PROBLEM 8.

Determine whether the block shown is in equilibrium and find the

magnitude and direction of the friction force when θ = 30° and P = 50
lb.

SOLUTION

Assume equilibrium:

S Fx = 0: F - (250 lb) sin 30 + (50 lb) cos 30 = 0

F = +81.699 lb
SFy = 0: N - (250 lb) cos 30 - (50 lb)sin 30 = 0

N = +241.51 lb
Maximum friction force:

Fm = ms N
= 0.30(241.51 lb)
= 72.5 lb
We note that F > Fm . Thus, block moves down◀

Actual friction force:

F = Fk = mk N = 0.20(241.51 lb) = 48.3 lb, F = 48.3 lb ◀

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1238
 PROBLEM 8.23

The 10-lb uniform rod AB is held in the position shown by the force P.
Knowing that the coefficient of static friction is 0.20 at A and B, determine
the smallest value of P for which equilibrium is maintained.

SOLUTION

Free-body diagram

SFy = 0: N A + 0.2 N B -10 lb = 0

N A = 10 - 0.2 N B

SM G = 0: N A (4 in.) - 0.2 N A (7.5 in.) - N B (7.5 in.) - 0.2 N B (4 in.) = 0

2.5N A - 8.3N B = 0
2.5(10 - 0.2 N B ) - 8.3N B = 0
N B = 2.8409 lb

SM A = 0: 10(4 in.) + P (7.5 in.) - N B (15 in.) - 0.2 N B (8 in.)=0

7.5P + 40 - (2.8409)(15) - 0.2 (2.8409)(8) = 0

Pmin = 0.955 lb ◀

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1265
 PROBLEM 5.3

Locate the centroid of the plane area shown.

SOLUTION

Area 1: Rectangle 120 mm by 100 mm

Area 2: Triangle b = 60 mm, h = 120 mm

A, mm2 x , mm y , mm xA,mm 3 yA,mm3

3 5 6
1 12x10 60 110 7.2x10 1.32x10
3 5 5
2 3.6x10 40 40 1.44x10 1.44x10
3 5 6
S 15.6x10 8.64x10 1.464x10

XSA= x A

X (15.6 ´103 mm2 ) = 8.64 ´105 mm3 X = 55.4 mm ◀

YSA=Sy A

Y (15.6 ´103 mm2 ) = 1.464 ´106 mm3 Y = 93.8 mm ◀

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561
 y PROBLEM 5.8

Locate the centroid of the plane area shown.

r = 38 in.
16 in.

x
20 in.

SOLUTION

A, in 2 x , in. y , in. xA, in3 yA, in3

p
1 (38)2 = 2268.2 0 16.1277 0 36,581
2

2 - 20 ´ 16 = 320 -10 8 3200 -2560

S 1948.23 3200 34,021

S xA 3200 in 3
Then X= = X = 1.643 in. ◀
SA 1948.23 in 2
S yA 34,021 in3
Y= = Y = 17.46 in. ◀
SA 1948.23 in 2

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566
 400 N/m PROBLEM 5.66
1600 N/m
For the beam and loading shown, determine (a) the magnitude
A B and location of the resultant of the distributed load, (b) the
6m reactions at the beam supports.

SOLUTION

1
(a) R I = (400 N/m)(6 m)
2
= 1200 N
R II = (1600 N/m)(6 m)
= 4800 N

+  SFy = 0 : - R = -R I - R II
R = 1200 N + 4800 N
R = 6000 N

SM A = 0 : - X (6000) = -2 (1200) - 4 ( 4800)

X = 3.60 m

R = 6000 N  , X = 3.60 m ◀

(b) Reactions SFx = 0 : A x = 0

SFy = 0 : A y - 6000 = 0 A y = 6000 N

A = 6000 N ◀

SM A = 0: M A - (6000 N)(3.60 m) = 0

MA = 21.6 kN ⋅ m ◀

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638
 PROBLEM 9.36

Determine the moments of inertia of the shaded area shown with respect to the
x and y axes.

SOLUTION

Assign area 1 to be a rectangle and area 2 to be a semicircle.

We have

Ix = (Ix )1 -(Ix )2
where

1
( I x )1 = (125 mm)(250 mm)3 = 651.04 ´106 mm 4
3
p p
( I x )2 = 8 (75 mm)4 + 2 (75 mm)2 (125 mm) = 150.484 ´106 mm 4
2

I x = 651.04 ´106 mm 4 -150.484 ´106 mm 4

I x = 501´10 6 mm 4 ◀

Also I y = ( I y )1 - ( I y )2

1
( I y )1 = (250 mm)(125 mm)3 = 162.760 ´106 mm4
3
p
( I y )2 = (75 mm)4 = 12.4252 ´106 mm 4
8
Then

I y = 162.760 ´106 mm 4 - 12.4252 ´106 mm 4

I y =150.3´106 mm4 ◀

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1466
PROBLEM 7.21

A force P is applied to a bent rod that is supported by a roller and a pin and bracket. For each of the three
cases shown, determine the internal forces at Point J.

SOLUTION

(a) FBD Rod: SMD = 0: aP - 2aA = 0

P
A=
2
P P
SFx = 0: V - =0 V= ◀
2 2

FBD AJ: SFy = 0: F=0◀

P aP
SM J = 0: M - a =0 M= ◀
2 2

(b) FBD Rod:

a æ3 ö
SM D = 0: aP - çç A÷÷÷ = 0
2 èç 5 ø÷

10 P
A=
3

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1046
 PROBLEM 7.21 (Continued)

4 10 P
FBD AJ: SFx = 0: -V = 0 V=
8P
◀
5 3 3

3 10 P
SFy = 0: -F = 0 F = 2P ◀
5 3
8aP
M= ◀
3
(c) FBD Rod:

æ 4 A ö÷ æ 3A ö
SM D = 0: aP - 2 a çç ÷- ÷ 2 a çç ÷÷÷ = 0
èç 5 ø÷ çè 5 ÷ø

5P
A=
14
æ 4 5P ÷ö 2P
SFx = 0: V - çç ÷= 0 V= ◀
èç 5 14 ÷÷ø 7

3 5P 3P
SFy = 0: -F = 0 F= ◀
5 14 14
æ 4 5P ö÷ 2 aP
SM J = 0: M - a çç ÷= 0 M= ◀
çè 5 14 ÷÷ø 7

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1047