - Ay19002 = 0 Ay = 0.6667P+ ©MB = 0;

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The man has a weight of 200 lb, and the coefficient of static

friction between his shoes and the floor is

Determine where he should position his center of gravity G

at d in order to exert the maximum horizontal force on the

door What is this force?

ms = 0.5

d

3 ft

G

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The crate has a weight of , and the coefficients

of static and kinetic friction are and ,

respectively Determine the friction force on the floor if

and

SOLUTION

Equations of Equilibrium: Referring to the FBD of the crate shown in Fig a,

Friction Formula: Here, the maximum frictional force that can be developed is

Since , the crate will slide Thus the frictional force

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The crate has a weight of , and the coefficients

of static and kinetic friction are and ,

respectively Determine the friction force on the floor if

and

SOLUTION

Equations of Equilibrium: Referring to the FBD of the crate shown in Fig a,

Friction Formula: Here, the maximum frictional force that can be developed is

Since the crate will not slide Thus, the frictional force

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Equations of Equilibrium:

(1) (2)

Friction: If the crate is on the verge of moving, slipping will have to occur Hence,

Substituting this value into Eqs (1)and (2) and solving, we have

In order to obtain the minimum P,

N + P sin u - W = 0+ c ©Fy = 0;

The crate has a weight W and the coefficient of static

friction at the surface is Determine the

orientation of the cord and the smallest possible force P

that has to be applied to the cord so that the crate is on the

verge of moving

ms = 0.3

θ

P

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If the coefficient of static friction between the man’s shoes

and the pole is , determine the minimum coefficient

of static friction required between the belt and the pole at A

in order to support the man The man has a weight of 180 lb

and a center of gravity at G.

SOLUTION

Free-Body Diagram: The man’s shoe and the belt have a tendency to slip

downward Thus, the frictional forces and must act upward as indicated on the

free-body diagram of the man shown in Fig a Here, is required to develop to its

FA = 101.12 lb

FA + 0.6(131.46) - 180 = 0+ c ©Fy = 0;

NA = 131.46 lb131.46 - NA= 0

©Fx = 0;

:+

NC= 131.46 lb

NC(4) + 0.6NC(0.75) - 180(3.25) = 0+ ©MA = 0;

4 ft

2 ft0.75 ft 0.5 ft

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The friction pawl is pinned at A and rests against the wheel

at B It allows freedom of movement when the wheel is

rotating counterclockwise about C Clockwise rotation is

prevented due to friction of the pawl which tends to bind

the wheel If determine the design angle

which will prevent clockwise motion for any value of

applied moment M Hint: Neglect the weight of the pawl so

that it becomes a two-force member

u1ms2B = 0.6,

SOLUTION

Friction: When the wheel is on the verge of rotating, slipping would have to occur.

Hence, From the force diagram ( is the force developed in

the two force member AB)

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If determine the minimum coefficient of static

friction at A and B so that equilibrium of the supporting

frame is maintained regardless of the mass of the cylinder C

Neglect the mass of the rods

uu

SOLUTION

Free-Body Diagram: Due to the symmetrical loading and system, ends A and B of

the rod will slip simultaneously Since end B tends to move to the right, the friction

force FB must act to the left as indicated on the free-body diagram shown in Fig a.

Equations of Equilibrium: We have

Therefore, to prevent slipping the coefficient of static friction ends A and B must be

FB = 0 5FBC

FBC sin30° - FB = 0

©Fx = 0;

:+

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If the coefficient of static friction at A and B is

determine the maximum angle so that the frame remains

in equilibrium, regardless of the mass of the cylinder

Neglect the mass of the rods

SOLUTION

Free-Body Diagram: Due to the symmetrical loading and system, ends A and B of

the rod will slip simultaneously Since end B is on the verge of sliding to the right, the

friction force FBmust act to the left such that as indicated on

the free-body diagram shown in Fig a.

Equations of Equilibrium: We have

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Solving Eqs (1), (2) and (3) yields

Ans.

Friction: The maximum friction force that can be developed between the

semicylinder and the inclined plane is

Since Fmax 7 F = 1.703m,the semicylinder will not slide down the plane.F max = mN = 0.3 9.661m = 2.898m.Ans.

u = 24.2°

N = 9.661m F = 1.703m

+ c ©Fy = 0 F sin 10° + N cos 10° - 9.81m = 0

F cos 10° - N sin 10° = 0:+ ©Fx = 0;

F1r2 - 9.81m sin ua34rpb = 0+ ©MO = 0;

The semicylinder of mass m and radius r lies on the rough

inclined plane for which and the coefficient of

static friction is Determine if the semicylinder

slides down the plane, and if not, find the angle of tip of its

base AB.

u

ms = 0.3.f = 10°

r A

B

θ

φ

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The semicylinder of mass m and radius r lies on the rough

inclined plane If the inclination determine the

smallest coefficient of static friction which will prevent the

semicylinder from slipping

F = msN

N - 9.81m cos 15° = 0 N = 9.476ma+ ©Fy¿ = 0;

F - 9.81m sin 15° = 0 F = 2.539m+Q©Fx¿ = 0;

r A

B

θ

φ

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30

The coefficient of static friction between the 150-kg crate

and the ground is , while the coefficient of static

friction between the 80-kg man’s shoes and the ground is

Determine if the man can move the crate

msœ = 0.4

ms = 0.3

SOLUTION

Free - Body Diagram: Since P tends to move the crate to the right, the frictional

force FC will act to the left as indicated on the free - body diagram shown in Fig a.

Since the crate is required to be on the verge of sliding the magnitude of FCcan be

computed using the friction formula, i.e As indicated on the

free - body diagram of the man shown in Fig b, the frictional force F macts to the

right since force P has the tendency to cause the man to slip to the left.

Equations of Equilibrium: Referring to Fig a,

Solving,

Using the result of P and referring to Fig b, we have

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Free - Body Diagram: Since force P tends to move the crate to the right, the

frictional force FCwill act to the left as indicated on the free - body diagram shown

in Fig a Since the crate is required to be on the verge of sliding,

As indicated on the free - body diagram of the man shown in

Fig b, the frictional force F macts to the right since force P has the tendency to cause

the man to slip to the left

Equations of Equilibrium: Referring to Fig a,

Solving yields

Using the result of P and referring to Fig b,

Thus, the required minimum coefficient of static friction between the man’s shoes

and the ground is given by

If the coefficient of static friction between the crate and the

ground is , determine the minimum coefficient of

static friction between the man’s shoes and the ground so

that the man can move the crate

ms = 0.3

30

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The thin rod has a weight W and rests against the floor and

wall for which the coefficients of static friction are and

, respectively Determine the smallest value of for

which the rod will not move

Friction: If the rod is on the verge of moving, slipping will have to occur at points A

and B Hence, and Substituting these values into Eqs (1),

(2), and (3) and solving we have

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4 ft

1 ft

The 80-lb boy stands on the beam and pulls on the cord with

a force large enough to just cause him to slip If the

coefficient of static friction between his shoes and the beam

is , determine the reactions at A and B The

beam is uniform and has a weight of 100 lb Neglect the size

of the pulleys and the thickness of the beam

(ms)D = 0.4

SOLUTION

Equations of Equilibrium and Friction: When the boy is on the verge of slipping,

+ ©MB = 0; 100(6.5) + 96.0(8) - 41.6a5

13b(13)

FD = 0.4(96.0) = 38.4 lb

T = 41.6 lb ND = 96.0 lb:+ ©Fx = 0; 0.4ND - Ta1213b = 0

+ c ©Fy = 0; ND- Ta135b - 80 = 0

FD = (ms)DND = 0.4ND

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SOLUTION

Equations of Equilibrium and Friction: From FBD (a),

Therefore, the friction force developed is

+ ©MB = 0; 100(6.5)+ 95.38(8) - 40a 5

13b(13)

FD = 36.92 lb = 36.9 lb(FD)max = (ms)ND = 0.4(95.38) = 38.15 lb 7 FD

:+ ©Fx = 0; FD - 40a1213b = 0 FD = 36.92 lb

+ c ©Fy = 0; ND - 40a135b - 80 = 0 ND = 95.38 lb

The 80-lb boy stands on the beam and pulls with a force of

40 lb If , determine the frictional force between

his shoes and the beam and the reactions at A and B The

beam is uniform and has a weight of 100 lb Neglect the size

of the pulleys and the thickness of the beam

4 ft

1 ft

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Determine the smallest force the man must exert on the

rope in order to move the 80-kg crate Also, what is the

angle at this moment? The coefficient of static friction

between the crate and the floor is u ms = 0.3

uSOLUTION

Crate:

(1) (2)

+ c ©Fy = 0; T sin 30° + T sin 45° - T¿ cos u = 0

:+ ©Fx = 0; -T cos 30° + T cos 45° + T¿ sin u = 0

+ c ©Fy = 0; NC + T¿ cos u - 80(9.81) = 0

:+ ©Fx = 0; 0.3NC - T¿ sin u = 0

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From FBD (b),

(3) (4)

Friction: If block A and B are on the verge to move, slipping would have to occur

Substituting these values into Eqs (1), (2),(3) and (4) and solving, we have

Two blocks A and B have a weight of 10 lb and 6 lb,

respectively They are resting on the incline for which the

coefficients of static friction are and

Determine the incline angle for which both blocks begin

to slide Also find the required stretch or compression in the

connecting spring for this to occur The spring has a stiffness

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