7.1 Friction: Basic Applications

7.1 Friction: Basic Applications Example 1, page 1 of 2 1. The uniform ladder is 2-m long and makes an angle of = 60 with the floor. If the wall at B is smooth and the coefficient of static friction at A is A = 0.3, determine if the ladder can remain in the position shown.

B Mass of ladder = 10 kg

1

Free-body diagram

2

No friction force is present because the wall is smooth.

A

NB

+

Fx = 0: NB

fA = 0

+

(2 m)(sin 60°) = 1.732 m

Fy = 0: NA

98.1 N = 0

MA = 0: (98.1 N)(0.5 m)

1m

A Weight = mg 2 = (10 kg)(9.81 m/s ) 60° = 98.1 N

fA

4 Impending motion

NA 5

(1 m)(cos 60°) = 0.5 m

Equations of equilibrium:

6

+

3

B 1m Because the ladder is uniform, the weight acts through the center.

The friction force must be drawn in a direction opposing the impending motion.

7

NB(1.732 m) = 0

Results of solving the above equations of equilibrium: NA = 98.1 N NB = 28.3 N f A = 28.3 N

7.1 Friction: Basic Applications Example 1, page 2 of 2 8

Compute the maximum possible friction force that the surfaces can develop at point A. f A-max

A NA

= (0.3)(98.1 N) = 29.4 N

To determine whether or not the ladder will stay in the original position, the friction force found from the equilibrium equations, f A = 28.3 N, must be compared with the maximum force that the surfaces at A can develop: f A-max = 29.4 N Since f A = 28.3 N < 29.4 N = f A-max the surfaces are able to develop enough friction force and the ladder will stay in equilibrium.

7.1 Friction: Basic Applications Example 2, page 1 of 2 2. The uniform ladder is 2-m long and the wall at B is smooth. If the coefficient of static friction at A is A = 0.2, determine the smallest angle for which the ladder can remain in the position shown.

2 1

B

Free-body diagram

Mass of ladder = 10 kg

NB

No friction force is present because the wall is smooth.

1m (2 m) sin 1m

A

5

Equations of equilibrium: Fx = 0: NB

fA = 0

+

Fy = 0: NA

98.1 N = 0

+

+

MA = 0: (98.1 N)(1 m) cos

6

Weight = mg = (10 kg)(9.81 m/s2) = 98.1 N (1)

fA NA

3

Impending motion

(2) NB(2 m) sin

=0

Three equations, but four unknowns: NB, f A, NA, and An additional equation is needed.

(3)

4 (1 m) cos

The friction force must be drawn in a direction opposing the motion.

7.1 Friction: Basic Applications Example 2, page 2 of 2 The fourth equation comes from the condition of impending slip at point A, because if slip is just about to occur, then the friction force, f A, is at its maximum value, which is ANA: fA

f A-max

A NA

= (0.2)NA

8

Using f A = 19.62 N in Eq. 1 gives NB = f A = 19.62 N

(4) Using NB = 19.62 N in Eq. 3 gives

Three of the four equations are linear but the moment equation, Eq. 3, is nonlinear because cos and sin appear.

(98.1 N) cos

(19.62 N)(2) sin

Dividing through by cos

+

7

MA = 0: (98.1 N)(1m) cos - NB(2 m) sin = 0

(Eq. 3 repeated)

The easiest way to solve these equations is to use the general equation solver on a calculator. Alternatively, manipulate the equations as follows. First note that Eq. 2 implies that NA = 98.1 N Then using this value for NA in Eq. 4 gives f A = NA = (0.2)(98.1 N) = 19.62 N

sin cos

=

=0

and rearranging gives

98.1 = 2.5 (19.62)(2)

Replacing the left-hand side of this equation by tan gives tan

= 2.5

which implies 68.2

Ans.

7.1 Friction: Basic Applications Example 3, page 1 of 2 3. The uniform ladder is 2-m long. The coefficient of static friction at A is A = 0.6 and at B is B = 0.4. Determine the smallest angle, , for which the ladder can remain in the position shown.

3 1

Free-body diagram

Friction force, f B, is present and opposes the possible motion.

B NB

Mass of ladder = 10 kg

1m

2

Impending motion (2 m) sin

fB 1m Weight = mg = (10 kg)(9.81 m/s2) = 98.1 N

A

5

fA

Equations of equilibrium: +

NA Fx = 0: NB

fA = 0

(1)

4

+

Fy = 0: NA + f B

+

(1 m) cos 98.1 N = 0

MA = 0: (98.1 N)(1 m) cos NB(2 m) sin = 0

(2) f B(2 m) cos

6 Three equilibrium equations but five unknowns: NA, NB, f A, f B, and two more equations are needed.

(3)

(2 m) cos

Friction force opposes the possible motion.

7.1 Friction: Basic Applications Example 3, page 2 of 2 7

The two additional equations come from the condition of impending slip at points A and B, because if slip is just about to occur, then the friction forces, f A and f B, are at their maximum values, ANA and NB fA

f A-max

A NA

fB

f B-max

BNB

= 0.6NA

(4)

= 0.4NB

(5)

Four of the five equations are linear but the moment equation Eq. 3 is nonlinear (sin and cos are present). To solve these equations, use the general equation solver on your calculator or manipulate the equation algebraically and use a trig identity such as tan = sin /cos . Results of solving the five equations (three equilibrium and two friction equations) given above: NA = 79.1 N NB = 47.5 N f A = 47.5 N f B = 19.0 N = 32.3°

Ans.

7.1 Friction: Basic Applications Example 4, page 1 of 3 4. Four round pegs A, B, C, and D are attached to the bracket and loosely straddle the vertical pole. When a 100-N force is applied as shown, the bracket rotates slightly and friction forces develop between pegs B, C, and the pole. If the coefficient of static friction between the pegs and the pole is determine the smallest value of for which the bracket will support the load. Neglect the effect of the rotation of the bracket on the distances shown.

1

Movement of bracket exaggerated for clarity

100 N Bracket rotates a small amount

Peg D loses contact with the pole 50 mm 100 N D

B 100 mm D

C 300 mm

B

A C

A

Peg A loses contact with the pole

7.1 Friction: Basic Applications Example 4, page 2 of 3 Free-body diagram

4

The friction forces, f B and f C, resist the motion by pushing the bracket up.

P = 100 N

B Impending motion of bracket

7

Equations of equilibrium:

NB 100 mm

fB C

NC 5

fC 300 mm

As the bracket inclines slightly, the pegs at A and D lose contact with the pole. That is why no forces appear at A and D on the free-body diagram.

50 mm

The normal forces, NB and NC, are directed from the pole to the pegs .

+

Fx = 0: NB

NC = 0

+

3

6

Fy = 0: f B + f C

+

2

100 N = 0

MC = 0: (100 N)(300 mm) + f B(50 mm) NB(100 mm) = 0

(1) (2)

(3)

7.1 Friction: Basic Applications Example 4, page 3 of 3 8

There are only three equations of equilibrium but four unknowns (f B, NB, f C, and NC), so at least one more equation is needed. The additional equation comes from the condition of impending slip at B, but if the bracket is going to slip at B, it will also slip at C. So we have two additional equations and one additional unknown, : f B = f B-max

NB

(4)

f C = f C-max

NC

(5)

Solving Eqs. 1-5 gives the results below (Note that Eqs. 4 and 5 are nonlinear because multiplies NB and NC): f B = 50 N NB = 325 N f C = 50 N NC = 325 N = 0.154

Ans.

7.1 Friction: Basic Applications Example 5, page 1 of 2 5. Arm ABC acts as a brake on the wheel. The coefficient of static friction at B is B = 0.4. Determine the largest moment M that can act on the wheel without causing it to slip. C

100 N

M O

Free-body diagram of wheel.

M Ox

300 mm

NB

O

B A

Impending motion of point on outer surface of wheel

2

Oy

400 mm

3 The friction force f B opposes the motion.

fB 200 mm

200 mm

4

+

Radius = 200 mm

1

Equation of moment equilibrium for the wheel (Since we were not asked to compute the reactions Ox and Oy, we do not need to write the force-equilibrium equations.): MO = 0: f B(200 mm)

M=0

(1)

7.1 Friction: Basic Applications Example 5, page 2 of 2 5 6

The sense of the friction force on the brake can be determined by Newton's Third Law (equal and opposite to the force on the wheel).

Free-body diagram of arm ABC. fB

100 N 300 mm

Ay NB

400 mm A Ax 200 mm

Equation of moment equilibrium for the brake (Since we were not asked to compute the reactions Ax and Ay, we do not need to write the force-equilibrium equations.):

+

7

MA = 0: f B(200 mm) NB(400 mm) + (100 N)(300 mm + 400 mm) = 0 (2)

Thus far we have two equations but three unknowns (M, f B, and NB), so another equation is needed.

8 The third equation follows from the condition that slip impends at B: fB

f B-max

BNB

= 0.4NB

(3)

Solving Eqs. 1-3 simultaneously yields f B = 87.5 N NB = 218.8 N M = 17 500 N·mm = 17.5 N·m

Ans.

7.1 Friction: Basic Applications Example 6, page 1 of 4 6. The uniform block is initially at rest when a 10-lb force is applied. The coefficient of static friction between the block and the plane is = 0.6. Determine if the block will move. 1 ft

10 lb

20 lb (weight)

A

2 ft

7.1 Friction: Basic Applications Example 6, page 2 of 4 1

Free-body diagram

4

Free-body diagram showing resultant forces 10 lb

1 ft 10 lb

20 lb 2 ft 20 lb

fd

5

The resultant of the distributed force f d is f.

A

B

f A

0.5 ft

B 3 The distributed friction force f d opposes possible slip to the right; the distributed normal force Nd opposes possible tipping of the block.

2

Nd As the block is pushed to the right by the 10-lb force, the floor opposes the possible motion by providing a distributed reaction force. The component of this reaction force parallel to the floor is the distributed friction force f d, and the component normal to the floor is the distributed normal force Nd.

x N

6

The resultant of the distributed force Nd is N. Because Nd opposes possible tipping of the block, it is not uniform but is greater near the right-hand side of the base of the block to balance the tendency to tip. Thus the resultant N does not act at the middle of the base but instead acts at some unknown distance, x, from the middle.

7.1 Friction: Basic Applications Example 6, page 3 of 4 7

Equations of equilibrium: +

+

Fy = 0: N

+

Fx = 0: 10 lb

MA = 0:

8

f=0

20 lb = 0 (20 lb)(0.5 ft)

(10 lb)(2 ft) + N(0.5 ft + x) = 0

Solving these equations gives f = 10 lb N = 20 lb x = 1 ft

9 These are the values required if the system is to stay in equilibrium, that is, not move. To determine if the system can produce the 10-lb friction force f required to keep the system in equilibrium, we have to compare f with the maximum possible value of the friction force: f max

N = (0.6)(20 lb) = 12 lb

Because f = 10 lb is less than the 12 lb maximum possible force, the surfaces can develop enough force to balance forces in the x direction (thus the block will not slide to the right).

7.1 Friction: Basic Applications Example 6, page 4 of 4 10 lb

10 We next consider whether or not the block will tip. Recall that solving the equilibrium equations gave the result x = 1 ft. That is, to maintain equilibrium, the normal force N must act at the location shown, 0.5 ft to the right of the block. But this is impossible because N is the normal force from the ground acting up on the block; the farthest N can act is at the right hand corner, B. Thus the block will tip because N cannot act far enough to the right to prevent it.

20 lb

A

B

f 0.5 ft

0.5 ft x = 1 ft

N (impossible location because outside the base of the block)

7.1 Friction: Basic Applications Example 7, page 1 of 6 7. The uniform block is initially at rest when the force P is applied. The coefficient of static friction between the block and the plane is = 0.6. Determine the minimum value of P that will cause the block to move. 1 ft

P

20 lb (weight)

A

2 ft

B

7.1 Friction: Basic Applications Example 7, page 2 of 6 1

3

Free-body diagram

Free-body diagram showing resultant forces P

1 ft P

20 lb 2 ft 20 lb

fd

A

4

The resultant of the distributed force f d is f. A f

B 0.5 ft

B

x N

5

Nd 2

As the block is pushed to the right by the force P, the floor opposes the possible motion by providing a distributed reaction force. The component of this reaction force parallel to the floor is the distributed friction force, f d, and the component normal to the floor is the distributed normal force Nd.

The resultant of the distributed force Nd is N. Because Nd opposes possible tipping of the block, it is not uniform but is greater near the right-hand side of the base of the block to balance the tendency to tip. Thus the resultant N does not act at the middle of the base but instead acts at some unknown distance, x, from the middle.

7.1 Friction: Basic Applications Example 7, page 3 of 6 6 +

Fx = 0: P

f=0

(1)

+

Fy = 0: N

20 lb = 0

(2)

+

Equations of equilibrium:

MA = 0:

(20 lb)(0.5 ft) P(2 ft) + N(0.5 ft + x) = 0

9

Case 1: Sliding 1 ft

(3) P

7

Three equations but four unknowns (P, f, N and x), so one more equation is needed.

8

The fourth equation comes from considering possible impending motion. There are two cases to consider: sliding and tipping. 20 lb (weight)

A

B

7.1 Friction: Basic Applications Example 7, page 4 of 6 10 Case 2: Tipping

11 We have to analyze each case separately. Let's (arbitrarily) choose Case 1 first. If sliding impends, then

P

f = f max

N = 0.6N

12 Solving Eqs. 1-4 simultaneously gives P = 12 lb N = 20 lb

20 lb (weight)

f = 12 lb x = 1.2 ft

A

B

(4)

7.1 Friction: Basic Applications Example 7, page 5 of 6 13 Free body diagram for Case 1 (Sliding impends) P = 12 lb 14 But this diagram shows that the only way the equilibrium equations for Case 1 can be satisfied is if the normal force N lies to the right of the block (x = 1.2 ft). Since this is impossible, the Case 1 assumption that sliding impends must be incorrect.

20 lb

B

A

N = 20 lb

f = 12 lb 0.5 ft x = 1.2 ft

7.1 Friction: Basic Applications Example 7, page 6 of 6 15 Free body diagram for Case 2 (Tipping impends) 0.5 ft

P

0.5 ft 16 Since the block is just about to tip, it loses contact with the floor except at the corner B, where the normal force N is concentrated. Since N acts at the corner, we know x = 0.5 ft

(5)

Solving the equilibrium equations, Eqs. 1, 2, and 3, simultaneously with Eq. 5 gives

20 lb

f = 5 lb N = 20 lb P = 5 lb B

A f

x

N

Ans.

Since there were only two possibilities, sliding and tipping, and we eliminated sliding, we know that the above result P = 5 lb is correct. However, we can also check our work by verifying that the friction force f is less than the maximum possible value: f = 5 lb < f max

N = (0.6)(20 lb) = 12 lb. (OK)

7.1 Friction: Basic Applications Example 8, page 1 of 3 8. The cylinder is initially at rest when a horizontal force P is applied. The coefficients of static friction at A and B are A = 0.3 and B = 0.6. Determine the minimum value of P that will cause the cylinder to move. Radius = 0.2 m P 1 0.3 m

Free-body diagram

B 20 kg A

Weight = mg = (20 kg)(9.81 m/s2) = 196.2 N

2

P NB

B

0.3 m

fB

A fA 5 The friction force from the floor opposes the motion of point A on the cylinder.

NA

4

Possible motion of point B on cylinder. Force P tends to rotate the cylinder clockwise.

0.2 m 3 The friction force from the wall opposes the motion of point B on the cylinder.

Possible motion of point A.

7.1 Friction: Basic Applications Example 8, page 2 of 3 6

7

Equilibrium equations

+

+

+

Fx = 0: P + f A Fy = 0:

NB = 0

(1)

196.2 N + f B + NA = 0

(2)

MA = 0: f B(0.2 m) + NB(0.2 m)

P(0.3 m) = 0

Case 2

The cylinder spins about its center.

f A = f A-max

A NA

= 0.3NA

(4)

f B = f B-max

BNB

= 0.6NB

(5)

(3)

There are three equations and five unknowns (P, f A, NA, f B, NB), so two more equations are needed. The two additional equations come from considering possible impending motion. There are two cases to consider: Case 1

We have to analyze each case separately. Let's (arbitrarily) choose Case 1 first. Thus if the cylinder is about to slip about its center, then slip impends simultaneously at points A and B, so

Solving Eqs. 1-5 simultaneously gives P = 554 N f A = 34.6 N

The cylinder rolls up the wall without slipping.

f B = 312 N NA = 115 N NB = 519 N

B A

B A

A negative normal force, NA, is impossible (The floor can't pull down on the cylinder), so the assumption of slip at both A and B must be wrong.

7.1 Friction: Basic Applications Example 8, page 3 of 3 8

Next consider Case 2 the cylinder is about to roll up the wall. Thus the cylinder is about to lose contact with the floor at point A, and so the friction and normal forces there are zero: fA = 0

(6)

NA = 0

(7)

Solving the equilibrium equations, Eqs. 1, 2, and 3, simultaneously with Eqs. 6 and 7 gives fA = 0 NA = 0 f B = 196 N NB = 392 N P = 392 N

Ans.

Since there were only two possibilities, spinning about the cylinder center or rolling up the wall, and we eliminated spinning, the above result P = 392 N must be correct. However, we can also check our work by comparing the friction force, f B, with the maximum possible value: f B = 196 N < f B-max

BNB

= (0.6)(392 N) = 235 N (OK)

7.1 Friction: Basic Applications Example 9, page 1 of 4 9. The small block B rests on top of the large block A. The coefficients of static friction are shown in the figure. Determine the smallest value of applied force P that will keep block A from sliding down the inclined plane. Frictionless pulley Cord B

= 0.3 = 0.2

A

B 10 kg 60 kg P

30°

7.1 Friction: Basic Applications Example 9, page 2 of 4 1

Free-body diagram of block B Weight = mg = (10 kg)(9.81 m/s2) = 98.1 N y

Tension in cord, T

3 4

Impending motion of B block B relative to block A (If block A moves down the plane, block B must move up the plane.) N Normal force B from block A 5

The numerical value of will be calculated later.

P fB

2

30° x

The friction force from block A opposes the impending motion of block B up the incline.

Equations of equilibrium for block B. We assume that the blocks will not tip because they are much longer than they are high; thus no moment equation is needed (Since no dimensions are given, we could not write a moment equation even if we wanted to).

6

+

+

Fx =0: P Fy = 0: NB

T + f B + (98.1 N) sin (98.1 N) cos

=0

It's convenient to use an inclined xy coordinate system.

=0

(1) (2)

7.1 Friction: Basic Applications Example 9, page 3 of 4 7

Geometry = 90° y

8

60° = 30°

Free-body diagram of block A 13 Weight = mg = (60 kg)(9.81 m/s2) = 588.6 N

60° 10 Friction force from block B opposes motion of block A. 30°

30° x

T

fB A

[Weight of block A alone (Note that the weight of block B is not included because block B is not part of this free-body. The effect of the weight of block B is = 30° transmitted through the normal force, NB.)] y NB 9 Impending motion of block A relative to block B.

fA 12 Friction force from inclined plane opposes NA motion of block A. Normal force from inclined plane 11 Impending motion of block A relative to inclined plane.

30° x

7.1 Friction: Basic Applications Example 9, page 4 of 4 Free body diagram of block A repeated

14 Equilibrium equations for block A +

Fx =0: (588.6 N) sin 30°

fA

fB

T=0

(3)

fB

T +

Fy = 0:

(588.6 N) cos 30° + NA

NB = 0

(4)

Four equations in six unknowns (T, P, f A, NA, f B, NB). Two more equations come from the condition of impending sliding between the blocks and between block A and the plane: fA

f A-max

A NA

= 0.2NA

fB

f B-max

BNB = 0.3NB

fA

588.6 N

= 30° y NB

A

(5) NA (6)

x

Solving Eqs. 1-6 simultaneously gives f A = 119 N NA = 595 N f B = 25 N NB = 85 N T = 150 N P = 75 N

30°

Ans.

7.1 Friction: Basic Applications Example 10, page 1 of 8 10. The three blocks are stationary when the force P is applied. The coefficients of static friction for each pair of surfaces are given in the figure. Determine the smallest value of P for which motion will occur. The blocks are sufficiently long that tipping will not occur.

1

10 kg

A

AB

= 0.8

10 kg

B

BC

= 0.3

10 kg

C

C

= 0.15

Free-body diagram of block A Weight = mg = (10 kg)(9.81 m/s2) = 98.1 N

P

4 +

Fx = 0: P

Fy = 0: NAB

A f AB 2 NAB

Impending motion of block A relative to B

3 Friction opposes the motion

Equilibrium equations for block A:

+

P

f AB = 0

(1)

98.1 N = 0

The last equation gives NAB = 98.1 N

(2)

7.1 Friction: Basic Applications Example 10, page 2 of 8 5

8

Free-body diagram of block B

Impending motion of block B relative to block A (An observer on A would see B moving in this direction.)

NAB = 98.1 N

9 Friction force opposes relative motion

f AB B 6

f BC

Impending motion of block B relative to C

Weight = 98.1 N NBC

+

Fx = 0: f AB

+

10 Equilibrium equations for block B:

Fy = 0:

f BC = 0

98.1 N

(3)

98.1 N + NBC = 0

The last equation gives NBC = 196.2 N

(4)

7

Friction force opposes relative motion

7.1 Friction: Basic Applications Example 10, page 3 of 8 11 Free-body diagram of block C 14 Impending motion of block C relative to block B (An observer on B would see C moving in this direction.)

NBC = 196.2 N 15 Friction force opposes relative motion f BC C fC

Weight = 98.1 N NC

+

Fx = 0: f BC

+

16 Equilibrium equations for block C:

Fy = 0:

fC = 0

98.1 N

(5)

196.2 N + NC = 0

The last equation gives NC = 294.3 N

(6)

12 Impending motion of block C relative to floor 13 Friction force opposes relative motion

7.1 Friction: Basic Applications Example 10, page 4 of 8 17

We now have six equilibrium equations but seven unknowns (P, f AB, NAB, f BC, NBC, f C, NC), so another equation is needed.

18 The seventh equation comes from the condition of impending slip. We have to consider three cases:

Case 1 Impending motion

A B

Stationary

C

Case 2 A

Impending motion: blocks A and B move together

B C

Stationary Case 3 A B C

Impending motion: blocks A, B and C move together

7.1 Friction: Basic Applications Example 10, page 5 of 8 19 Analyze each case separately. 22 For block B, Eq. 3 is f AB

f BC = 0

20 Case 1 So for equilibrium,

A B

Stationary

C

= 78.5 N Let's compare this with the maximum possible friction force:

21 Slip impends so f AB = f AB-max ABNAB

f BC = f AB

f BC-max

by Eq. 2 = (0.8)(98.1 N) = 78.5 N

(7)

We have to check to see if the surfaces of contact between blocks B and C develop enough friction force to keep block B stationary.

BCNBC

= (0.3)(196.2 N) = 58.9 N

(8)

So the surfaces can develop only 58.9 N while 78.5 N are needed for equilibrium. Thus block B will move, contrary to our assumption for Case 1.

7.1 Friction: Basic Applications Example 10, page 6 of 8 23 Case 2 A

Impending motion together

f BC

B C

25 For block C, Eq. 5 is

Stationary

fC = 0

So for equilibrium, f C = f BC = 58.9 N

24 Slip impends so f BC = f BC-max by Eq. 8 = 58.9 N We have to check to see if the surfaces of contact between block C and the ground develop enough friction force to keep block C stationary.

Compare this with the maximum possible friction force f C-max

CNC

by Eq. 6 = (0.15)(294.3 N) = 44.1 N

(9)

So the surfaces can develop only 44.1 N while 58.9 N are needed for equilibrium. Thus block C will move, contrary to our assumption for Case 2.

7.1 Friction: Basic Applications Example 10, page 7 of 8 28 Eq. 5 gives f BC:

26 Case 3 A

Impending motion together

B

f BC

fC = 0

Thus

C f BC = f C

by Eq. 10

= 44.1 N

(11)

27 Slip impends so and so f C = f C-max = 44.1 N

by Eq. 8

(10)

44.1 N = f BC < f BC-max = 58.9 N

by Eq. 9 We don't have to check that the surfaces of contact between blocks A and B and between B and C develop enough friction to keep A and B in equilibrium, since there were only three cases of possible motion, and we showed that the first two cases were impossible. Nonetheless, we can verify that our work is correct by showing that the friction forces acting between A and B and between B and C are less than their maximum possible values.

(OK)

Eq. 3 gives f AB: f AB

f BC = 0

Thus f AB = f BC = 44.1 N

(12)

and so by Eq. 7 44.1 N = f AB < f AB-max = 78.5 N

(OK)

7.1 Friction: Basic Applications Example 10, page 8 of 8 29 Thus the surfaces of contact between blocks A and B and between B and C can develop enough friction to keep blocks A, B, and C moving together as a unit. 30 Finally, we can calculate P from Eq. 1: P or,

f AB = 0 by Eq. 12

P = f AB = 44.1 N

Ans.

31 Why didn't we consider a case like this? Case 4 A

32 Answer: No matter what the impending motion is, there are only seven unknown forces (f AB, NAB, f BC, NBC, f C, NC, and P). Since these seven unknowns must satisfy the six equations of equilibrium, the unknowns can be chosen to satisfy only one additional equation a friction equation. In the unlikely event that the masses and 's just happen to have values such that the seven forces simultaneously satisfy the six equilibrium equations and two friction equations, then one of the eight equations must be redundant. Applying this reasoning to Case 4, we see that if forces exist that satisfy Case 4's equations, then these forces must be identical to the forces satisfying the equations for Case 1 (slip between A and B) and Case 2 (slip between B and C). Since solving Case 4 would give the same answer as solving Case 1 (or Case 2), we don't have to consider Case 4. A similar argument can be made for other possible motions. Impending motion of A relative to B

B

Impending motion of B relative to C

C

Stationary

7.1 Friction: Basic Applications Example 11, page 1 of 9 11. The two cylinders shown are initially at rest when horizontal forces of magnitude P/2 are applied to the ends of the axle in the lower cylinder. The coefficients of static friction for each pair of surfaces are given in the figure. Determine the largest value of P that can be applied without moving the cylinders up the inclined plane. P/2 Radius of each cylinder = 300 mm B

= 0.4

C

B

= 0.5 Mass of each cylinder = 50 kg

C A A

= 0.6

P/2 25°

7.1 Friction: Basic Applications Example 11, page 2 of 9 1

Impending motion of point C on lower cylinder relative to upper cylinder (An observer on the upper cylinder would see this motion as the lower cylinder moves).

5

Free-body diagram of lower cylinder y NC

3

Weight = mg = (50 kg)(9.81 m/s2) = 490.5 N

C 6 The friction force from f C the upper cylinder opposes the relative motion of point C on the lower cylinder. fA

P

O

4 The numerical values of calculated later.

and

will be

A 25° x

Radius = 300 mm 8

The friction force from the plane opposes the motion of point A on the cylinder.

2

NA 7

Impending motion of point A on cylinder. The x component of the applied force, P cos , is pushing the cylinder up the plane.

It is convenient to use an inclined xy coordinate-system.

7.1 Friction: Basic Applications Example 11, page 3 of 9 9

Equilibrium equations for cylinder:

+

+

+

Fx = 0: (490.5 N) sin Fy = 0:

90.5 N) cos

MO = 0: f A(300 mm)

P cos P sin

Free-body diagram of lower cylinder repeated + f A + NC = 0

(1)

+ f C + NA = 0

f C(300 mm) = 0

(2)

y NC

490.5 N

(3) C fC

10 Geometry

= 90° = 25°

65°

O y A

fA 65°

P

25° x Radius = 300 mm

NA = 25°

25°

x

7.1 Friction: Basic Applications Example 11, page 4 of 9 11 Free-body diagram of upper cylinder y

Radius = 300 mm

Weight = 490.5 N

= 25°

O

P fC C

fB

15 The friction force from the plane opposes the motion up the plane.

B

13 The friction force from the lower cylinder opposes the relative motion of point C on the upper cylinder. . NC x

12 Impending motion of point C on upper cylinder relative to lower cylinder (An NB observer on the lower cylinder would see this motion as the upper cylinder moves). 14 Impending motion of point B as normal force NC pushes the upper cylinder up the plane.

7.1 Friction: Basic Applications Example 11, page 5 of 9 Free-body diagram of upper cylinder repeated

16 Equilibrium equations

+

+

+

Fx = 0: (490.5 N) sin 25° + f B Fy = 0:

(490.5 N) cos 25°

MO' = 0: f B(300 mm)

NC = 0

(4)

f C + NB = 0

(5)

f C(300 mm) = 0

y

490.5 N

(6)

17 Thus far we have six equation but seven unknowns (P, f A, NA, f B, NB, f C, NC), so another equation is needed. The seventh equation comes from the condition of impending slip. We have to consider only two cases: 1. slip occurs at point B (and simultaneously rolling occurs about points A and C). 2. slip occurs at point C (and simultaneously rolling occurs about points A and B). Slip at point A will be discussed later.

O

25°

P C

fB

B

NC x

Radius = 300 mm NB

fC

7.1 Friction: Basic Applications Example 11, page 6 of 9 18 Case 1 Before motion O' C

B

O

After motion A O' B 20 Slip

C

For impending slip at B, f B = f B-max

BNB

= 0.4NB

O

Displacement of point O (Point O moves up the plane) A

(7)

19 Rolling without slipping (The radial line OA on the lower cylinder rotates through the same angle, , as the radial line O'C on the upper cylinder.)

7.1 Friction: Basic Applications Example 11, page 7 of 9 21 Solving Eqs. 1-7 simultaneously gives f A = 296 N

NA = 618 N

f B = 296 N

NB = 741 N

f C = 296 N

NC = 504 N

P = 1111 N

22 We must check that the surfaces at A and C can provide enough friction force to prevent slip and allow rolling: fA

296

f A-max

A NA

= (0.6)(618 N) = 371 N (OK)

f C-max CNC = (0.5)(504 N) = 252 N (Not enough! We need f C = 296 N for equilibrium.) So the assumption of impending slip at B is wrong.

7.1 Friction: Basic Applications Example 11, page 8 of 9 23 Case 2 (Slip at C, rolling at A and B) Before motion O' C

B

O

After motion A O'

C O

B

25 Rolling without slipping

Displacement of point O A

26 Slip 24 Rolling without slipping

For impending slip at C, f C = f C-max

CNC

= 0.5NC

(8)

7.1 Friction: Basic Applications Example 11, page 9 of 9 27 Solving the six equilibrium equations, Eqs. 1-6, plus Eq. 8 yields f A = 207 N

NA = 624 N

f B = 207 N

NB = 652 N

f C = 207 N

NC = 415 N P = 915 N

Ans.

28 The above answers must be correct since we eliminated the only other possible case where slip impends. But we can check our results by verifying that the friction forces at A and B are less than their maximum possible values. f A = 207 N

f A-max

A NA

= (0.6)(624 N)

= 374 N (OK) f B = 207 N

f B-max

BNB

= (0.4)(652 N)

= 261 N (OK)

29 What about slip occurring at point A only? Well if the lower cylinder moves, then the upper cylinder must also move. But the only way that the upper cylinder can move is if either 1) it slips at point B, or 2) it slips at point C. Thus the case of slip impending at point A alone is impossible and does not have to be considered. What about simultaneous slip at A and B? Answer: we have already found values of the seven unknowns in the problem that satisfy the six equilibrium equations and the equation for slip at B. In the unlikely case that the seven values happen to satisfy an eighth equation (slip at A), then that equation must be redundant, and the solution for the eight equations is the same as we have already found for the seven equations. An analogous statement can be made for the case of simultaneous slip at A and C.