978-0133915426 Chapter 17 Part 3

17–41.

The smooth 180-lb pipe has a length of 20 ft and a negligible

diameter. It is carried on a truck as shown. If the truck

accelerates at a

=

5 ft

>

s

2

, determine the normal reaction at

A and the horizontal and vertical components of force

which the truck exerts on the pipe at B.

SOLUTION

S

+Σ

F

x=

ma

x

; Bx–NA

a5

13 b

=

180

32.2

(5)

+

c

Σ

Fy=0; By–180 +NA

a12

13 b

=0

a+

ΣMB=Σ(Mk)B

; –180(10)

a

12

13 b

+NA(13) =–

180

32.2

(5)(10)

a

5

13 b

Solving,

Bx=73.9 lb

Ans.

B

y

=69.7 lb

Ans.

NA=120 lb

Ans.

B

A

20 ft

5 ft

12 ft

Bx=73.9 lb

17–42.

The uniform crate has a mass of 50 kg and rests on the cart

having an inclined surface. Determine the smallest

acceleration that will cause the crate either to tip or slip

relative to the cart. What is the magnitude of this

acceleration? The coefficient of static friction between the

crate and the cart is .ms=0.5

SOLUTION

a

(1)

(2)

(3)

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

Ans.

Since , then crate will not tip.Thus, the crate slips.Ans.x

6

0.3 m

a=2.01 m>s2

N=447.81 N

x=0.250 m

R+©F

x¿=m(aG)x¿; 50(9.81) sin 15° –0.5N=-50acos 15°

+Q©F

y¿=m(aG)y¿;N–50(9.81) cos 15° =-50asin 15°

=50acos 15°(0.5) +50asin 15°(x)

+©MA=©(Mk)A; 50(9.81) cos 15°(x)–50(9.81) sin 15°(0.5)

1m

0.6 m

F

17–43.

1 ft 1 ft

P

4 ft

G

Determine the acceleration of the 150-lb cabinet and the

normal reaction under the legs Aand Bif .The

coefficients of static and kinetic friction between the

cabinet and the plane are and ,

respectively.The cabinet’s center of gravity is located at G.

mk=0.15ms=0.2

P=35 lb

*17–44.

L

A

a

u

The uniform bar of mass mis pin connected to the collar,

which slides along the smooth horizontal rod. If the collar is

given a constant acceleration of a, determine the bar’s

inclination angle Neglect the collar’s mass. u.

SOLUTION

17–45.

The drop gate at the end of the trailer has a mass

of 1.25 Mg and mass center at G. If it is supported by the

cable AB and hinge at C, determine the tension in the cable

when the truck begins to accelerate at 5 m

>

s2. Also, what

are the horizontal and vertical components of reaction at

the hinge C?

B

C

$

30

1.5 m

1 m

45

G

SOLUTION

17–46.

The drop gate at the end of the trailer has a mass of 1.25 Mg

and mass center at G. If it is supported by the

cable AB and hinge at C, determine the maximum

deceleration of the truck so that the gate does not begin to

rotate forward. What are the horizontal and vertical

components of reaction at the hinge C?

SOLUTION

B

C

$

30

1.5 m

1 m

45

G

17–47.

SOLUTION

a

Ans.

Writing the force equations of motion along the xand yaxes,

Ans.

Ans.N

A=400 lb

+c©F

y=m(aG)y;N

A–250 –150 =0

F

A=248.45 lb =248 lb

;

+©F

x=m(aG)x;F

A=150

32.2 (20) +250

32.2 (20)

hmax =3.163 ft =3.16 ft

+©MA=(Mk)A; 250(1.5) +150(0.5) =150

32.2(20)(hmax)+250

32.2 (20)(1)

The snowmobile has a weight of 250 lb,centered at ,while

the rider has a weight of 150 lb,centered at .If the

acceleration is ,determine the maximum height h

of of the rider so that the snowmobile’s front skid does not

lift off the ground. Also,what are the traction (horizontal)

force and normal reaction under the rear tracks at A?

G2

a=20 ft>s2

G2

G1

a

0.5 ft

G1

G2

1ft

h

A

*17–48.

The snowmobile has a weight of 250 lb,centered at ,while

the rider has a weight of 150 lb,centered at .If ,

determine the snowmobile’s maximum permissible

acceleration aso that its front skid does not lift off the

ground. Also,find the traction (horizontal) force and the

normal reaction under the rear tracks at A.

h=3ftG2

G1

SOLUTION

a

Ans.

Writing the force equations of motion along the xand yaxes and using this result,

we have

Ans.

Ans.N

A=400 lb

+c©F

y=m(aG)y;N

A–150 –250 =0

F

A=257.14 lb =257 lb

;

+©F

x=m(aG)x;F

A=150

32.2 (20.7) +250

32.2 (20.7)

amax =20.7 ft>s2

+©MA=(Mk)A; 250(1.5) +150(0.5) =a150

32.2 amax b(3) +a250

32.2 amax b(1)

a

0.5 ft

G1

G2

1ft

h

A

17–49.

P

30

1 m

C

B

A

If the cart’s mass is 30 kg and it is subjected to a horizontal

force of , determine the tension in cord AB and

the horizontal and vertical components of reaction on end

Cof the uniform 15-kg rod BC.

P=90 N

SOLUTION

17–50.

P

30

1 m

C

B

A

If the cart’s mass is 30 kg, determine the horizontal force P

that should be applied to the cart so that the cord AB just

becomes slack. The uniform rod BC has a mass of 15 kg.

SOLUTION

17–51.

The pipe has a mass of 800 kg and is being towed behind the

truck. If the acceleration of the truck is

determine the angle and the tension in the cable.The

coefficient of kinetic friction between the pipe and the

ground is mk=0.1.

u

at=0.5 m>s2,

SOLUTION

a

Ans.

u

=45° –

f

=18.6°

sin f=0.1(6770.9)

1523.24

f=26.39°

T=1523.24 N =1.52 kN

NC=6770.9N

+©MG=0; –0.1NC(0.4) +Tsin f(0.4) =0

+c©Fy=may;NC–800(9.81) +Tsin 45° =0

:

45

0.4 m

G

A

B

a

t

u

*17–52.

SOLUTION

a

Ans.

Ans.a=1.33 m>s2

T=2382 N =2.38 kN

NC=6161 N

+©MG=0; Tsin 15°(0.4) –0.1NC(0.4) =0

+c©Fy=may;NC–800(9.81) +Tsin 45° =0

:

T

h

e p

i

pe

h

as a mass of 800

k

g an

d

i

s

b

e

i

ng towe

d

b

e

hi

n

d

a

truck. If the angle determine the acceleration of the

truck and the tension in the cable.The coefficient of kinetic

friction between the pipe and the ground is mk=0.1.

u=30°,

45

0.4 m

G

A

B

a

t

u

843

17–53.

Ans:

a

=9.67 rad>s2

The crate C has a weight of 150 lb and rests on the truck

elevator for which the coefficient of static friction is

m

s=0.4.

Determine the largest initial angular acceleration

a,

starting from rest, which the parallel links AB and DE

can have without causing the crate to slip. No tipping occurs.

SOLUTION

150

B

AC

D

2 ft

E

30



a

17–54.

The crate C has a weight of 150 lb and rests on the truck

elevator. Determine the initial friction and normal force of

the elevator on the crate if the parallel links are given an

angular acceleration a

=

2 rad

>

s

2

starting from rest.

SOLUTION

B

AC

D

2 ft

E

30



a

845

17–55.

The 100-kg uniform crate C rests on the elevator floor

where the coefficient of static friction is m

s=0.4.

Determine the largest initial angular acceleration

a,

starting from rest at u

=90°,

without causing the crate

to slip. No tipping occurs.

Ans:

a

=

2.62 rad

>

s

2

SOLUTION

1.2 m

0.6 m

1.5 m

C

B

D

E

A

a

uu

846

*17–56.

The two uniform 4-kg bars DC and EF are fixed (welded)

together at E. Determine the normal force

NE,

shear force

VE,

and moment

ME,

which DC exerts on EF at E if at the

instant

u=60°

BC has an angular velocity v

=2 rad>s

and

an angular acceleration a

=

4 rad

>

s

2

asshown.

a

 4 rad/s2

E

F

2 m

D

C

1.5 m

SOLUTION

NE=27.53 N =27.5 N

Ans:

VE=43.7 N

NE=27.5 N

ME=32.8 N #m

17–57.

The 10-kg wheel has a radius of gyration If

the wheel is subjected to a moment where t

is in seconds, determine its angular velocity when

starting from rest. Also, compute the reactions which the

fixed pin Aexerts on the wheel during the motion.

t=3s

M=15t2N#m,

k

A

=200 mm.

SOLUTION

c

Ans.

Ans.Ay

=

98.1 N

Ax=0

v=56.2 rad>s

v=L3

0

12.5tdt=12.5

2(3)2

a=dv

dt =12.5t

:

A

M

848

17–58.

The uniform 24-kg plate is released from rest at the position

shown. Determine its initial angular acceleration and the

horizontal and vertical reactions at the pin A.

a

=

14.7 rad

>

s

2

Ax=88.3 N

A

y

=147 N

SOLUTION

0.5 m

A

0.5 m

849

17–59.

A

B

L

3

2

3L

u

The uniform slender rod has a mass m. If it is released from

rest when , determine the magnitude of the reactive

force exerted on it by pin Bwhen .u=90°

u=0°

SOLUTION

2

*17–60.

The bent rod has a mass of 2 kg

>

m. If it is released from rest

in the position shown, determine its initial angular

acceleration and the horizontal and vertical components of

reaction at A.

1.5 m

A

C

SOLUTION