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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–1.
Locate the center of mass of the homogeneous rod bent
into the shape of a circular arc.

y

30Њ
300 mm
x

SOLUTION
dL = 300 d u
30Њ

'
x = 300 cos u
'
y = 300 sin u

x =

L

2p
3

'
x dL

=

L

L-2p3

300 cos u (300du)
2p
3

dL

L-2p3

300d u

(300)2 C sin u D -3 2p3
2p

=

4
300 a p b
3

Ans.

= 124 mm
y = 0


Ans.

(By symmetry)

Ans:
x = 124 mm
y = 0
879


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–2.
y

Determine the location (x, y) of the centroid of the wire.

2 ft

4 ft

SOLUTION
Length and Moment Arm: The length of the differential element is dL =
2dx2 + dy2 = ¢

B

1 + a


y = x2

2

dy
dy
b ≤ dx and its centroid is ∼
= 2x.
y = y = x2. Here,
dx
dx

x

Centroid: Due to symmetry
∼

Ans.

x = 0

Applying Eq. 9–7 and performing the integration, we have
2 ft

∼
∼

y =

LL


ydL
=
dL

LL

L-2 ft

x 21 + 4x2 dx

2 ft

L-2 ft
=

2

21 + 4x2 dx

16.9423
= 1.82 ft
9.2936

Ans.

Ans:
x = 0
y = 1.82 ft
880



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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–3.
y

Locate the center of gravity x of the homogeneous rod. If
the rod has a weight per unit length of 100 N>m, determine
the vertical reaction at A and the x and y components of
reaction at the pin B.

1m
B

y ϭ x2
A

1m

x

Solution
Length

And

Arm. The length of the differential element is
dy

dy 2
dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~
x = x. Here
= 2x.
A
dx
dx
Perform the integration
L =

Moment

LL

dL =

L0

= 2

1m

L0

= cx

21 + 4x2 dx

1m


A

A

x2 +

x2 +

= 1.4789 m
LL

x~ dL =

L0

= 2

1
dx
4

1
1
1 1m
+ ln ax +
x2 + b d
4
4
A
4 0


1m

x21 + 4x2 dx

L0

1m

x

A

x2 +

1
dx
4

2
1 3>2 1 m
= c ax2 + b d
3
4
0

= 0.8484 m2
Centroid.
x =


~
0.8484 m2
1L x dL
= 0.5736 m = 0.574 m
=
1.4789 m
1L dL

Ans.

Equations of Equilibrium. Refering to the FBD of the rod shown in Fig. a
+ ΣFx = 0;
S

Ans.

Bx = 0

a+ΣMB = 0;  100(1.4789) (0.4264) - Ay(1) = 0
Ans.

Ay = 63.06 N = 63.1 N
a+ΣMA = 0;  By(1) - 100(1.4789) (0.5736) = 0

Ans.

By = 84.84 N = 84.8 N

Ans:
x = 0.574 m

Bx = 0
Ay = 63.1 N
By = 84.8 N
881


© 2016 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.

*9–4.
y

Locate the center of gravity y  of the homogeneous rod.

1m
B

y ϭ x2
A

1m

x

Solution
Length

And

Arm. The length of the differential element is

dy
dy 2
dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~
y = y. Here
= 2x.
A
dx
dx
Perform the integration,
L =

Moment

LL

L0

dL =

= 2

1m

21 + 4x2 dx

L0

= cx

1m


A

A

x2 +

= 1.4789 m
LL

y~ dL =

L0

= 2

1
dx
4

1
1
1 1m
+ ln ax +
x2 + b d
4
4
A
4 0


1m

x2 21 + 4x2 dx

L0

= 2c

x2 +

1m

x

2

A

x2 +

1
dx
4

x
1
1
1
1 1m
1 3

2
x x2 + ln ax + x2 + b d
ax + b 4A
32 A
4 128
A
4 0
4

= 0.6063 m2
Centroid.

0.6063 m2
1L y dL
y =
= 0.40998 m = 0.410 m
=
1.4789 m
1L dL
~



Ans.

Ans:
y = 0.410 m
882



© 2016 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.

9–5.
y

Determine the distance y  to the center of gravity of the
homogeneous rod.

1m

2m

Solution
Length

And

y ϭ 2x3

x

Moment

Arm. The length of the differential element is

dy 2
y = y. Here
dL = 2dx2 + dy2 = a 1 + a b b dx and its centroid is at ~
A

dx
dy
= 6x2. Evaluate the integral numerically,
dx
L =
LL

~

LL

dL =

y dL =

L0

L0

1m

1m

21 + 36x4 dx = 2.4214 m

3

2x 21 + 36x4 dx = 2.0747 m2

Centroid. Applying Eq. 9–7,

y =

~
2.0747 m2
1L y dL
= 0.8568 = 0.857 m
=
2.4214 m
1L dL

Ans.

Ans:
y = 0.857 m
883


© 2016 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.

9–6.
Locate the centroid y of the area.

y

y

1
1 – – x2
4


1m
x
2m

SOLUTION
Area and Moment Arm: The area of the differential element is
y
1
1
1
'
= a1 - x 2 b .
dA = ydx = a1 - x2 b dx and its centroid is y =
4
2
2
4
Centroid: Due to symmetry
Ans.

x = 0
Applying Eq. 9–4 and performing the integration, we have
2m

'
ydA

y =


LA

=

1 2
1
1
¢ 1 - x ≤ ¢ 1 - x2 ≤ dx
2
4
4
L- 2m
2m

dA

LA

L- 2m

=

¢

¢1 -

1 2
x ≤ dx
4


x
x3
x 5 2m
+
≤`
2
12
160 - 2m
x
¢x ≤`
12 - 2m
3

2m

=

2
m
5

Ans.

Ans:
y =
884

2
m
5



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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–7.
Determine the area and the centroid x of the parabolic area.

y

h

SOLUTION

y

Differential Element:The area element parallel to the x axis shown shaded in Fig. a
will be considered. The area of the element is
dA = x dy =

a
h1>2

h x2
––
a2
x

a


y 1>2 dy

x
a
'
'
Centroid: The centroid of the element is located at x =
=
y 1>2 and y = y.
2
2h1>2
Area: Integrating,
h

A =

LA

LA

=
dA

LA

h

'
xdA


x =

dA =

L0

¢

a
1>2

L0 h

y 1>2 dy =

2a
3h

a
y1>2 ≤ ¢ 1>2 y 1>2 dy ≤
1>2
2h
h
a

2
ah
3

1>2


A y3>2 B 2 =
h
0

2
ah
3

Ans.

a2 y2 h
a
¢ ≤`
y dy
2h 2 0
3
L0 2h
=
=
= a Ans.
2
2
8
ah
ah
3
3
h 2


Ans:
x =
885

3
a
8


© 2016 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.

*9–8.
Locate the centroid of the shaded area.

y

y ϭ a cospx
L
a

L
2

x

L
2

Solution

Area And Moment Arm. The area of the differential element shown shaded in
y
p
a
p
Fig. a is dA = ydx = a cos x dx and its centroid is at y~ = = cos x.
2
2
2
L
Centroid. Perform the integration

y =

~
1A y dA

1A dA

L>2

=

p
p
a
a cos xbaa cos x dxb
L
L
L-L>2 2

L>2

L-L>2

L>2

=

=

L-L>2

p
x dx
L

a2
2p
x + 1b dx
acos
4
L
L>2

L-L>2

a cos

p
x dx

L

L>2
a2 L
2p
a
sin
x + xb `
4 2p
L
-L>2

a

=

a cos

L>2
aL
p
sin xb `
p
L
-L>2

a2 L>4

2aL>p


=

p
a
8

Ans.

x = 0

Ans.

Due to Symmetry,


Ans:
p
a
8
x = 0
y =

886


© 2016 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.

9–9.
y


Locate the centroid x of the shaded area.

4m
yϭ

1 2
x
4
x

4m

Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a
1
is dA = x dy and its centroid is at ~
x = x. Here, x = 2y1>2
2
Centroid. Perform the integration

x =

~
1A x dA

1A dA

=


=

L0

4m

1
a2y1>2 ba2y1>2 dyb
2
L0

4m

2y1>2 dy

3
m
2

Ans.

Ans:
x =
887

3
m
2



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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–10.
y

Locate the centroid y of the shaded area.

4m
yϭ

1 2
x
4
x

4m

Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a
y = y. Here, x = 2y1>2.
is dA = x dy and its centroid is at ∼
Centroid. Perform the integration

y =

~
1A y dA

1A dA


=

=

=

L0

4m

L0

y a2y1>2 dyb

4m

2y1>2 dy

4m
4
a y 5>2 b `
5
0
4m
4
a y3>2 b `
3
0


12
m
5

Ans.

Ans:
y =
888

12
m
5


© 2016 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.

9–11.
y

Locate the centroid x of the area.

h
y ϭ —2 x2
b

SOLUTION

h


dA = y dx
'
x = x

x =

LA

b

'
x dA

LA

=
dA

h 3
x dx
2
L0 b
b

h 2
x dx
2
L0 b


=

B

h 4
x R
4b2
0

h
B 2 x3 R
3b
0

b

x

b

b

=

3
b
4

Ans.


Ans:
x =
889

3
b
4


© 2016 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.

*9–12.
y

Locate the centroid y of the shaded area.

h
y ϭ —2 x2
b

SOLUTION

h

dA = y dx
y
'
y =
2


y =

LA

b

'
y dA

LA

=
dA

2

h 4
x dx
4
L0 2b
b

h 2
x dx
2
b
L0

=


h2 5 b
x R
B
10b4
0

B

h 3
x R
3b2
0

b

x

b

=

3
h
10

Ans.

Ans:
y =

890

3
h
10


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–13.
y

Locate the centroid x of the shaded area.

SOLUTION

4m

1
dA = 14 - y2dx = a x 2 b dx
16
'
x = x

x =

LA

8


'
xdA

LA
x = 6m

=
dA

L0
L0

8

xa
a

y

4

1 2
––
x
16
8m

x2
b dx

16

1 2
x b dx
16
Ans.

Ans:
x = 6m
891

x


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–14.
y

Locate the centroid y of the shaded area.

SOLUTION

4m

dA = 14 - y2dx = a
y =

y =


1 2
x b dx
16

y

4

1 2
––
x
16
8m

4 + y
2

LA

8

'
ydA

LA

=
dA


x2
x2
1
b a b dx
¢8 2 L0
16
16
8

L0

a

1 2
x b dx
16
Ans.

y = 2.8 m

Ans:
y = 2.8 m
892

x


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


9–15.
Locate the centroid x of the shaded area. Solve the problem by
evaluating the integrals using Simpson’s rule.

y
y = 0.5ex2

SOLUTION
At x = 1 m
2

y = 0.5e1 = 1.359 m
1

LA

dA =

L0

1

(1.359 - y) dx =

L0

x

2


a 1.359 = 0.5 ex b dx = 0.6278 m2

1m

x = x
1

x dA =

LA

L0

2

x a 1.359 - 0.5 ex b dx

= 0.25 m3

x =

x dA
LA
LA

dA

=

0.25

= 0.398 m
0.6278

Ans.

Ans:
x = 0.398 m
893


© 2016 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.

*9–16.
Locate the centroid y of the shaded area. Solve the problem by
evaluating the integrals using Simpson’s rule.

y
y = 0.5ex2

SOLUTION
1

dA =

L0

LA

y =


1

(1.359 - y) dx =

L0

a 1.359 - 0.5ex b dx = 0.6278 m

2

2

1.359 + y
2
1

y dA =

LA

L0

x
1m

a

x2


1.359 + 0.5 e
2
b A 1.359 - 0.5 ex B dx
2

1

=

y =

1
2
a 1.847 - 0.25 e2x b dx = 0.6278 m3
2 L0

y dA
LA
dA

=

0.6278
= 1.00 m
0.6278

Ans.

LA


Ans:
y = 1.00 m
894


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–17.
y

Locate the centroid y of the area.

SOLUTION

y
4 in.

Area: Integrating the area of the differential element gives
A =

LA

dA =

8 in.

8 in.

x


L0

2
––

x3

2>3

3
dx = c x 5>3 d 2
0
5

x

= 19.2 in.2

8 in.

'
1
Centroid: The centroid of the element is located at y = y>2 = x2>3. Applying
2
Eq. 9–4, we have
8 in.
'

y =


y dA
LA
LA
c

=

L0

=

dA

8 in.

1 2>3 2>3
1 4>3
x A x B dx
x dx
L
2
2
0
=
19.2
19.2

8 in.


3 7>3 2
x d
14
0
19.2

Ans.

= 1.43 in.

Ans:
y = 1.43 in.
895


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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9–18.
y

Locate the centroid x of the area.

y

SOLUTION

h

h n

—
nx
a

h

dA = y dx
'
x = x

x =

LA

x =

=

L0

dA

h
2

B hx -

h n+1
x ≤ dx
an


a

h n
x ≤ dx
an

¢h -

h1xn + 22

a 1n + 22
n

h1x

2

n+1

a n1n + 12

h
h
b a2
2
n + 2

ah -


¢ hx -

L0

B x2 -

a

a

'
x dA

LA

=

x

a

h
ba
n + 1

=

R

R


a
0

a
0

a(1 + n)
2(2 + n)

Ans.

Ans:
x =
896

a(1 + n)
2(2 + n)


© 2016 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.

9–19.
y

Locate the centroid y of the area.

y


SOLUTION

h

h n
—
x
an

h

dA = y dx

y =

LA

y =

a

'
y dA

LA

=

x


a

y
'
y =
2

=

h2
1
h2
¢ h2 - 2 n xn + 2n x2n ≤ dx
2 L0
a
a
a

dA
L0

¢h -

h n
x ≤ dx
an

2h21xn + 12
h21x 2n + 12 a
1 2

+ 2n
Bh x - n
R
2
a 1n + 12
a 12n + 12 0

B hx -

h1x n + 12
n

a 1n + 12

2n2
h
21n + 1212n + 12
n
n + 1

=

R

a
0

hn
2n + 1


Ans.

Ans:
y =
897

hn
2n + 1


© 2016 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.

*9–20.
Locate the centroid y of the shaded area.

y

y

h xn
––
an
h

SOLUTION
dA = y dx

x
a


y
y =
2
'
ydA

y =

LA

1
2

=
dA

LA

a

L0
L0

h2
a 2n

x 2n dx
=


a
h
an

xn dx

h2(a 2n + 1)
2a 2n(2n + 1)
h(a n + 1)
a n(n + 1)

=

hn + 1
2(2n + 1)

Ans.

Ans:
y =
898

hn + 1
2(2n + 1)


© 2016 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.

9–21.

y

Locate the centroid x of the shaded area.

1

y ϭ (4 Ϫ x2 )2

16 ft

4 ft
x

4 ft

Solution
Area And Moment Arm. The area of the differential element shown shaded
in  Fig.  a is dA = y dx = ( 4 - x1>2 ) 2 dx = (x - 8x1>2 + 16)dx and its centroid is
~ = x.
at x
Centroid. Perform the integration

x =

~
1A x dA

1A dA

=


=

L0

4 ft

L0
a

a
= 1

x(x - 8x1>2 + 16)dx

4 ft

( x - 8x1>2 + 16) dx

4 ft
x3
16 5>2
x + 8x2 b `
3
5
0

4 ft
16 3>2
x2

x + 16xb `
2
3
0

3
ft
5

Ans.

Ans:

3
x = 1 ft
5

899


© 2016 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.

9–22.
y

Locate the centroid y of the shaded area.

1


y ϭ (4 Ϫ x2 )2

16 ft

4 ft
x

4 ft

Solution
Area And Moment Arm. The area of the differential element shown shaded in
1

Fig.  a is dA = y dx = ( 4 - x 2 ) 2 dx = (x - 8x1>2 + 16)dx and its centroid is at
y
1
= ( 4 - x1>2 ) 2.
y~ =
2
2
Centroid. Perform the integration,

y =

~
1A y dA

1A dA

=


=

=

L0

4 ft

1
( 4 - x1>2 ) 2 ( x - 8x1>2 + 16 ) dx
2
L0

L0

4 ft

4 ft

(x - 8x1>2 + 16)dx

1
a x2 - 8x3>2 + 48x - 128x1>2 + 128bdx
2
L0

a

= 4


4 ft

( x - 8x1>2 + 16 ) dx

4 ft
x3
16 5>2
256 3>2
x + 24x2 x + 128xb `
6
5
3
0

8
ft
55

a

4 ft
16 3>2
x2
x + 16xb `
2
3
0




Ans.

Ans:
y = 4
900

8
ft
55


© 2016 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.

9–23.
y

Locate the centroid x of the shaded area.

y ϭ Ϫ h2 x2ϩh
a
h

x

a

Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a

h
is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~
x = x.
a
Centroid. Perform the integration,

x =

~
1A x dA

1A dA

=

L0

a

L0

x aa

a-

h 2
x + hbdx
a2

h


a

2

2

x + hbdx

h 4
h 2 a
x
+
x b`
2
4a2
0
=
a
h 3
a - 2 x + hxb `
3a
0
a-

=

3
a
8


Ans.



Ans:
x =
901

3
a
8


© 2016 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.

*9–24.
y

Locate the centroid y of the shaded area.

y ϭ Ϫ h2 x2ϩh
a
h

x

a


Solution
Area And Moment Arm. The area of the differential element shown shaded in Fig. a
y
1
h2
h
is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~
y = = a - x2 + hb.
2
2
a
a
Centroid. Perform the integration,

y =

a

~
1A y dA

1
h
h
a - 2 x2 + hba - 2 x2 + hbdx
a
a
L0 2
=
a

h
dA
2
1A
a - 2 x + hbdx
L0
a
a
1 h2 5
2h2 3
2
a 4x x
+
h
xb
`
2 5a
3a2
0
=
a
h 3
a - 2 x + hxb `
3a
0

=

2
h

5



Ans.

Ans:
y =
902

2
h
5


© 2016 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.

9–25.
z

The plate has a thickness of 0.25 ft and a specific weight of
g = 180 lb>ft3. Determine the location of its center of
gravity. Also, find the tension in each of the cords used to
support it.

B

SOLUTION
1

2

Area and Moment Arm: Here, y = x - 8x + 16. The area of the differential
1
'
element is dA = ydx = 1x - 8x2 + 162dx and its centroid is x = x and
1
1
'
y = 1x - 8x2 + 162. Evaluating the integrals, we have
2
16 ft

A =

16 ft

16 ft

LA

dA =

L0

A

1
y2


ϩ

1
x2

C
y

ϭ4

x

1x - 8x2 + 162dx
1

16 ft
1
16 3
= a x2 x2 + 16x b `
= 42.67 ft2
2
3
0

LA

'
xdA =

16 ft


1

x31x - 8x 2 + 162dx4

L0

16 ft
1
16 5
= a x3 x2 + 8x2 b `
= 136.53 ft3
3
5
0

LA

'
ydA =

=

16 ft

L0

1
1
1

1x - 8x2 + 16231x - 8x2 + 162dx4
2

16 ft
1 1 3
32 5
512 3
a x x2 + 48x2 x2 + 256xb `
2 3
5
3
0

= 136.53 ft3
Centroid: Applying Eq. 9–6, we have

x =

LA

'
xdA

LA

y =

LA

=


136.53
= 3.20 ft
42.67

Ans.

=

136.53
= 3.20 ft
42.67

Ans.

dA

'
ydA

LA

dA

Equations of Equilibrium: The weight of the plate is W = 42.6710.25211802 = 1920 lb.
©Mx = 0;

192013.202 - TA1162 = 0 TA = 384 lb

Ans.


©My = 0;

TC1162 - 192013.202 = 0 TC = 384 lb

Ans.

©Fz = 0;

TB + 384 + 384 - 1920 = 0
Ans.

TB = 1152 lb = 1.15 kip
903

Ans:
x = 3.20 ft
y = 3.20 ft
TA = 384 lb
TC = 384 lb
TB = 1.15 kip