PROBLEM 4.

27
For the frame and loading shown, determine the reactions at A and E
when (a) α = 30o , (b) α = 45o.

SOLUTION

(a) Given α = 30°

ΣM A = 0: − ( 90 N )( 0.2 m ) − ( 90 N )( 0.06 m )

+ ( E cos 60° )( 0.160 m ) + ( E sin 60° )( 0.100 m ) = 0

∴ E = 140.454 N

or E = 140.5 N 60°

ΣFx = 0: Ax − 90 N + (140.454 N ) cos 60° = 0

∴ Ax = 19.7730 N

or A x = 19.7730 N

ΣFy = 0: Ay − 90 N + (140.454 N ) sin 60° = 0

∴ Ay = −31.637 N

or A y = 31.6 N

Then A= Ax2 + Ay2 = (19.7730 )2 + ( 31.637 )2

 Ay  −1  −31.637 
and θ = tan −1   = tan  
 Ax   19.7730 

= −57.995°

or A = 37.3 N 58.0°
 PROBLEM 4.27 CONTINUED

(b) (b) Given α = 45°

+ ( E cos 45° )( 0.160 m ) + ( E sin 45° )( 0.100 m ) = 0

∴ E = 127.279 N
or E = 127.3 N 45°

ΣFx = 0: Ax − 90 + (127.279 N ) cos 45° = 0

∴ Ax = 0

ΣFy = 0: Ay − 90 + (127.279 N ) sin 45° = 0

∴ Ay = 0

or A = 0
 PROBLEM 4.28
A lever AB is hinged at C and is attached to a control cable at A. If the
lever is subjected to a 300-N vertical force at B, determine
(a) the tension in the cable, (b) the reaction at C.

SOLUTION
First
x AC = ( 0.200 m ) cos 20° = 0.187 939 m

y AC = ( 0.200 m ) sin 20° = 0.068 404 m

Then
yDA = 0.240 m − y AC
= 0.240 m − 0.068404 m
= 0.171596 m
yDA 0.171 596
and tan α = =
x AC 0.187 939

∴ α = 42.397°
and β = 90° − 20° − 42.397° = 27.603°
(a) From f.b.d. of lever AB
ΣM C = 0: T cos 27.603° ( 0.2 m )

− 300 N ( 0.3 m ) cos 20° = 0

∴ T = 477.17 N or T = 477 N
(b) From f.b.d. of lever AB

ΣFx = 0: C x + ( 477.17 N ) cos 42.397° = 0

∴ C x = −352.39 N

or C x = 352.39 N

ΣFy = 0: C y − 300 N − ( 477.17 N ) sin 42.397° = 0

∴ C y = 621.74 N

or C y = 621.74 N
PROBLEM 4.28 CONTINUED

Then C = C x2 + C y2 = ( 352.39 )2 + ( 621.74 )2 = 714.66 N

 Cy  −1  621.74 
and θ = tan −1   = tan   = −60.456°
 x
C  −352.39 

or C = 715 N 60.5°
 PROBLEM 4.29
Neglecting friction and the radius of the pulley, determine the tension
in cable BCD and the reaction at support A when d = 80 mm.

SOLUTION

First
 60 
α = tan −1   = 12.0948°
 280 

 60 
β = tan −1   = 36.870°
 80 
From f.b.d. of object BAD
ΣM A = 0: ( 40 N )( 0.18 m ) + (T cosα )( 0.08 m )
+ (T sin α )( 0.18 m ) − (T cos β )( 0.08 m )

− (T sin β )( 0.18 m ) = 0

 7.2 N ⋅ m 
∴ T =  = 128.433 N
 0.056061 
or T = 128.4 N

ΣFx = 0: (128.433 N )( cos β − cos α ) + Ax = 0

∴ Ax = 22.836 N

or A x = 22.836 N

ΣFy = 0: Ay + (128.433 N )( sin β + sin α ) + 40 N = 0

∴ Ay = −143.970 N

or A y = 143.970 N

Then A= Ax2 + Ay2 = ( 22.836 )2 + (143.970 )2 = 145.770 N

 Ay  −1  −143.970 
and θ = tan −1   = tan   = −80.987°
 Ax   22.836 

or A = 145.8 N 81.0°
 PROBLEM 4.30
Neglecting friction and the radius of the pulley, determine the tension in
cable BCD and the reaction at support A when d = 144 mm.

SOLUTION

First note
 60 
α = tan −1   = 15.5241°
 216 

 60 
β = tan −1   = 22.620°
 144 
From f.b.d. of member BAD
ΣM A = 0: ( 40 N )( 0.18 m ) + (T cosα )( 0.08 m )
+ (T sin α )( 0.18 m ) − (T cos β )( 0.08 m )

− (T sin β )( 0.18 m ) = 0

 7.2 N ⋅ m 
∴ T =  = 404.04 N
 0.0178199 m 
or T = 404 N

ΣFx = 0: Ax + ( 404.04 N )( cos β − cos α ) = 0

∴ Ax = 16.3402 N

or A x = 16.3402 N

ΣFy = 0: Ay + ( 404.04 N )( sin β + sin α ) + 40 N = 0

∴ Ay = −303.54 N

or A y = 303.54 N

Then A= Ax2 + Ay2 = (16.3402 )2 + ( 303.54 )2 = 303.98 N

 Ay  −1  −303.54 
and θ = tan −1   = tan   = −86.919°
 x
A  16.3402 

or A = 304 N 86.9°
 PROBLEM 4.31
Neglecting friction, determine the tension in cable ABD and the reaction
at support C.

SOLUTION

From f.b.d. of inverted T-member

ΣM C = 0: T ( 25 in.) − T (10 in.) − ( 30 lb )(10 in.) = 0

∴ T = 20 lb

or T = 20.0 lb W

ΣFx = 0: Cx − 20 lb = 0

∴ C x = 20 lb

or C x = 20.0 lb

ΣFy = 0: C y + 20 lb − 30 lb = 0

∴ C y = 10 lb

or C y = 10.00 lb

Then C = C x2 + C y2 = ( 20 )2 + (10 )2 = 22.361 lb

 Cy  −1  10 
and θ = tan −1   = tan   = 26.565°
 Cx   20 

or C = 22.4 lb 26.6° W
 PROBLEM 4.32
Rod ABC is bent in the shape of a circular arc of radius R. Knowing
that θ = 35o , determine the reaction (a) at B, (b) at C.

SOLUTION

For θ = 35°

(a) From the f.b.d. of rod ABC

ΣM D = 0: Cx( R ) − P( R ) = 0

∴ Cx = P

or Cx = P

ΣFx = 0: P − B sin 35° = 0

P
∴ B= = 1.74345P
sin 35°

or B = 1.743P 55.0° W

(b) From the f.b.d. of rod ABC

ΣFy = 0: C y + (1.74345P ) cos 35° − P = 0

∴ C y = −0.42815P

or C y = 0.42815P

Then C = C x2 + C y2 = ( P )2 + ( 0.42815P )2 = 1.08780 P

 Cy  −1  −0.42815P 
and θ = tan −1   = tan   = −23.178°
 Cx   P 

or C = 1.088P 23.2° W
 PROBLEM 4.33
Rod ABC is bent in the shape of a circular arc of radius R. Knowing
that θ = 50o , determine the reaction (a) at B, (b) at C.

SOLUTION

For θ = 50°

(a) From the f.b.d. of rod ABC

ΣM D = 0: Cx ( R ) − P ( R ) = 0

∴ Cx = P

or Cx = P

ΣFx = 0: P − B sin 50° = 0

P
∴ B= = 1.30541P
sin 50°

or B = 1.305P 40.0° W

(b) From the f.b.d. of rod ABC

ΣFy = 0: C y − P + (1.30541P ) cos 50° = 0

∴ C y = 0.160900P

or C y = 0.1609 P

Then C = C x2 + C y2 = ( P )2 + ( 0.1609P )2 = 1.01286 P

 Cy  −1  0.1609 P 
and θ = tan −1   = tan   = 9.1405°
 x
C  P 

or C = 1.013P 9.14° W
 PROBLEM 4.34
Neglecting friction and the radius of the pulley, determine (a) the
tension in cable ABD, (b) the reaction at C.

SOLUTION

First note
 15 
α = tan −1   = 22.620°
 36 

 15 
β = tan −1   = 36.870°
 20 
(a) From f.b.d. of member ABC
ΣM C = 0: ( 30 lb )( 28 in.) − (T sin 22.620° )( 36 in.)
− (T sin 36.870° )( 20 in.) = 0

∴ T = 32.500 lb
or T = 32.5 lb W
(b) From f.b.d. of member ABC

ΣFx = 0: Cx + ( 32.500 lb )( cos 22.620° + cos 36.870° ) = 0

∴ C x = −56.000 lb

or C x = 56.000 lb

ΣFy = 0: C y − 30 lb + ( 32.500 lb )( sin 22.620° + sin 36.870° ) = 0

∴ C y = −2.0001 lb

or C y = 2.0001 lb

Then C = C x2 + C y2 = ( 56.0 )2 + ( 2.001)2 = 56.036 lb

 Cy  −1  −2.0 
and θ = tan −1   = tan   = 2.0454°
 Cx   −56.0 

or C = 56.0 lb 2.05° W