Mechanical Engineering Chapter 8 Homework The coefficient of static friction between the belt

848

8–102.

50 mm

20 mm

A

B

C

D

50 mm

45°

30°

M= 0.8 N⋅m

The belt on the portable dryer wraps around the drum D,

idler pulley A, and motor pulley B. If the motor can develop

a maximum torque of determine the

smallest spring tension required to hold the belt from

slipping.The coefficient of static friction between the belt

and the drum and motor pulley is ms=0.3.

M=0.80 N #m,

SOLUTION

a

–T

110.022+T

210.022–0.8 =0+©MB=0;

Ans:

8–103.

SOLUTION

For block A and B: Assuming block B does not slip

For block B:

Solving Eqs. (1) and (2) yields:

FB=40.32 lb NB=17.25 lb

Since FB=40.32 lb 7mNB=0.6(17.25) =10.35 lb, slipping does occur between

A and B. Therefore, the assumption is no good.

Blocks A and B weigh 50 lb and 30 lb, respectively. Using

the coefﬁ cients of static friction indicated, determine the

greatest weight of block D without causing motion.

A

B

C

D

m0.5

mBA 0.6

mAC 0.4

20

Ans:

*8–104.

SOLUTION

Equations of Equilibrium: From FBD (a),

a(1)

From FBD (b),

T

211002+T

112002–196.211002=0+©MC=0;

The 20-kg motor has a center of gravity at Gand is pin-

connected at Cto maintain a tension in the drive belt.

M

100 mm

B

Ans:

8–105.

A

B

D

E

C

uu

A 10-kg cylinder D, which is attached to a small pulley B,is

placed on the cord as shown. Determine the largest angle

so that the cord does not slip over the peg at C.The cylinder

at Ealso has a mass of 10 kg, and the coefficient of static

friction between the cord and the peg is

SOLUTION

Since pully Bis smooth, the tension in the cord between pegs Aand Cremains

constant. Referring to the free-body diagram of the joint Bshown in Fig.a, we have

In the case where cylinder Eis on the verge of ascending, and

Here,,Fig.b.Thus,

Solving by trial and error, yields

In the case where cylinder E is on the verge of descending,T2= 10(9.81) N and

T1= . Here,.Thus,

Solving by trial and error, yields

Thus, the range of at which the wire does not slip over peg Cis

u

p

2+u

49.05

sin u

p

2+uT

1=10(9.81) N.

T

2=T=49.05

sin u

ms=0.1.

X

Ans:

852

8–106.

SOLUTION

Equations of Equilibrium: From FBD (a),

From FBD (b),

A conveyer belt is used to transfer granular material and

the frictional resistance on the top of the belt is

Determine the smallest stretch of the spring attached to the

moveable axle of the idle pulley Bso that the belt does not

slip at the drive pulley Awhen the torque Mis applied.

What minimum torque Mis required to keep the belt

moving? The coefficient of static friction between the belt

and the wheel at Ais ms=0.2.

F=500 N.

0.1 m 0.1 m

F= 500 N

k=4kN/m

A

B

M

Ans:

853

8–107.

The collar bearing uniformly supports an axial force of

P

=

5 kN. If the coefficient of static friction is m

s=0.3

,

determine the smallest torque M required to overcome

friction.

Ans:

SOLUTION

s=0.3

200 mm

150 mm

P

M

854

*8–108.

The collar bearing uniformly supports an axial force of

P

=

8 kN. If a torque of M

=

200 N

#

m is applied to the

shaft and causes it to rotate at constant velocity, determine

the coefficient of kinetic friction at the surface of contact.

Ans:

SOLUTION

200 mm

150 mm

P

M

855

8–109.

The floor-polishing machine rotates at a constant angular

velocity. If it has a weight of 80 lb. determine the couple

forces Fthe operator must apply to the handles to hold the

machine stationary.The coefficient of kinetic friction

between the floor and brush is Assume the brush

exerts a uniform pressure on the floor.

mk=0.3.

SOLUTION

1.5 ft

Ans:

856

8–110.

The double-collar bearing issubjectedto an axial force

Assuming that collar Asupports 0.75Pand

collar Bsupports 0.25P,both with auniform distribution of

pressure,determine the maximum frictional moment M

that may be resisted by the bearing.Take for both

collars.

ms=0.2

P=4 kN.

SOLUTION

Ans:

8–111.

The double-collar bearing is subjected to an axial force

P

=

16 kN. Assuming that collar A supports 0.75P and

collar B supports 0.25P, both with a uniform distribution of

pressure, determine the smallest torque M that must be

applied to overcome friction. Take m

s=0.2

for both collars.

P

30 mm

50 mm

A

B

M

75 mm

100 mm

SOLUTION

858

*8–112.

The pivot bearing is subjected to a pressure distribution at

its surface of contact which varies as shown. If the

coefficient of static friction is determine the torque M

required to overcome friction if the shaft supports an axial

force P.

m,

SOLUTION

P

M

R

Ans:

859

8–113.

SOLUTION

The conical bearing is subjected to a constant pressure

distribution at its surface of contact. If the coefficient of

static friction is determine the torque Mrequired to

overcome friction if the shaft supports an axial force P.

ms,

P

M

R

Ans:

8–114.

T

h

e4–

i

n.-

di

ameter s

h

aft

i

s

h

e

ld i

nt

h

e

h

o

l

esuc

h

t

h

at t

h

e

normal pressure acting around the shaft varies linearly with

its depth as shown. Determine the frictional torque that

must be overcome to rotate the shaft. Take ms=0.2.

SOLUTION

861

8–115.

200 mm

F

A

D

P

100 mm

125 mm

150 mm

S

SOLUTION

The plate clutch consists of a flat plate A that slides over the

rotating shaft S. The shaft is fixed to the driving plate gear B.

If the gear C, which is in mesh with B, is subjected to a

torque of M = 0.8N # m, determine the smallest force P, that

must be applied via the control arm, to stop the rotation.

The coefficient of static friction between the plates A and D

is ms = 0.4. Assume the bearing pressure between A and D to

be uniform.

Ans:

*8–116.

The collar fits loosely around a fixed shaft that has a radius

of 2 in. If the coefficient of kinetic friction between the shaft

and the collar is determine the force Pon the

horizontal segment of the belt so that the collar rotates

counterclockwise with a constant angular velocity. Assume

that the belt does not slip on the collar; rather, the collar

slips on the shaft. Neglect the weight and thickness of the

belt and collar.The radius, measured from the center

of the collar to the mean thickness of the belt, is 2.25 in.

mk=0.3,

SOLUTION

fk=tan–1mk=tan–10.3 =16.699°

20 lb

P

2in.

2.25 in.

Ans:

863

8–117.

SOLUTION

The collar fits loosely around a fixed shaft that has a radius of

2in. If the coefficient of kinetic friction between the shaft and

the collar is determine the force Pon the horizontal

segment of the belt so that the collar rotates clockwise with a

constant angular velocity.Assume that the belt does not slip

on the collar; rather,the collar slips on the shaft. Neglect the

weight and thickness of the belt and collar.The radius,

measured from the center of the collar to the mean thickness

of the belt, is 2.25 in.

mk=0.3,

P

2in.

2.25 in.