Cap 2 libro hibbeler estatica

Engineering Mechanics - Statics

Chapter 2

Problem 2-1 Determine the magnitude of the resultant force FR = F1 + F 2 and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 600 N F 2 = 800 N F 3 = 450 N

α = 45 deg β = 60 deg γ = 75 deg

Solution:

ψ = 90 deg − β + α
FR = F1 + F 2 − 2 F1 F 2 cos ( ψ)
2 2

F R = 867 N FR F2

sin ( ψ)

=

sin ( θ )sin ( ψ) ⎞ ⎟ FR ⎠

θ = asin ⎜F 2 θ = 63.05 deg φ = θ+α φ = 108 deg

⎛ ⎝

Problem 2-2 Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis. Given: F 1 = 80 lb F 2 = 60 lb
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© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyrightlaws 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.

Engineering Mechanics - Statics

Chapter 2

θ = 120 deg
Solution: FR = F1 + F 2 − 2 F1 F 2 cos ( 180 deg − θ )
2 2

F R = 72.1 lb

β = asin ⎜ F1 β = 73.9 deg

⎛ ⎝

sin ( 180 deg − θ ) ⎞ ⎟ FR ⎠

Problem 2-3 Determine themagnitude of the resultant force F R = F1 + F 2 and its direction, measured counterclockwise from the positive x axis. Given: F 1 = 250 lb F 2 = 375 lb

θ = 30 deg φ = 45 deg
Solution: FR = F1 + F 2 − 2 F1 F 2 cos ( 90 deg + θ − φ )
2 2

F R = 178 kg sin ( 90 deg + θ − φ ) FR = sin ( β ) F1

β = asin ⎜

⎛ F1 ⎝ FR

sin ( 90 deg + θ − φ )⎟

⎞

⎠

β = 37.89 deg
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© 2007 R. C.Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, 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.

Engineering Mechanics - Statics

Chapter 2

Angle measured ccw from x axis 360 deg − φ + β = 353 degProblem 2-4 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive u axis. Given: F 1 = 300 N F 2 = 500 N

α = 30 deg β = 45 deg γ = 70 deg
Solution: FR = F1 + F 2 − 2 F1 F 2 cos ( 180 deg − β − γ + α )
2 2

F R = 605 N sin ( 180 deg − β − γ + α ) FR = sin ( θ ) F2

θ = asin ⎜F 2 θ = 55.40 deg φ = θ+α φ = 85.4 deg
Problem2-5

⎛ ⎝

sin ( 180 deg − β − γ + α ) ⎞ ⎟ FR ⎠

Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 300 N

α = 30 deg
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© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics

Chapter 2

F 2 = 500 N

β = 45 deg

γ = 70 deg
Solution: F1u = F1

sin ( γ − α ) F 1u = F1

sin ( 180 deg − γ ) sin ( γ − α )

sin ( 180 deg − γ )

F 1u = 205 N sin ( α ) F 1v = sin ( 180 deg − γ ) sin ( α ) F1

F 1v = F 1sin ( 180 deg − γ )

F 1v = 160 N

Problem 2-6 Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. Given: F 1 = 300 N F 2 = 500 N

α = 30 deg β = 45 deg γ = 70 deg
Solution: F 2u = F2 ⎜

⎛ sin ( β ) ⎞ ⎟ ⎝ sin ( γ ) ⎠
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© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rightsreserved. 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.

Engineering Mechanics - Statics

Chapter 2

F 2u = 376.2 N F 2v = F 2 ⎢

⎡ sin ⎡180 deg − ( β + γ )⎤⎤ ⎣ ⎦ ⎥ sin ( γ ) ⎣ ⎦

F 2v = 482.2 N

Problem 2-7 Determine the magnitude of...