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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⋅
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Engineering Mechanics - Statics Chapter 1
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.
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Engineering Mechanics - Statics Chapter 1
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.
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Engineering Mechanics - Statics Chapter 1
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
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Engineering Mechanics - Statics Chapter 2
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.
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Engineering Mechanics - Statics Chapter 2
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.
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Engineering Mechanics - Statics Chapter 2
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.
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Engineering Mechanics - Statics Chapter 2
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.
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Engineering Mechanics - Statics Chapter 2
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.
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Engineering Mechanics - Statics Chapter 2
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?
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Engineering Mechanics - Statics Chapter 2
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.
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Engineering Mechanics - Statics Chapter 2
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.