Problem 7.29 Determine the coordinates of the
centroids.
Solution: Break into a rectangle, a triangle and a circular hole
xD
y
yD
2 in
8 in
5[108] C 12
108 C
1
1
2 86 4[108] C 13 8
108 C
2 86
1
2 86
4[22 ]
22
3[22 ]
1
2
2 86 2
D 6.97 in
D 3.79 in
x D 6.97 in
y D 3.79 in
3 in
x
4 in
6 in
10 in
Problem 7.30 Determine the coordinates of the
centroids.
Solution: The strategy is to find the centroid for the half circle
area, and use the result in the composite algorithm. The area: The
element of area is a vertical
strip y high and dx wide. From the equation
p
of the circle, y D š R2 x 2 . The
p height of the strip will be twice
the positive value, so that dA D 2 R2 x 2 dx, from which
dA D 2
AD
A
R
R2 x 2 1/2 dx
y
10 in
x
20 in
0
p
R
R2
x R2 x 2
R2
1 x
D
D2
C
sin
2
2
R
2
0
The x-coordinate:
x dA D 2
A
R
x
R2 x 2 dx
0
R
2R3
R2 x 2 3/2
D
D2 .
3
3
0
Divide by A: x D
4R
3
The y-coordinate: From symmetry, the y-coordinate is zero.
420
D 8.488 in. For
3
the inner half circle x2 D 4.244 in. The areas are
The composite: For a complete half circle x1 D
A1 D 628.32 in2
and A2 D 157.08 in2 .
c 2008 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.
525
Problem 10.10 If x D 4 ft, what are the internal forces
and moment at A?
y
600 lb/ft
x
x
A
600 lb/ft
3 ft
3 ft
Solution: Isolating the part of the beam to the right of A, we
represent the distributed load by an equivalent force.
We can obtain the magnitude of the distributed load by similar triangles:
2
600 lb/ft D 400 lb/ft.
3
If we represent the distributed load to the right of point A by a single
equivalent force, its magnitude is
1
400 lb/ft2 ft D 400 lb,
2
and it acts at the centroid of the distributed load to the right of point
A. The distance from A to the centroid is
1
2 ft D 0.667 ft.
3
From this free-body diagram, we write the equilibrium equations:
Fx : PA D 0,
Fy : VA C 400 lb D 0,
MA : MA C 400 lb0.667 ft D 0.
Solving yields
PA D 0, VA D 400 lb, MA D 267 ft-lb.
c 2008 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.
795
Problem 10.13 Determine the internal forces and
moment at A.
300 lb/ft
200 lb/ft
A
6 ft
8 ft
4 ft
Solution: Use the whole body to find the reactions
MC : B8 ft C 1600 lb4 ft
C 400 lb2.67 ft 600 lb1.33 ft D 0
) B D 833 lb
400 lb
600 lb
1600 lb
B
C
Now examine the section to the left of the cut
Fx : PA D 0
Fy : B 1200 lb 225 lb VA D 0
MA : B6 ft C 1200 lb3 ft
C 225 lb2 ft C MA D 0
Solving
PA D 0, VA D 592 lb, MA D 950 ft-lb
225 lb
1200 lb
275 lb/ft
200 lb/ft
MA
6 ft
B
798
PA
VA
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 9.3 A student pushes a 200-lb box of books
across the floor. The coefficient of kinetic friction
between the carpet and the box is k D 0.15.
(a) If he exerts the force F at angle ˛ D 25° , what is
the magnitude of the force he must exert to slide
the box across the floor?
(b) If he bends his knees more and exerts the force
F at angle ˛ D 10° , what is the magnitude of the
force he must exert to slide the box?
Solution:
a
F
F
200 lb
Fx : F cos ˛ f D 0
α
Fy : N 200 lb F sin ˛ D 0
f D 0.15 N
(a)
˛ D 25° ) F D 35.6 lb
f
(b)
˛D
10°
) F D 31.3 lb
N
Problem 9.4 The 2975-lb car is parked on a sloped
street. The brakes are applied to both its front and rear
wheels.
(a) If the coefficient of static friction between the car’s
tires and the road is s D 0.8, what is the steepest
slope (in degrees relative to the horizontal) on
which the car could remain in equilibrium?
(b) If the street were icy and the coefficient of static
friction between the car’s tires and the road was
s D 0.2, what is the steepest slope on which the
car could remain in equilibrium?
Solution: Let ˛ be the slope of the street in degrees. The equilibrium equations and impending slip friction equation are
Fx : W sin ˛ f D 0,
Fy : N W cos ˛ D 0,
f D s N
Solving, we find that
f D W sin ˛, N D W cos ˛, s D tan ˛.
˛ D tan1 s .
a ˛ D tan1 0.8 D 38.7° .
b ˛ D tan1 0.2 D 11.3° .
a ˛ D 38.7° , b ˛ D 11.3° .
686
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 9.14 The box is stationary on the inclined
surface. The coefficient of static friction between the
box and the surface is s .
(a)
(b)
If the mass of the box is 10 kg, ˛ D 20° , ˇ D 30° ,
and s D 0.24, what force T is necessary to start
the box sliding up the surface?
Show that the force T necessary to start the box
sliding up the surface is a minimum when
tan ˇ D s .
T
β
α
Solution:
T
mg
y
α
β
α
f
x
N
˛ D 20°
s D 0.24
m D 10 kg
g D 9.81 m/s2
(a)
Fx :
T cos ˇ C f C mg sin ˛ D 0
Fy :
N C T sin ˇ mg cos ˛ D 0
ˇ D 30° , f D s N
Substituting the known values and solving, we get
T D 56.5 N,
N D 64.0 N,
f D 15.3 N
Solving the 2nd equilibrium eqn for N and substituting for
f f D s N in the first eqn, we get
T cos ˇ C s mg cos ˛ s T sin ˇ C mg sin ˛ D 0
Differentiating with respect to ˇ, we get
dT
Tsin ˇ s cos ˇ
D
dˇ
cos ˇ C s sin ˇ
Setting
dT
D 0, we get
dˇ
tan ˇ D s
c 2008 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.
693