7.4 Flat Belts

7.4 Flat Belts Example 1, page 1 of 2

1. Determine the minimum number of turns of rope that will allow

the 5-lb force to support the 600-lb block, if the coefficient of

static friction is 0.15.

}n turns

600 lb

5 lb

5 lb

Tensions in the rope

Impending motion

(The motion can't be up, since the 5-lb force is too

small to lift the 600-lb block.)

600 lb

1

2

7.4 Flat Belts Example 1, page 2 of 2

Since = 0.15, Eq. 1 becomes

600 lb = (5 lb) e

Solving gives = 31.917 radians. If n is the

number of turns of rope, then the number of

radians is 2 n. Equating this to gives an

equation for n:

2 n =

= 31.917 rad

Solving for n gives

n = 5.08 turns

Rounding off to the next higher integer then gives

n = 6 Ans.

Apply the equation for belt friction:

T2 = T1e (1)

In this equation, T2 is the tension in the direction

of impending motion

T2 = 600 lb

The other tension, T1, is in the direction opposite

the impending motion, so

T1 = 5 lb

3 4

7.4 Flat Belts Example 2, page 1 of 2

2. If the coefficient of static friction between the rope and the

fixed circular drums A and B is 0.2, determine the largest value

of the force P that can be applied without moving the 150-lb

weight upwards.

P

150 lb

A

B

60°

Rope is horizontal.

Impending motion

(The rope must move to

the right if the 150-lb

weight moves up)

150 lb

T AB

=

A

Tensions in the rope on either side of drum A

1

Because slip impends between the rope and the drum, we can

apply the equation for belt friction:

T 2 = T

1e (1)

where T2 is the tension in the direction of impending motion.

In our particular problem,

T2 = TAB

and the tension opposing the impending motion is

T1 = 150 lb

Using these results and = 0.2 and = 2 in Eq. 1 gives

T AB = (150)e

= 205.4 lb

0.2( /2)

2

7.4 Flat Belts Example 2, page 2 of 2

B

60°

Impending

motion

P

T AB = 205.4 lb

Flat-belt friction equation:

T 2 = T

1e

so,

P = (205.4 lb)e (2)

4

60

60°

30°

60°

Geometry5

= 60° = rad

6 Using = 60° = rad in Eq. 2 gives

P = (205.4)e

= 253 lb Ans.

/

Tensions in the rope on

either side of drum B

3

7.4 Flat Belts Example 3, page 1 of 3

6 in.

3. Determine the smallest force P applied to the handle of the

band brake that will prevent the drum from rotating when the

15 lb·ft moment is applied. The coefficient of static friction

is 0.25, and the weight of lever arm ABC can be neglected.

AB C

P

D

x

D y

T B

T A

6 in.

D

Impending slip of

band relative to

drum (An observer

on the drum would

see the belt move

down.)

Free-body diagram of drum

3

+

Equilibrium equation for drum

M D = 0: T

A(6 in.) T B(6 in.) 180 lb·in. = 0 (1)

2

1

80

80

o

o

15 lb-ft

D

15 lb·ft = 15 lb·ft 12 in./ft

= 180 lb·in.

15 in.10 in.

7.4 Flat Belts Example 3, page 2 of 3

80°80°

10°

10°

Geometry5The belt-friction equation is in general

T 2 = T

1e (2)

where T2 is the tension in the direction

of impending slip. In our particular

problem,

T2 = TA

and the tension opposing the impending

motion is

T1 = TB

Eq. (2) becomes

T A = T

Be (3)

Eq. 3 becomes, with 3.316 and 0.25

TA = TBe0.25(3.316)

Solving Eqs. 1 and 4 simultaneously gives

TA = 53.237 lb

TB = 23.237 lb

4

= 180° + 10°

= 190°

= (190/180) rad

= 3.316 rad

7.4 Flat Belts Example 3, page 3 of 3 +

7

Ax

T A = 53.237 lb

T B cos 80°

6 Free-body diagram of lever arm ABC

15 in.10 in.

P

BA

Equilibrium equation for lever arm

M A = 0: [(23.237 lb) sin 80°](10 in.) P(10 in. + 15 in.) = 0 (5)

Solving gives

P = 9.15 lb Ans.

Ay

T B sin 80° = (23.237 lb) sin 80°

7.4 Flat Belts Example 4, page 1 of 3

4. The uniform beam ABC weighs 40 lb. The coefficient

of static friction between the cord and the fixed drum D is

0.3. Determine the smallest value of the weight W for

which the beam will remain horizontal.

Free-body diagram of block B

(Tension in the cord)

If end A of the beam

is about to move

down, then block B

is about to lose

contact with the

beam. Thus the

normal force N is zero.

Equilibrium equation for block B

F y = 0: T

B W = 0 (1)

D

A B W C

4 ft 4 ft

W

N = 0

TB

1

2

3+

7.4 Flat Belts Example 4, page 2 of 3

Free-body diagram of beam

T A is the tension in the cord.

The 40-lb weight of the beam acts

through the center of the span.

Equilibrium equation for the beam

M C = 0: (40 lb)(4 ft) T

A(4 ft + 4 ft) = 0 (2)

Solving gives

TA = 20 lbBecause the cord is on the verge of slipping, we can

apply the equation for belt friction:

T 2 = T

1e (3)

where T2 = 20 lb is the tension in the direction of

impending slip and T1 = TB is the tension opposite the

direction of impending motion

Tensions in

cord on either

side of drum

CBA

4 ft 4 ft

TA = 20 lb

Impending

motion

40 lb

N = 0 Cy

Cx

y

x

4

5

6

7

8

9

+

TB

7.4 Flat Belts Example 4, page 3 of 3

TB

10

TA

=

Geometry

With and 0.3, Eq. 3 becomes

T 2 = T

1e (Eq. 3 repeated)

20 lb 0.3

Solving gives

TB = 7.79 lb

Eq. 1 then gives

W = TB = 7.79 lb Ans.

11

7.4 Flat Belts Example 5, page 1 of 4

80 lb

A

B

W

70°50°

80 lb

A

TA

NA

fA

y

x

5. Determine the largest value W of the weight

of block B for which neither block will move.

The coefficients of static friction are 0.2 between

the blocks and the planes, and 0.25 between the

cord and the drum.

Free-body diagram of block A

Impending motion of

block A (Since we are

to determine the

"largest value of the

weight" of block B,

block B must be about

to move down the

plane. Thus block A

must be about to

move up the plane.)

70°

Equilibrium equations for block A:

Fx = 0: TA fA (80 lb) sin (1)

F y

= 0: NA (80 lb) cos = 0 (2)

Slip impends, so

fA = fA-max

= NA

= (0.2)NA (3)

+

+

1

2

3

7.4 Flat Belts Example 5, page 2 of 4

Geometry

Solving Eqs. 1, 2, and 3, with = 70° gives

T A = 80.65 lb

f A = 5.47 lb

N A = 27.36 lb

Tensions in cord on either side of drum

Impending motion

(Block B moves down plane)

Because the cord is about to slip over the drum, we can

apply the equation for belt friction:

T2 = T1 e (4)

where T2 = TB is the tension in the direction of impending

slip, and T1 = 80.65 lb is the tension opposite to the direction

of impending motion.

A

TA = 80.65 lb

TB

4

5

6

7

drum

20°

20°

= 70°

70°

7.4 Flat Belts Example 5, page 3 of 4

Substituting = 2 /3, T2 = TB , T1 = 80.65 lb,

and drum 0.25 in Eq. 4 gives

T B = (80.65 lb) e

Evaluating the right hand side yields

T B = 136.1 lb

9

8 Geometry

= 70° + 50° = 120°

= rad

70°

20°

70° 50°

50°70°

50°

40°

23

7.4 Flat Belts Example 5, page 4 of 4

Free-body diagram of block B Equilibrium equations for block B:

F x' =0: N

B W cos = 0 (5)

F y' = 0: 136.1 lb + f

B W sin = 0 (6)

f B = fB-max

= N B

= (0.2)N B (7)

Solving Eqs. 5, 6, and 7, with = 50° gives

f B = 27.4 lb

N B = 137.2 lb

W = 213 lb Ans.

Geometry

B

= 50°

40°

50°

B

50°

NB

TB = 136.1 lb

WImpending

motion

y

x

fB

10 12

11

13

+

+

7.4 Flat Belts Example 6, page 1 of 4

6. A motor attached to pulley A drives the pulley clockwise with a 200 lb-in. torque.

The flat belt then overcomes the resisting torque T at pulley B and rotates the pulley B

clockwise. Determine the minimum tension that can exist in the belt without causing

the belt to slip at pulley A. Also determine the corresponding resisting torque T. The

coefficient of static friction between the belt and the pulleys is 0.3.

22 in.

B

T

7 in.

Driven pulley

A

200 lb·in.

Driving pulley

4 in.

7.4 Flat Belts Example 6, page 2 of 4

1

2

Equilibrium equation for pulley A

M A = 0: TD(4 in.) TC(4 in.) 200 lb·in. = 0 (1)

+

3Free-body diagram of pulley A

Impending motion of belt relative to pulley

(Since pulley A is driven by a clockwise

torque, the pulley would slip in a clockwise

sense relative to the belt. Thus an observer

at point D on the pulley would see the belt

move in the direction shown.)

200 lb·in.

A

TC

TD

A y

A x

4 in.

C

D

Because the belt is about to slip on the pulley, the belt friction

equation applies:

T 2 = T

1e

where T2 = TD, the tension in the direction of impending motion,

and T1 = TC, so

TD = TC e (2)

7.4 Flat Belts Example 6, page 3 of 4

4 Geometry

4 in.

4 in.

22 in.

Radius = 7 in.

A B

C

Solving Eqs. 1 and 3 simultaneously gives

TC = 36.65 lb

TD = 86.65 lb Ans.

5

B x

B y

T

TC = 36.65 lb

TD = 86.65 lb

7 in.

Free-body diagram of pulley B6

Equilibrium equation for pulley B

M B = 0: T + (36.65 lb)7 in. (86.65 lb)7 in. = 0

Solving gives

T = 350 lb·in. Ans.

+

7

From triangle ABC,

cos ( 2 ) =

3 in.22 in.

TD = TC e0.3(2.868) (3)

Solving gives = 164.33° = 2.868 rad.

So Eq. 2, with = 0.3, becomes

7 in. 4 in. = 3 in.

7.4 Flat Belts Example 6, page 4 of 4

BA

Finally, we must check that pulley B does not slip. But this follows

from the observation that pulley B has a larger angle of wrap than

pulley A. Thus it must be able to carry a maximum possible tension

larger than the 86.5 lb maximum tension that pulley A carries.

8

'

86.65 lb tension

36.65 lb tension

More precisely, because

' >

we know that

e ' > e .

Multiplying through by 36.65 lb gives

(36.65 lb)e ' > (36.65 lb)e

9

Tmax-B , the maximum

possible tension that

pulley B could support

without slipping.

86.65 lb, by Eq. 2

Thus Tmax-B > 86.65 lb and pulley B won't slip.

7.4 Flat Belts Example 7, page 1 of 2

W

TETD

C

C

100 lb

50 lb

E

7. If the coefficient of static friction between the fixed

drums D and E and the ropes is 0.35, determine the

largest weight W that can be supported.

Free-body diagram of block C

Equilibrium equation for block C

F y

= 0: TE + TD W = 0 (1)

1

2

+

W

A

D

B

7.4 Flat Belts Example 7, page 2 of 2

D

E=

100 lbImpending

motion

50 lbTE

Rope tensions acting on drum D

Impending motion (Since we are to determine the largest

value of weight of block C that can be supported, block C

must be about to move down, so the rope connected to C

must also be about to move down.)

Because slip is about to occur, the belt-friction equation

applies:

T 2 = T

1e

Here, T2 = TD, T1 = 100 lb, 0.35, and so

T D = (100 lb)e

= 300.3 lb (2)

Using the results of Eqs. 2 and 3 in

Eq. 1 gives

W = 450 lb Ans.

Flat-belt friction equation:

T 2 = T

1e or,

T E = (50 lb)e0.35

= 150.1 lb (3)

3

4

7

8

5

Rope tensions acting on

drum E (not drum D)6

=

TD

7.4 Flat Belts Example 8, page 1 of 3

80 mm

80 mm

6 N·mT C

A

B

Driving pulleyDriven pulley

8. Pulley A is rotating under the action of a 6 N-m torque. This motion is transmitted

through a flat belt to drive pulley B, which is in turn acted upon by a resisting torque T

(the "load" on pulley B). The coefficient of static friction between the belt and the

pulleys is 0.45. Determine a) the maximum possible value of T, b) the maximum force

in the belt, and c) the corresponding force required in the spring C.

7.4 Flat Belts Example 8, page 2 of 3

6 N-m

Ay

Ax

= A

T1

T2

Free-body diagram of pulley A

Impending motion (Since the driving torque acting

on pulley A is clockwise, if slip is about to occur,

then pulley A will slip clockwise relative to the

belt. An observer located on point D on pulley A

would see the belt move in the direction shown.)

Equilibrium equations for pulley A

F x = 0: A

x T 1 T

2 = 0 (1)

Fy = 0: Ay = 0 (2)

M A = 0: T

2(0.08 m) T 1(0.08 m) 6 N·m = 0 (3)

Flat-belt friction equation:

T 2 = T

1e

= T 1e (4)

Solving Eqs. 1-4 simultaneously gives

A x = 123.21 N

Ay = 0

T 1 = 24.11 N

T 2 = 99.11 N Ans.

1

2

3

4

+

+80 mm

+

D

7.4 Flat Belts Example 8, page 3 of 3

B

T2 = 99.11 N

T1 = 24.11 N

By

Bx

FspringAx = 123.21 N

Free-body diagram of member AC

Equilibrium equations for member AC

F x = 0: F

spring 123.21 N = 0 (4)

Solving gives

F

spring = 123.21 N Ans.

Equilibrium equations for pulley B

M B = 0: T + (24.11 N)(0.08 m) 99 (0.08 m) = 0

Solving gives

T = 6 N·m Ans.

That is, the maximum resisting torque equals the driving

torque.

Free-body diagram of pulley B5

6

7

8

+

+

T80 mm