Engineering Mechanics - Statics Chapter 1 -...

Engineering Mechanics - Statics Chapter 1

Problem 1-1

Represent each of the following combinations of units in the correct SI form using anappropriate prefix:(a) m/ms (b) μkm (c) ks/mg (d) km μN⋅

Units Used:

μN 10 6− N=

μkm 10 6− km=

Gs 109 s=

ks 103 s=

mN 10 3− N=

ms 10 3− s=

Solution:

a( ) mms

1 103×ms

=

mms

1kms

=

b( ) μkm 1 10 3−× m=

μkm 1 mm=

c( ) ksmg

1 109×s

kg=

ksmg

1Gskg

=

d( ) km μN⋅ 1 10 3−× m N=

km μN⋅ 1 mm N⋅=

1

© 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

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Problem 1-2

Wood has a density d. What is its density expressed in SI units?

Units Used:

Mg 1000 kg=

Given:

d 4.70slug

ft3=

Solution:

1slug 14.594 kg=

d 2.42Mg

m3=

Problem 1-3

Represent each of the following combinations of units in the correct SI form using anappropriate prefix:

(a) Mg/mm (b) mN/μs (c) μm Mg⋅

Solution:

a( )Mgmm

103 kg

10 3− m=

106 kgm

=Ggm

=

Mgmm

Ggm

=

b( )mNμs

10 3− N

10 6− s=

103 Ns

=kNs

=

mNμs

kNs

=

c( ) μm Mg⋅ 10 6− m( ) 103 kg( )= 10 3− m kg⋅=

μm Mg⋅ mm kg⋅=

2




Problem 1-4

Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b)N/mm, (c) mN/( kg μs⋅ ).

Solution:

a( )Mgms

103 kg

10 3− s=

106 kgs

=Ggs

=

Mgms

Ggs

=

b( )N

mm1 N

10 3− m= 103 N

m=

kNm

=

Nmm

kNm

=

c( )mN

kg μs⋅

10 3− N

10 6− kg s⋅=

kNkg s⋅

=

mNkg μs⋅

kNkg s⋅

=

Problem 1-5

Represent each of the following with SI units having an appropriate prefix: (a) S1, (b) S2, (c) S3.

Units Used:

kg 1000 g= ms 10 3− s= kN 103 N=

Given:

S1 8653 ms=

S2 8368 N=

S3 0.893 kg=

Solution:

a( ) S1 8.653 s=

3




b( ) S2 8.368 kN=

c( ) S3 893 g=

Problem 1-6

Represent each of the following to three significant figures and express each answer in SI unitsusing an appropriate prefix: (a) x, (b) y, and (c) z.

Units Used:

MN 106 N=

μg 1 10 6−× gm=

kN 103 N=

Given:

x 45320 kN=

y 568 105×( ) mm=

z 0.00563 mg=

Solution:

a( ) x 45.3 MN=

b( ) y 56.8 km=

c( ) z 5.63 μg=

Problem 1-7

Evaluate ( a b⋅ )/c to three significant figures and express the answer in SI units using anappropriate prefix.

Units Used:

μm 10 6− m=

Given:

a 204 mm( )=

4




b 0.00457 kg( )=

c 34.6 N( )=

Solution:

la bc

= l 26.945μm kg⋅

N=

Problem 1-8

If a car is traveling at speed v, determine its speed in kilometers per hour and meters per second.

Given:

v 55mihr

=

Solution:

v 88.514kmhr

=

v 24.6ms

=

Problem 1-9

Convert: (a) S1 to N m⋅ , (b) S2 to kN/m3, (c) S3 to mm/s. Express the result to threesignificant figures. Use an appropriate prefix.

Units Used:

kN 103 N=

Given:

S1 200g lb ft⋅=

S2 350glb

ft3=

S3 8fthr

=

5




Solution:

a( ) S1 271 N m⋅=

b( ) S2 55.0kN

m3=

c( ) S3 0.677mm

s=

Problem 1-10

What is the weight in newtons of an object that has a mass of: (a) m1, (b) m2, (c) m3? Expressthe result to three significant figures. Use an appropriate prefix.

Units Used:

Mg 103 kg=

mN 10 3− N=

kN 103 N=

Given:

m1 10 kg=

m2 0.5 gm=

m3 4.50 Mg=

Solution:

a( ) W m1 g=

W 98.1 N=

b( ) W m2 g=

W 4.90 mN=

c( ) W m3 g=

W 44.1 kN=

6




Problem 1-11

If an object has mass m, determine its mass in kilograms.

Given:

m 40 slug=

Solution:

m 584 kg=

Problem 1-12

The specific weight (wt./vol.) of brass is ρ. Determine its density (mass/vol.) in SI units.Use an appropriate prefix.

Units Used:

Mg 103 kg=

Given:

ρ 520lb

ft3=

Solution:

ρ 8.33Mg

m3=

Problem 1-13

A concrete column has diameter d and length L. If the density (mass/volume) of concrete is ρ,determine the weight of the column in pounds.

Units Used:

Mg 103 kg=

kip 103 lb=

Given:

d 350 mm=

L 2 m=

7




ρ 2.45Mg

m3=

Solution:

V πd2⎛⎜⎝⎞⎟⎠

2L= V 192.423 L=

W ρ V= W 1.04 kip=

Problem 1-14

The density (mass/volume) of aluminum is ρ. Determine its density in SI units. Use anappropriate prefix.

Units Used:

Mg 1000 kg=

Given:

ρ 5.26slug

ft3=

Solution:

ρ 2.17Mg

m3=

Problem 1-15

Determine your own mass in kilograms, your weight in newtons, and your height in meters.

Solution:

Example

W 150 lb=

m W= m 68.039 kg=

W g 667.233 N=

h 72 in= h 1.829 m=

8




Problem 1-16

Two particles have masses m1 and m2, respectively. If they are a distance d apart, determinethe force of gravity acting between them. Compare this result with the weight of each particle.

Units Used:

G 66.73 10 12−×m3

kg s2⋅=

nN 10 9− N=

Given:

m1 8 kg=

m2 12 kg=

d 800 mm=

Solution:

FG m1 m2

d2=

F 10.0 nN=

W1 m1 g= W1 78.5 N=W1F

7.85 109×=

W2 m2 g= W2 118 N=W2F

1.18 1010×=

Problem 1-17

Using the base units of the SI system, show that F = G(m1m2)/r2 is a dimensionallyhomogeneous equation which gives F in newtons. Compute the gravitational force actingbetween two identical spheres that are touching each other. The mass of each sphere is m1, andthe radius is r.

Units Used:

μN 10 6− N= G 66.73 10 12− m3

kg s2⋅=

9




Given:

m1 150 kg=

r 275 mm=

Solution:

FG m1

2

2r( )2=

F 4.96 μN=

Since the force F is measured in Newtons, then the equation is dimensionally homogeneous.

Problem 1-18

Evaluate each of the following to three significant figures and express each answer in SI unitsusing an appropriate prefix: (a) x, (b) y, (c) z.

Units Used:

MN 106 N=

kN 103 N=

μm 10 6− m=

Given:

x 200 kN( )2=

y 0.005 mm( )2=

z 400 m( )3=

Solution:

a( ) x 0.040 MN2=

b( ) y 25.0 μm2=

c( ) z 0.0640 km3=

10




Problem 1-19

Evaluate each of the following to three significant figures and express each answer in SI unitsusing an appropriate prefix: (a) a1/b1, (b) a2b2/c2, (c) a3b3.

Units Used:

μm 10 6− m= Mm 106 m=

Mg 106 gm= kg 103 gm=

ms 10 3− s=

Given:

a1 684 μm=

b1 43 ms=

a2 28 ms=

b2 0.0458 Mm=

c2 348 mg=

a3 2.68 mm=

b3 426 Mg=

Solution:

a( )a1b1

15.9mm

s=

b( )a2 b2

c23.69 Mm

skg

=

c( ) a3 b3 1.14 km kg⋅=

Problem 1-20

Evaluate each of the following to three significant figures and express each answer in SI unitsusing an appropriate prefix: (a) a1/b1

2 (b) a22b2

3.

Units Used:

Mm 106 m=

11




Given:

a1 0.631 Mm=

b1 8.60 kg=

a2 35 mm=

b2 48 kg=

Solution:

a( )a1

b12

8.532km

kg2=

b( ) a22 b2

3 135.48 kg3 m2⋅=

12




Problem 2-1

Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measuredcounterclockwise from the positive x axis.

Given:

F1 600 N=

F2 800 N=

F3 450 N=

α 45 deg=

β 60 deg=

γ 75 deg=

Solution:

ψ 90 deg β− α+=

FR F12 F2

2+ 2 F1 F2 cos ψ( )−=

FR 867 N=

FR

sin ψ( )F2

sin θ( )=

θ asin F2sin ψ( )

FR

⎛⎜⎝

⎞⎟⎠

=

θ 63.05 deg=

φ θ α+=

φ 108 deg=

Problem 2-2

Determine the magnitude of the resultant force and its direction measured counterclockwisefrom the positive x axis.

Given:

F1 80 lb=

F2 60 lb=

13




θ 120 deg=

Solution:

FR F12 F2

2+ 2 F1 F2 cos 180 deg θ−( )−=

FR 72.1 lb=

β asin F1sin 180 deg θ−( )

FR

⎛⎜⎝

⎞⎟⎠

=

β 73.9 deg=

Problem 2-3

Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measuredcounterclockwise from the positive x axis.

Given:

F1 250 lb=

F2 375 lb=

θ 30 deg=

φ 45 deg=

Solution:

FR F12 F2

2+ 2 F1 F2 cos 90 deg θ+ φ−( )−=

FR 178 kg=

FR

sin 90 deg θ+ φ−( )F1

sin β( )=

β asinF1FR

sin 90 deg θ+ φ−( )⎛⎜⎝

⎞⎟⎠

=

β 37.89 deg=

14




Angle measured ccw from x axis

360 deg φ− β+ 353 deg=

Problem 2-4

Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measuredcounterclockwise from the positive u axis.

Given:

F1 300 N=

F2 500 N=

α 30 deg=

β 45 deg=

γ 70 deg=

Solution:

FR F12 F2

2+ 2 F1 F2 cos 180 deg β− γ− α+( )−=

FR 605 N=

FR

sin 180 deg β− γ− α+( )F2

sin θ( )=

θ asin F2sin 180 deg β− γ− α+( )

FR

⎛⎜⎝

⎞⎟⎠

=

θ 55.40 deg=

φ θ α+=

φ 85.4 deg=

Problem 2-5

Resolve the force F1 into components acting along the u and v axes and determine themagnitudes of the components.

Given:

F1 300 N= α 30 deg=

15




F2 500 N= β 45 deg=

γ 70 deg=

Solution:

F1u

sin γ α−( )F1

sin 180 deg γ−( )=

F1u F1sin γ α−( )

sin 180 deg γ−( )=

F1u 205 N=

F1v

sin α( )F1

sin 180 deg γ−( )=

F1v F1sin α( )

sin 180 deg γ−( )=

F1v 160 N=

Problem 2-6

Resolve the force F2 into components acting along the u and v axes and determine themagnitudes of the components.

Given:

F1 300 N=

F2 500 N=

α 30 deg=

β 45 deg=

γ 70 deg=

Solution:

F2u F2sin β( )sin γ( )

⎛⎜⎝

⎞⎟⎠

=

16




F2u 376.2 N=

F2v F2sin 180 deg β γ+( )−⎡⎣ ⎤⎦

sin γ( )⎡⎢⎣

⎤⎥⎦

=

F2v 482.2 N=

Problem 2-7

Determine the magnitude of the resultant force FR = F1 + F2 and its direction measuredcounterclockwise from the positive u axis.

Given:

F1 25 lb=

F2 50 lb=

θ1 30 deg=

θ2 30 deg=

θ3 45 deg=

Solution:

α 180 deg θ3 θ1+( )−=

FR F22 F1

2+ 2 F1 F2 cos α( )−=

FR 61.4 lb=

sin θ'( )F2

sin α( )FR

= θ' asin sin α( )F2FR

⎛⎜⎝

⎞⎟⎠

=

θ' 51.8 deg=

θ θ' θ3−=

θ 6.8 deg=

Problem 2-8

Resolve the force F1 into components acting along the u and v axes and determine the components.

17




Given:

F1 25 lb=

F2 50 lb=

θ1 30 deg=

θ2 30 deg=

θ3 45 deg=

Solution:

Fu−

sin θ3 θ2−( )F1

sin θ2( )=

FuF1− sin θ3 θ2−( )

sin θ2( )=

Fu 12.9− lb=

Fv

sin 180 deg θ3−( )F1

sin θ2( )=

FvF1 sin 180 deg θ3−( )

sin θ2( )=

Fv 35.4 lb=

Problem 2-9

Resolve the force F2 into components acting alongthe u and v axes and determine the components.

Given:

F1 25 lb=

F2 50 lb=

θ1 30 deg=

18




θ2 30 deg=

θ3 45 deg=

Solution:

Fu

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦

F2

sin θ2( )=

FuF2 sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦

sin θ2( )=

Fu 86.6 lb=

Fv−

sin θ1( )F2

sin θ2( )=

FvF2− sin θ1( )sin θ2( )=

Fv 50− lb=

Problem 2-10

Determine the components of the F force acting along the u and v axes.

Given:

θ1 70 deg=

θ2 45 deg=

θ3 60 deg=

F 250 N=

Solution:

Fu

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦

F

sin θ2( )=

FuF sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦

sin θ2( )= Fu 320 N=

19




Fv

sin θ1( )F

sin θ2( )=

FvF sin θ1( )sin θ2( )= Fv 332 N=

Problem 2-11

The force F acts on the gear tooth. Resolve this force into two components acting along thelines aa and bb.

Given:

F 20 lb=

θ1 80 deg=

θ2 60 deg=

Solution:

Fsin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦

Fa

sin θ1( )=

FaF sin θ1( )

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦= Fa 30.6 lb=

Fsin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦

Fb

sin θ2( )=

FbF sin θ2( )

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦= Fb 26.9 lb=

Problem 2-12

The component of force F acting along line aa is required to be Fa. Determine the magnitudeof F and its component along line bb.

20




Given:

Fa 30 lb=

θ1 80 deg=

θ2 60 deg=

Solution:

Fa

sin θ1( )F

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦=

F Fasin 180 deg θ1− θ2−( )

sin θ1( )⎛⎜⎝

⎞⎟⎠

= F 19.6 lb=

Fa

sin θ1( )Fb

sin θ2( )=

FbFa sin θ2( )

sin θ1( )= Fb 26.4 lb=

Problem 2-13

A resultant force F is necessary to hold the ballon in place. Resolve this force into componentsalong the tether lines AB and AC, and compute the magnitude of each component.

Given:

F 350 lb=

θ1 30 deg=

θ2 40 deg=

Solution:

FAB

sin θ1( )F

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦=

21




FAB Fsin θ1( )

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦

⎡⎢⎣

⎤⎥⎦

=

FAB 186 lb=

FAC

sin θ2( )F

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦=

FAC Fsin θ2( )

sin 180 deg θ1 θ2+( )−⎡⎣ ⎤⎦

⎡⎢⎣

⎤⎥⎦

=

FAC 239 lb=

Problem 2-14

The post is to be pulled out of the ground using two ropes A and B. Rope A is subjected to forceF1 and is directed at angle θ1 from the horizontal. If the resultant force acting on the post is to beFR, vertically upward, determine the force T in rope B and the corresponding angle θ.

Given:

FR 1200 lb=

F1 600 lb=

θ1 60 deg=

Solution:

T F12 FR

2+ 2 F1 FR cos 90 deg θ1−( )−=

T 744 lb=

sin θ( )FR

sin 90 θ1−( )T

=

θ asin sin 90 deg θ1−( )F1T

⎛⎜⎝

⎞⎟⎠

= θ 23.8 deg=

22




Problem 2-15

Resolve the force F1 into components acting along the u and v axes and determine the magnitudesof the components.

Given:

F1 250 N=

F2 150 N=

θ1 30 deg=

θ2 30 deg=

θ3 105 deg=

Solution:

F1v

sin θ1( )F1

sin θ3( )=

F1v F1sin θ1( )sin θ3( )

⎛⎜⎝

⎞⎟⎠

=

F1v 129 N=

F1u

sin 180 deg θ1− θ3−( )F1

sin θ3( )=

F1u F1sin 180 deg θ1− θ3−( )

sin θ3( )⎛⎜⎝

⎞⎟⎠

=

F1u 183 N=

Problem 2-16

Resolve the force F2 into components acting along the u and v axes and determine the magnitudesof the components.

Given:

F1 250 N=

23




F2 150 N=

θ1 30 deg=

θ2 30 deg=

θ3 105 deg=

Solution:

F1v

sin θ1( )F2

sin 180 deg θ3−( )=

F1v F2sin θ1( )

sin 180 deg θ3−( )⎛⎜⎝

⎞⎟⎠

=

F1v 77.6 N=

F2u

sin 180 deg θ3−( )F2

sin 180 deg θ3−( )=

F2u F2sin 180 deg θ3−( )sin 180 deg θ3−( )

⎛⎜⎝

⎞⎟⎠

=

F2u 150 N=

Problem 2-17

Determine the magnitude and direction of the resultant force FR. Express the result in terms ofthe magnitudes of the components F1 and F2 and the angle φ.

Solution:

FR2 F1

2 F22+ 2F1 F2 cos 180 deg φ−( )−=

24




Since cos 180 deg φ−( ) cos− φ( )= ,

FR F12 F2

2+ 2 F1 F2 cos φ( )+=

From the figure,

tan θ( )F1 sin φ( )

F2 F1 cos φ( )+=

Problem 2-18

If the tension in the cable is F1, determine the magnitude and direction of the resultant force actingon the pulley. This angle defines the same angle θ of line AB on the tailboard block.

Given:

F1 400 N=

θ1 30 deg=

Solution:

FR F12 F1

2+ 2F1 F1 cos 90 deg θ1−( )−=

FR 400 N=

sin 90 deg θ−( )FR

sin θ1( )F1

=

θ 90 deg asinFRF1

sin θ1( )⎛⎜⎝

⎞⎟⎠

−=

θ 60 deg=

25




Problem 2-19

The riveted bracket supports two forces. Determine the angle θ so that the resultant forceis directed along the negative x axis. What is the magnitude of this resultant force?

Given:

F1 60 lb=

F2 70 lb=

θ1 30 deg=

Solution:

sin θ( )F1

sin θ1( )F2

=

θ asin sin θ1( )F1F2

⎛⎜⎝

⎞⎟⎠

=

θ 25.4 deg=

φ 180 deg θ− θ1−=

φ 124.6 deg=

R F12 F2

2+ 2F1 F2 cos φ( )−=

R 115 lb=

Problem 2-20

The plate is subjected to the forces acting on members A and B as shown. Determine the magnitudeof the resultant of these forces and its direction measured clockwise from the positive x axis.

Given:

FA 400 lb=

FB 500 lb=

26




θ1 30 deg=

θ 60 deg=

Solution:

Cosine law:

FR FB2 FA

2+ 2FB FA cos 90 deg θ− θ1+( )−=

FR 458 lb=

Sine law:

sin 90 deg θ− θ1+( )FR

sin θ α−( )FA

=

α θ asin sin 90 deg θ− θ1+( )FAFR

⎛⎜⎝

⎞⎟⎠

−=

α 10.9 deg=

Problem 2-21

Determine the angle θ for connecting member B to the plate so that the resultant of FA and FBis directed along the positive x axis. What is the magnitude of the resultant force?

Given:

FA 400 lb=

FB 500 lb=

θ1 30 deg=

27




Solution:

Sine law:

sin θ( )FA

sin 90 deg θ1−( )FB

=

θ asin sin 90 deg θ1−( )FAFB

⎛⎜⎝

⎞⎟⎠

=

θ 43.9 deg=

FR

sin 90 deg θ1+ θ−( )FA

sin θ( )=

FR FAsin 90 deg θ− θ1+( )

sin θ( )=

FR 561 lb=

Problem 2-22

Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forcesby first finding the resultant F' = F1 + F2 and then forming FR = F' + F3.

Given:

F1 30 N=

F2 20 N=

F3 50 N=

θ 20 deg=

c 3=

d 4=

28




Solution:

α atancd

⎛⎜⎝

⎞⎟⎠

=

F' F12 F2

2+ 2F1 F2 cos 90 deg θ+ α−( )−=

F' 30.9 N=

F'sin 90 deg θ− α+( )( )

F1

sin 90 deg θ− β−( )=

β 90 deg θ− asin F1sin 90 deg θ− α+( )

F'⎛⎜⎝

⎞⎟⎠

−=

β 1.5 deg=

Now add in force F3.

FR F'2 F32+ 2F' F3 cos β( )−=

FR 19.2 N=

FR

sin β( )F'

sin φ( )=

φ asin F'sin β( )

FR

⎛⎜⎝

⎞⎟⎠

=

φ 2.4 deg=

Problem 2-23

Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces byfirst finding the resultant F' = F2 + F3 and then forming FR = F' + F1.

Given:

F1 30 N=

F2 20 N=

29




F3 50 N=

θ 20 deg=

c 3=

d 4=

Solution:

F' F22 F3

2+ 2F2 F3 cos 90 deg θ−( )( )−=

F' 47.07 N=

F2

sin β( )F'

sin 90 deg θ−( )=

β asin F2sin 90 deg θ−( )

F'⎛⎜⎝

⎞⎟⎠

=

β 23.53 deg=

α atancd

⎛⎜⎝

⎞⎟⎠

=

γ α β−=

FR F'2 F12+ 2F' F1 cos γ( )−=

FR 19.2 N=

FR

sin γ( )F1

sin φ( )=

φ asin F1sin γ( )

FR

⎛⎜⎝

⎞⎟⎠

=

φ 21.16 deg=

ψ β φ−=

ψ 2.37 deg=

30



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