Dynamics 14th edition by r c hibbeler chapter 08

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-b

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) - P2 = 0

NB = 2427.78 lb = 2.43 kip+ ©MC = 0 2NB (9) + 400(2.5) - 7500(5) - 600(12) = 0

The tractor exerts a towing force Determine

the normal reactions at each of the two front and two rear

tires and the tractive frictional force F on each rear tire

needed to pull the load forward at constant velocity The

tractor has a weight of 7500 lb and a center of gravity

located at An additional weight of 600 lb is added to its

front having a center of gravity at Take 0.4 The

front wheels are free to roll

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

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?

N A = 16.5 kN

N B = 42.3 kN

It does not move

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

The automobile has a mass of 2 Mg and center of mass at G

Determine the towing force F required to move the car if

the back brakes are locked, and the front wheels are free to

C

B

1.50 m

1 m

The automobile has a mass of 2 Mg and center of mass at G

Determine the towing force F required to move the car

Both the front and rear brakes are locked Take ms = 0.3

C

B

1.50 m

1 m

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

Ans:

NoYes

No

Yes

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 holdthe wheel stationary when the force applied to the lever is (a)P = 30 N, (b) P = 70 N

Lever,a

P = 30 N 6 34.26 N No

PReqd. = 34.26 N+ ©MA = 0; PReqd.(0.6) - 111.1(0.2) + 33.333(0.05) = 0

5 N m

50 mm

Ans:

NoYes

The pipe of weight W is to be pulled up the inclined plane of

slope using a force P If P acts at an angle , show that for

of static friction;u = tan-1 ms

u

a + u)>cos(f - u

P = W

fa

SOLUTION

Q.E.D.

= W(cos ucos fcos usin a+ sin f+ sin usin ucos a) = Wcos(fsin(a- u)+ u)

P = W(sin acos f + tan u+ tan usin fcos a)+Q©Fx¿ = 0; Pcos f - Wsin a - tan u(Wcos a - Psin f) = 0

+a©Fy¿ = 0; N + Psin f - Wcos a = 0 N = Wcos a - Psin f

P

α

φ

P = W(sin acos f + tan u sin f+ tan u cos a)

+Q©Fx ¿ = 0; P cos f - W sin a - tan u(W cos a - P sin f) = 0

+a©Fy ¿ = 0; N + P sin f - W cos a = 0 N = W cos a - P sin f

Determine the angle at which the applied force P should

act on the pipe so that the magnitude of P is as small as

possible for pulling the pipe up the incline What is the

corresponding value of P? The pipe weighs W and the slope

is known Express the answer in terms of the angle of

kinetic friction,u = tan-1 mk

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

Ans:

W = 318 lb

W = 360 lb

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

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

Ans:

F CD= 3.05 kN

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

Ans:

u = 21.8°

The log has a coefficient of static friction of with

the ground and a weight of 40 lb/ft If a man can pull on the

rope with a maximum force of 80 lb, determine the greatest

length l of log he can drag.

l

Ans:

l = 26.7 ft

Free - Body Diagram Since the weight of the man tends to cause the friction pad A

to slide to the right, the frictional force FAmust act to the left as indicated on the

free - body diagram of the ladder, Fig a Here, the ladder is on the verge of slipping.

c u - 0.4 sin u = 0.3+ ©MB = 0; 180(10 cos u°) - 0.4(180)(10 sin u°) - 180(3) = 0

+ c ©Fy = 0; NA- 180 = 0 NA = 180 lb

FA = msNA

The 180-lb man climbs up the ladder and stops at the position

shown after he senses that the ladder is on the verge of

slipping Determine the inclination of the ladder if the

coefficient of static friction between the friction pad A and the

ground is .Assume the wall at B is smooth.The center

of gravity for the man is at G Neglect the weight of the ladder.

Ans:

u = 52.0°

The 180-lb man climbs up the ladder and stops at the position

shown after he senses that the ladder is on the verge of

slipping Determine the coefficient of static friction between

the friction pad at A and ground if the inclination of the ladder

is and the wall at B is smooth.The center of gravity for

the man is at G Neglect the weight of the ladder.

u = 60°

SOLUTION

Free - Body Diagram Since the weight of the man tends ot cause the friction pad A

to slide to the right, the frictional force FAmust act to the left as indicated on the

free - body diagram of the ladder, Fig a Here, the ladder is on the verge of slipping.

Ans:

ms = 0.231

The spool of wire having a weight of 300 lb rests on the

ground at B and against the wall at A Determine the force

P required to begin pulling the wire horizontally off the

spool The coefficient of static friction between the spool

and its points of contact is ms = 0.25

The spool of wire having a weight of 300 lb rests on the

ground at B and against the wall at A Determine the

normal force acting on the spool at A if P = 300 lb

The coefficient of static friction between the spool and the

ground at B is m s = 0.35 The wall at A is smooth.

Friction Since F B 6 (F B)max = ms N B = 0.35(300) = 105 lb, slipping will not occur

at B Thus, the spool will remain at rest.

The ring has a mass of 0.5 kg and is resting on the surface of

the table In an effort to move the ring a normal force P from

the finger is exerted on it If this force is directed towards the

ring’s center O as shown, determine its magnitude when the

ring is on the verge of slipping at A The coefficient of static

P

60

A

A man attempts to support a stack of books horizontally by

applying a compressive force of to the ends of

the stack with his hands If each book has a mass of 0.95 kg,

determine the greatest number of books that can be

supported in the stack The coefficient of static friction

between the man’s hands and a book is and

between any two books (ms)b = 0.4 (ms)h = 0.6

F = 120 N

SOLUTION

Equations of Equilibrium and Friction: Let be the number of books that are on

the verge of sliding together between the two books at the edge Thus,

From FBD (a),

Let n be the number of books are on the verge of sliding together in the stack

Thus, the maximum number of books can be supported in stack is

P

G

The tongs are used to lift the 150-kg crate, whose center of

mass is at G Determine the least coefficient of static

friction at the pivot blocks so that the crate can be lifted

SOLUTION

Free - Body Diagram Since the crate is suspended from the tongs, P must be equal

to the weight of the crate; i.e., as indicated on the free - body

diagram of joint H shown in Fig.a Since the crate is required to be on the verge of

slipping downward,FAand FBmust act upward so that and

as indicated on the free - body diagram of the crate shown in Fig c.

Equations of Equilibrium Referring to Fig a,

Ans:

ms = 0.595

The beam is supported by a pin at A and a roller at B which

has negligible weight and a radius of 15 mm If the coefficient

of static friction is mB = mC = 0.3, determine the largest

angle u of the incline so that the roller does not slip for any

force P applied to the beam.

The term in parentheses is zero when

From Eq (3), N C (cos 33.4° + 0.3 sin 33.4°) = N B

N C = N B

Since Eq (4) is satisfied for any value of N C , any value of P can act on the beam

Also, the roller is a “two-force member.”

FA= 6.25 lb:+ ©Fx = 0; 6.25 - FA = 0

NB = 6.25 lb+ ©MA= 0; 30 (5) - NB(24) = 0

The uniform thin pole has a weight of 30 lb and a length of

26 ft If it is placed against the smooth wall and on the rough

floor in the position , will it remain in this position

when it is released? The coefficient of static friction is

The uniform pole has a weight of 30 lb and a length of 26 ft.

Determine the maximum distance d it can be placed from

the smooth wall and not slip The coefficient of static

friction between the floor and the pole is ms = 0.3

Ans:

d = 13.4 ft

The block brake is used to stop the wheel from rotating

when the wheel is subjected to a couple moment M0 =

360 N#m If the coefficient of static friction between the

wheel and the block is ms = 0.6, determine the smallest

force P that should be applied.

A worker walks up the sloped roof that is defined by the

curve y = (5e 0.01x ) ft, where x is in feet Determine how

high h he can go without slipping The coefficient of static

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)

Equations of Equilibrium: Using the spring force formula, , from

FBD (a),

(1) (2)

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

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 angle which will cause motion of one of

the blocks What is the friction force under each of the

blocks when this occurs? The spring has a stiffness of

and is originally unstretched

Equations of Equilibrium: Since neither block A nor block B is moving yet,

the spring force From FBD (a),

(1) (2)

From FBD (b),

(3) (4)

Friction: Assuming block A is on the verge of slipping, then

(5)

Solving Eqs (1),(2),(3),(4), and (5) yields

Therefore, the above assumption is correct Thus

Determine the smallest force P that must be applied in

order to cause the 150-lb uniform crate to move The

coefficent of static friction between the crate and the floor

The man having a weight of 200 lb pushes horizontally on

the crate If the coefficient of static friction between the

450-lb crate and the floor is ms= 0.3 and between his shoes

and the floor is m′s = 0.6, determine if he can move the

F m = P = 135 lb x = 0.9 ft Since x < 1 ft, the crate indeed slides before tipping as assumed.

Also, since F m > (F m)max = ms ′N C = 0.6(200) = 120 lb, the man slips.

Thus he is not able to move the crate.

The uniform hoop of weight W is subjected to the horizontal

force P Determine the coefficient of static friction between

the hoop and the surface of A and B if the hoop is on the

If P = 12W, the quadratic term drops out, and then

W - P > 0 W > P Also, P > 0 Thus

Note: Choosing the larger value of ms in the quadratic solution leads to N A , F A < 0,

which is nonphysical Also, (ms)max = 1 For ms > 1, the hoop will tend to climb the

wall rather than rotate in place

Determine the maximum horizontal force P that can be

applied to the 30-lb hoop without causing it to rotate The coefficient of static friction between the hoop and the

surfaces A and B is m s = 0.2 Take r = 300 mm.

Determine the minimum force P needed to push the tube E

up the incline The force acts parallel to the plane, and the

coefficients of static friction at the contacting surfaces are

mA = 0.2, mB = 0.3, and mC = 0.4 The 100-kg roller and

40-kg tube each have a radius of 150 mm

Since F B 6 (F B)max = mB N B = 0.3(388.88) = 116.66 N and F C < (F C)max = mC N C

= 0.4(1061.15) = 424.46 N, slipping indeed will not occur at B and C Thus, the

assumption was correct

The coefficients of static and kinetic friction between the

drum and brake bar are and , respectively

and vertical components of reaction at the pin O Neglect

the weight and thickness of the brake The drum has a mass

point B and rotates Therefore, the coefficient of kinetic friction should be used.

FB = mkNB = 0.3NB

FB 7 (FB)max = msNB = 0.4(407.14) = 162.86 N

NB= 407.14 N+ ©MA = 0; 85(1.00) + 400(0.5) - NB(0.7) = 0

+ ©MO = 0 50 - FB(0.125) = 0 FB = 400 N

A

M P

Friction: When the drum is on the verge of rotating,

Substituting into Eq [1] yields

+ ©MO = 0 35 - FB(0.125) = 0 FB = 280 N

The coefficient of static friction between the drum and

determine the smallest force P that needs to be applied to

the brake bar in order to prevent the drum from rotating

Also determine the corresponding horizontal and vertical

components of reaction at pin O Neglect the weight and

thickness of the brake bar The drum has a mass of 25 kg

M= 35 N#m

ms = 0.4

A

M P

Determine the smallest coefficient of static friction at both

A and B needed to hold the uniform 100-lb bar

in  equilibrium Neglect the thickness of the bar