BME 201 Biomechanics I Exam 2 Review

1. The spring has a stiffness of k = 800 N/m and an unstretched length of 200 mm. Determine the force in cables BC and BD when the spring is held in the position shown.

2.The lamp has a mass of 15 kg and is supported by a pole AO and cables AB and AC. If force in the poles acts along its axis, determine the forces in AO, AB, and AC for equilibrium.

3. Determine the magnitude of the force at the pin A and in the cable BC needed to support the 500-lb load. Neglect the weight of the boom AB.

4.The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.

245

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4–7.

Determine the magnitude of force at the pin A and in the cable BC needed to support the 500-lb load. Neglect the weight of the boom AB.

35!22!

8 ft

C

B

A

Determine the magnitude of force at the pin and in thecable needed to support the 500-lb load. Neglect theweight of the boom .

SOLUTION

Equations of Equilibrium: The force in cable can be obtained directly bysumming moments about point .

Ans.

Thus, Ans.FA = Ax = 2060.9 lb = 2.06 kip

Ay = 0

Ay + 1820.7 sin 13° - 500 cos 35° = 0a + ©Fy = 0;

Ax = 2060.9 lb

Ax - 1820.7 cos 13° - 500 sin 35° = 0+Q ©Fx = 0;

FBC = 1820.7 lb = 1.82 kip

FBC sin 13°(8) - 500 cos 35°(8) = 0a+ ©MA = 0;

ABC

ABBC

A

35!22!

8 ft

C

B

A

Determine the magnitude of force at the pin and in thecable needed to support the 500-lb load. Neglect theweight of the boom .

SOLUTION

Equations of Equilibrium: The force in cable can be obtained directly bysumming moments about point .

Ans.

Thus, Ans.FA = Ax = 2060.9 lb = 2.06 kip

Ay = 0

Ay + 1820.7 sin 13° - 500 cos 35° = 0a + ©Fy = 0;

Ax = 2060.9 lb

Ax - 1820.7 cos 13° - 500 sin 35° = 0+Q ©Fx = 0;

FBC = 1820.7 lb = 1.82 kip

FBC sin 13°(8) - 500 cos 35°(8) = 0a+ ©MA = 0;

ABC

ABBC

A

35!22!

8 ft

C

B

A

Ans:FBC = 1.82 kip FA = 2.06 kip

BME 201 Biomechanics I Exam 2 Review

5. Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N forces are parallel to the x, y, and z axes, respectively.

6. 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 from and rear brakes are locked. Take us = 0.3

7. Determine the force in each member of the truss and state if the members are in tension or compression.

266

© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

SOLUTIONEquations of Equilibrium. Referring to the FBD of the rod shown in Fig. a,

ΣFx = 0; Ax - 400 = 0 Ax = 400 N Ans.

ΣFy = 0; 500 - Ay = 0 Ay = 500 N Ans.

ΣFz = 0; Az - 600 = 0 Az = 600 N Ans.

ΣMx = 0; (MA)x - 500(1.25) - 600(1) = 0

(MA)x = 1225 N # m = 1.225 kN # m Ans.

ΣMy = 0; (MA)y - 400(0.75) - 600(0.75) = 0

(MA)y = 750 N # m Ans.

ΣMz = 0; (MA)z = 0 Ans.

*4–28.

Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N forces are parallel to the x, y, and z axes, respectively.

y

400 N

600 N

500 N

1 m

0.5 m

0.75 m

z

x

A

0.75 m

Ans:Ax = 400 NAy = 500 NAz = 600 N(MA)x = 1.225 kN # m(MA)y = 750 N # m(MA)z = 0

288

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SOLUTIONEquations of Equilibrium. Referring to the FBD of the car shown in Fig. a,

S+ ΣFx = 0; FA + FB - F cos 30° = 0 (1)

+ cΣFy = 0; F sin 30° + NA + NB - 2000(9.81) = 0 (2)

a+ΣMA = 0; F cos 30°(0.3) - F sin 30°(0.75) +

NB (2.5) - 200(9.81)(1) = 0 (3)

Friction. It is required that both the front and rear wheels are on the verge to slip. Thus,

FA = ms NA = 0.3 NA (4)

FB = ms NB = 0.3 NB (5)

Solving Eqs. (1) to (5),

F = 5793.16 N = 5.79 kN Ans.

NB = 8114.93 N NA = 8608.49 N FA = 2582.55 N FB = 2434.48 N

4–47.

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.

F

0.75 m

30!

0.3 m 0.6 mG

AC

B1.50 m1 m

Ans:F = 5.79 kN

311

© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

5–3.

Determine the force in each member of the truss and state if the members are in tension or compression.

3 ft 3 ft 3 ft

125

13

130 lb

A B

CE

D

F

4 ft 4 ft

Determine the force in each member of the truss, and stateif the members are in tension or compression. Set .

SOLUTION

Support Reactions: From the free-body diagram of the truss, Fig. a, and applyingthe equations of equilibrium, we have

a

Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze the equilibrium of joint B.

Joint C: From the free-body diagram in Fig. b, we can write

Ans.

Ans.

Joint A: From the free-body diagram in Fig. c, we can write

Ans.

Ans.

Joint B: From the free-body diagram in Fig. d, we can write

Ans.

Note: The equilibrium analysis of joint D can be used to check the accuracy of thesolution obtained above.

2.362 - 2.362 = 0 (check!)©Fx = 0;:+FBD = 4 kN (T)

FBD - 4 = 0+ c ©Fy = 0;

FAB = 2.362 kN = 2.36 kN (T)

FAB - 1.458a45b - 1.196 = 0©Fx = 0;:+

FAD = 1.458 kN = 1.46 kN (C)

0.875 - FAD a35b = 0+ c ©Fy = 0;

FCB = 2.362 kN = 2.36 kN (T)

5.208a45b - 3.608 sin 30° - FCB = 0©Fx = 0;:+

FCD = 5.208 kN = 5.21 kN (C)

3.608 cos 30° - FCD a 35b = 0+ c ©Fy = 0;

Ay = 0.875 kN

Ay + 3.608 cos 30° - 4 = 0+ c ©Fy = 0;

Ax = 1.196 kN

3 - 3.608 sin 30° - Ax = 0©Fx = 0;:+NC = 3.608 kN

NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0+ ©MA = 0;

u = 30°

A C

B

D

2 m

4 kN

3 kN

2 m

1.5 m

uSOLUTION

Joint A:

Ans.

Ans.

Joint B:

Ans.

Ans.

Joint C:

Ans.

Ans.

Joint D:

Ans.

Ans.

Joint E:

Ans.FEF = 300 lb (C)

:+ ©Fx = 0; 180 - 35

(FEF) = 0

FDF = 230 lb (T)

:+ ©Fx = 0; FDF - 140 - 35

(150) = 0

FDE = 120 lb (C)

+ c ©Fy = 0; FDE - 45

(150) = 0

FCE = 180 lb (C)

:+ ©Fx = 0; -FCE + 35

(150) + 35

(150) = 0

FCD = 150 lb (T)

+ c ©Fy = 0; a 45b FCD - a 4

5b 150 = 0

+ c ©Fy = 0; FBC = 0

FBD = 140 lb (T)

:+ ©Fx = 0; FBD - 140 = 0

FAB = 140 lb (T)

:+ ©Fx = 0; FAB - 35

(150) - 513

(130) = 0

FAC = 150 lb (C)

+ c ©Fy = 0;45

(FAC) - 1213

(130) = 0

Determine the force in each member of the truss. State ifthe members are in tension or compression.

3 ft 3 ft 3 ft

125

13

130 lb

A B

CE

DF

4 ft 4 ft

SOLUTION

Joint A:

Ans.

Ans.

Joint B:

Ans.

Ans.

Joint C:

Ans.

Ans.

Joint D:

Ans.

Ans.

Joint E:

Ans.FEF = 300 lb (C)

:+ ©Fx = 0; 180 - 35

(FEF) = 0

FDF = 230 lb (T)

:+ ©Fx = 0; FDF - 140 - 35

(150) = 0

FDE = 120 lb (C)

+ c ©Fy = 0; FDE - 45

(150) = 0

FCE = 180 lb (C)

:+ ©Fx = 0; -FCE + 35

(150) + 35

(150) = 0

FCD = 150 lb (T)

+ c ©Fy = 0; a45b FCD - a4

5b 150 = 0

+ c ©Fy = 0; FBC = 0

FBD = 140 lb (T)

:+ ©Fx = 0; FBD - 140 = 0

FAB = 140 lb (T)

:+ ©Fx = 0; FAB - 35

(150) - 513

(130) = 0

FAC = 150 lb (C)

+ c ©Fy = 0;45

(FAC) - 1213

(130) = 0

Determine the force in each member of the truss. State ifthe members are in tension or compression.

3 ft 3 ft 3 ft

125

13

130 lb

A B

CE

DF

4 ft 4 ft

Ans: FAC = 150 lb (C)FAB = 140 lb (T)FBD = 140 lb (T)FBC = 0FCD = 150 lb (T)FCE = 180 lb (C)FDE = 120 lb (C)FDF = 230 lb (T)FEF = 300 lb (C)

245

© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

4–7.

Determine the magnitude of force at the pin A and in the cable BC needed to support the 500-lb load. Neglect the weight of the boom AB.

35!22!

8 ft

C

B

A

Determine the magnitude of force at the pin and in thecable needed to support the 500-lb load. Neglect theweight of the boom .

SOLUTION

Equations of Equilibrium: The force in cable can be obtained directly bysumming moments about point .

Ans.

Thus, Ans.FA = Ax = 2060.9 lb = 2.06 kip

Ay = 0

Ay + 1820.7 sin 13° - 500 cos 35° = 0a + ©Fy = 0;

Ax = 2060.9 lb

Ax - 1820.7 cos 13° - 500 sin 35° = 0+Q ©Fx = 0;

FBC = 1820.7 lb = 1.82 kip

FBC sin 13°(8) - 500 cos 35°(8) = 0a+ ©MA = 0;

ABC

ABBC

A

35!22!

8 ft

C

B

A

Determine the magnitude of force at the pin and in thecable needed to support the 500-lb load. Neglect theweight of the boom .

SOLUTION

Equations of Equilibrium: The force in cable can be obtained directly bysumming moments about point .

Ans.

Thus, Ans.FA = Ax = 2060.9 lb = 2.06 kip

Ay = 0

Ay + 1820.7 sin 13° - 500 cos 35° = 0a + ©Fy = 0;

Ax = 2060.9 lb

Ax - 1820.7 cos 13° - 500 sin 35° = 0+Q ©Fx = 0;

FBC = 1820.7 lb = 1.82 kip

FBC sin 13°(8) - 500 cos 35°(8) = 0a+ ©MA = 0;

ABC

ABBC

A

35!22!

8 ft

C

B

A

Ans:FBC = 1.82 kip FA = 2.06 kip

266

© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

SOLUTIONEquations of Equilibrium. Referring to the FBD of the rod shown in Fig. a,

ΣFx = 0; Ax - 400 = 0 Ax = 400 N Ans.

ΣFy = 0; 500 - Ay = 0 Ay = 500 N Ans.

ΣFz = 0; Az - 600 = 0 Az = 600 N Ans.

ΣMx = 0; (MA)x - 500(1.25) - 600(1) = 0

(MA)x = 1225 N # m = 1.225 kN # m Ans.

ΣMy = 0; (MA)y - 400(0.75) - 600(0.75) = 0

(MA)y = 750 N # m Ans.

ΣMz = 0; (MA)z = 0 Ans.

*4–28.

Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N forces are parallel to the x, y, and z axes, respectively.

y

400 N

600 N

500 N

1 m

0.5 m

0.75 m

z

x

A

0.75 m

Ans:Ax = 400 NAy = 500 NAz = 600 N(MA)x = 1.225 kN # m(MA)y = 750 N # m(MA)z = 0

288

© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

SOLUTIONEquations of Equilibrium. Referring to the FBD of the car shown in Fig. a,

S+ ΣFx = 0; FA + FB - F cos 30° = 0 (1)

+ cΣFy = 0; F sin 30° + NA + NB - 2000(9.81) = 0 (2)

a+ΣMA = 0; F cos 30°(0.3) - F sin 30°(0.75) +

NB (2.5) - 200(9.81)(1) = 0 (3)

Friction. It is required that both the front and rear wheels are on the verge to slip. Thus,

FA = ms NA = 0.3 NA (4)

FB = ms NB = 0.3 NB (5)

Solving Eqs. (1) to (5),

F = 5793.16 N = 5.79 kN Ans.

NB = 8114.93 N NA = 8608.49 N FA = 2582.55 N FB = 2434.48 N

4–47.

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.

F

0.75 m

30!

0.3 m 0.6 mG

AC

B1.50 m1 m

Ans:F = 5.79 kN

311

© 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

5–3.

Determine the force in each member of the truss and state if the members are in tension or compression.

3 ft 3 ft 3 ft

125

13

130 lb

A B

CE

D

F

4 ft 4 ft

Determine the force in each member of the truss, and stateif the members are in tension or compression. Set .

SOLUTION

Support Reactions: From the free-body diagram of the truss, Fig. a, and applyingthe equations of equilibrium, we have

a

Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze the equilibrium of joint B.

Joint C: From the free-body diagram in Fig. b, we can write

Ans.

Ans.

Joint A: From the free-body diagram in Fig. c, we can write

Ans.

Ans.

Joint B: From the free-body diagram in Fig. d, we can write

Ans.

Note: The equilibrium analysis of joint D can be used to check the accuracy of thesolution obtained above.

2.362 - 2.362 = 0 (check!)©Fx = 0;:+FBD = 4 kN (T)

FBD - 4 = 0+ c ©Fy = 0;

FAB = 2.362 kN = 2.36 kN (T)

FAB - 1.458a45b - 1.196 = 0©Fx = 0;:+

FAD = 1.458 kN = 1.46 kN (C)

0.875 - FAD a35b = 0+ c ©Fy = 0;

FCB = 2.362 kN = 2.36 kN (T)

5.208a45b - 3.608 sin 30° - FCB = 0©Fx = 0;:+

FCD = 5.208 kN = 5.21 kN (C)

3.608 cos 30° - FCD a35b = 0+ c ©Fy = 0;

Ay = 0.875 kN

Ay + 3.608 cos 30° - 4 = 0+ c ©Fy = 0;

Ax = 1.196 kN

3 - 3.608 sin 30° - Ax = 0©Fx = 0;:+NC = 3.608 kN

NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0+ ©MA = 0;

u = 30°

A C

B

D

2 m

4 kN

3 kN

2 m

1.5 m

uSOLUTION

Joint A:

Ans.

Ans.

Joint B:

Ans.

Ans.

Joint C:

Ans.

Ans.

Joint D:

Ans.

Ans.

Joint E:

Ans.FEF = 300 lb (C)

:+ ©Fx = 0; 180 - 35

(FEF) = 0

FDF = 230 lb (T)

:+ ©Fx = 0; FDF - 140 - 35

(150) = 0

FDE = 120 lb (C)

+ c ©Fy = 0; FDE - 45

(150) = 0

FCE = 180 lb (C)

:+ ©Fx = 0; -FCE + 35

(150) + 35

(150) = 0

FCD = 150 lb (T)

+ c ©Fy = 0; a45b FCD - a4

5b 150 = 0

+ c ©Fy = 0; FBC = 0

FBD = 140 lb (T)

:+ ©Fx = 0; FBD - 140 = 0

FAB = 140 lb (T)

:+ ©Fx = 0; FAB - 35

(150) - 513

(130) = 0

FAC = 150 lb (C)

+ c ©Fy = 0;45

(FAC) - 1213

(130) = 0

Determine the force in each member of the truss. State ifthe members are in tension or compression.

3 ft 3 ft 3 ft

125

13

130 lb

A B

CE

DF

4 ft 4 ft

SOLUTION

Joint A:

Ans.

Ans.

Joint B:

Ans.

Ans.

Joint C:

Ans.

Ans.

Joint D:

Ans.

Ans.

Joint E:

Ans.FEF = 300 lb (C)

:+ ©Fx = 0; 180 - 35

(FEF) = 0

FDF = 230 lb (T)

:+ ©Fx = 0; FDF - 140 - 35

(150) = 0

FDE = 120 lb (C)

+ c ©Fy = 0; FDE - 45

(150) = 0

FCE = 180 lb (C)

:+ ©Fx = 0; -FCE + 35

(150) + 35

(150) = 0

FCD = 150 lb (T)

+ c ©Fy = 0; a45b FCD - a4

5b 150 = 0

+ c ©Fy = 0; FBC = 0

FBD = 140 lb (T)

:+ ©Fx = 0; FBD - 140 = 0

FAB = 140 lb (T)

:+ ©Fx = 0; FAB - 35

(150) - 513

(130) = 0

FAC = 150 lb (C)

+ c ©Fy = 0;45

(FAC) - 1213

(130) = 0

Determine the force in each member of the truss. State ifthe members are in tension or compression.

3 ft 3 ft 3 ft

125

13

130 lb

A B

CE

DF

4 ft 4 ft

Ans: FAC = 150 lb (C)FAB = 140 lb (T)FBD = 140 lb (T)FBC = 0FCD = 150 lb (T)FCE = 180 lb (C)FDE = 120 lb (C)FDF = 230 lb (T)FEF = 300 lb (C)