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Pacis, R. M.- Engineering Mechanics problems with solutions
Engineering Mechanics (University of Northern Philippines)
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PACIS, ROSAVINNE M.
BSCE 2B
10/30/21
PLATE NO. 1
Plate No. 1 Part I:
1. A 300-N box is held at rest on a smooth plane by a force P inclined at an angle θ with the plane as
shown in Fig. P-310. If θ = 45°, determine the value of P and the normal pressure N exerted by the plane.
�� = 0
Solution:
Pcosθ = Wsin30o
Pcos45o = 300sin30o
P = 212.13 N
�� = 0
N = Psinθ + Wcos30o
N = 212.13sin45o + 300cos30o
N = 409.81 N
2. Determine the magnitude of P and F necessary to keep the concurrent force system in Fig. P-312 in
equilibrium.
Solution:
�� = 0
Fcos60o+ 200cos45o = 300 + Pcos30o
F = 317.16 + 1.7320P
�� = 0
Fsin60o = 200sin45o + Psin30o
(317.16 + 1.7320P) sin60o = 200sin45o + Psin30o
274.67 +1.5P = 141.42 + 0.5P
P = - 133.25 N
F = 317.16 + 1.7320 (-133.25)
F = 86.37 N
3. Figure P-313 represents the concurrent force system acting at a joint of a bridge truss. Determine
the value of P and E to maintain equilibrium of the forces.
Solution:
�� = 0
Fcos60o + 300 = Pcos15o + 400cos30o
F = 1.9318P + 92.82
�� = 0
Fsin60o + Psin15o = 200 + 400sin30o
(1.9318P+ 92.82) sin60o + Psin15o = 200 + 400sin30o
1.6730P + 80.38 + 0.2588P = 200 + 200
1.9318P = 319.62
P = 165.45 lb
F = 1.9318 (165.45) + 92.82
F = 412.44 lb
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4. The five forces shown in Fig. P-314 are in equilibrium. Compute the values of P and
F.
�� = 0
Solution:
Fsin30o + 40cos15o = 30sin30o + 20sin60o
0.5F = -6.3165
F = -12.63 kN
�� = 0
P + 20cos60o + 40sin15o = 30cos30o + Fcos30o
P + 10 + 10.35 = 25.98 + (-12.63)(0.8660)
P = -5.31 kN
5. The 300-lb force and the 400-lb force shown in Fig. P-315 are to be held in equilibrium by a third
force F acting at an unknown angle θ with the horizontal. Determine the values of F and θ.
Solution:
By Cosine Law
F2 = 4002 + 3002 − 2(400)(300)cos30o
F2 = 42153.90 lb
F= 205.31 lb
4002 = 3002 + F2 − 2(300F)cosθ
2(300F)cosθ = 3002 + F2 − 4002
600(205.31)cosθ = 3002 + 42153.90 − 4002
123186cosθ = −27846.1
Cosθ = −0.2260446244
θ= 103.06o
6. Determine the values of α and θ so that the forces shown in Fig. P-316 will be in equilibrium.
Solution:
By Cosine Law
302 = 202 + 402 − 2(20)(40)cosα
2(20)(40)cosα = 202 + 402 − 302
1600cosα = 1100
Cosα = 0.6875
α = 46.57o
202 = 302 + 402 − 2(30)(40)cosθ
2(30)(40)cosθ = 302 + 402 − 202
2400cosθ = 2100
Cosθ = 0.875
θ = 28.96o
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7. The system of knotted cords shown in Fig. P-317 support the indicated weights. Compute the
tensile force in each cord.
Solution:
From the knot where 400-lb load is hanging
ΣFH=0
Dsin75o = Csin30o
D= 0.5176C
ΣFV=0
Dcos75o + Ccos30o = 400
(0.5176C)cos75o + Ccos30o = 400
C= 400 lb
D=0.5176(400)
D= 207.04 lb
From the knot where 300-lb load is hanging:
ΣFV=0
o
Bsin45 = 300 + Ccos30
o
Bsin45o = 300+ 400cos30o
B= 914.16 lb
ΣFH=0
A= Bcos45o + Csin30o
A= 914.16cos45o + 400sin30o
A= 846.41 lb
8. Three bars, hinged at A and D and pinned at B and C as shown in Fig. P-318, form a four-link
mechanism. Determine the value of P that will prevent motion.
Solution:
At joint B
ΣFy=0
o
FABcos30 =20sin45
FAB=16.33 kN
o
ΣFx=0
FBC=20cos45o+FABsin30o
FBC=20cos45o+16.33sin30o
FBC=22.31 kN
At joint C
ΣFy=0
FCDcos15o = Psin60o
FCD = 0.8966P
ΣFx=0
Pcos60o + FCDsin15o = FBC
Pcos60o + (0.8966P)sin15o = 22.31
0.7321P = 22.31
P = 30.47 kN
9. A 300-lb box is held at rest on a smooth plane by a force P inclined at an angle θ with
the plane as shown in Fig. P-310. If θ = 45°, determine the value of P and the normal
pressure N exerted by the plane.
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Solution:
�� = 0
Pcosθ = Wsin30o
Pcos45o = 300sin30o
P = 212.13 lb
�� = 0
N = Psinθ + Wcos30o
N = 212.13sin45o + 300cos30o
N = 409.81 lb
Plate No. 1 Part II:
10. The Fink truss shown in Fig. P-322 is supported by a roller at A and a hinge at B. The given
loads are normal to the inclined member. Determine the reactions at A and B. Hint: Replace the loads
by their resultant.
Solution:
R=2(1000)+3(2000)
R=8000 lb
Rx = Rsin30O
Rx = 8000sin30O
Rx = 4000 lb
RB =
�2� + �2�
RB = 40002 + 2309.402
RB = 4618.80 lb
TanθBx =
TanθBx =
BV
ΣMB=0
60RA = 40Ry
60RA = 40(6928.20)
RA=4618.80 lb
Ry = Rcos30O
Ry = 8000cos30O
Ry = 6928.20 lb
ΣMA = 0
60BV = 20Ry
60BV = 20(6928.20)
BV = 2309.40 lb
ΣFH = 0
��
BH =Rx
2309.40
4000
BH = 4000 lb
θBx = 30O
Thus,
RB = 4618.80 lb at 30° with horizontal
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11. The truss shown in Fig. P-323 is supported by a hinge at A and a roller at B. A load of
is applied at C. Determine the reactions at A and B.
20 kN
Solution:
ΣMA = 0
9RB = (3+1.5)(20cos30o) + (9+3)(20sin30o)
9RB = 197.94
RB = 21.99 kN
ΣFH = 0
RA =
RA =
AH = 20cos30o
�2� + �2�
17.322 + 11.992
RA = 21.07 kN
TanθBx =
TanθBx =
ΣMB = 0
9AV = 1.5AH + 3(20cos30o) + 3(20sin30o)
9AV = 1.5(17.32) + 3(20cos30o) + 3(20sin30o)
AV
��
9AV = 107.94
11.99
AV = 11.99 kN
17.32
θBx = 34.7
AH = 17.32 kN
O
Thus,
RA = 21.07 kN down to the left at 34.7° with the horizontal.
12. The wheel loads on a jeep are given in Fig. P-342. Determine the distance x so that the reaction
of the beam at A is twice as great as the reaction at B.
Solution:
The reaction at A is twice as the reaction at B
RA = 2RB
ΣFV = 0
RA+RB = 600+200
2RB+RB = 800
3RB= 800
R= 266.67 lb
ΣMA = 0
600x + 200 (x+4) = 15RB
600x + 200x + 800 = 15(266.67)
800x = 3200
x = 4 ft
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13. A wheel of 10-in radius carries a load of 1000 lb, as shown in Fig. P-324. (a) Determine the
horizontal force P applied at the center which is necessary to start the wheel over a 5-in. block. Also find
the reaction at the block. (b) If the force P may be inclined at any angle with the horizontal, determine
the minimum value of P to start the wheel over the block; the angle P makes with the horizontal; and the
reaction at the block.
Solution:
Part (a)
tan30o =
Part (b)
�
1000
P =1732.05 lb
sin30o =
=
���60°
�
1000���60°
P=
1000
��
�
��
R = 2000 lb
1000
���(30°+�)
���(30°+�)
−1000���60° ���(30°+�)
=
���2(30°+�)
=0
−1000sin60ocos(30o+θ) = 0
cos(30o+θ)=0
30o + θ = 90o
Pmin =
α=180o−60o−(30o+θ)
1000���60°
���(30°+ 60°)
θ = 60o
α=180o−60o−(30o+60o)
Pmin = 866.03lb
α=30o
�
sinα
�
=
sin30°
1000
���(30°+�)
=
1000
���(30°+60°)
R = 500 lb
14. Determine the amount and direction of the smallest force P required to start the wheel in Fig.
P-325 over the block. What is the reaction at the block?
Solution:
1.5
cosβ =
2
β = 41.41o
30o + β = 71.41o
∅ = 18.59o + α
θ = 90o − α
�
��� 71.41°
��
��
=
=
P=
2000
��� ∅
2000��� 71.41°
��� 18.59° +�
−2000��� 71.41° ���(18.59° +�)
���2 (18.59° +�)
=0
−2000��� 71.41° ���(18.59° + �) = 0
���(18.59° + �) = 0
(18.59° + �) = 0
α = 71.41°
Pmin =
2000��� 71.41°
��� (18.59° +71.41°)
Pmin = 1895.65 lb
�
��� θ
�
=
��� 18.59°
2000
��� ∅
=
∅ = 18.59o + 71.41o = 90o
θ = 90o − 71.41o = 18.59o
2000
��� 90°
R=637.59 lb
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15. The cylinders in Fig. P-326 have the indicated weights and dimensions. Assuming smooth
contact surfaces, determine the reactions at A, B, C, and D on the cylinders.
Solution:
cos θ =
2.6
2+1
θ = 29.93o
From the FBD of 400 kN cylinder
From the FBD of 200 kN cylinder
ΣFH = 0
ΣFV = 0
RA = RC cos θ
RC sin θ = 200
RA = 400.85 cos 29.93o
o
RC sin 29.93 = 200
RA = 347.39 kN
RC = 400.85 kN
ΣFV = 0
ΣFH = 0
RD = RC cos θ
RD = 400.85 cos 29.93
RD = 347.39 kN
RB = 400 + RC sin θ
RB = 400 + 400.85 sin 29.93o
o
RB = 600 kN
16. Forces P and F acting along the bars shown in Fig. P-327 maintain equilibrium of pin
Determine the values of P and F.
A.
Solution:
ΣFH = 0
1
3
F ( ) = P ( ) + 30
5
5
F=
3
5
� + 50
ΣFV = 0
→
Equation (1)
4
2
P ( 5) + F (5) = 18
5
3
� + F = 22.5
Substitute F of Equation (1)
5
3
5 5
6
5
� +( 3 � + 50) = 22.5
�= -27.5
From Equation (1)
F=
+ 50
5
3
� + 50 =
5
3
(−14.76)
F = 39 kN
P = −14.76 kN
17. Two weightless bars pinned together as shown in Fig. P-328 support a load of 35 kN.
Determine the forces P and F acting respectively along bars AB and AC that maintain equilibrium of pin
A.
Solution:
ΣFH = 0
F(
5
41
P(
)=P(
2
13
)
F = 0.7104P
ΣFV = 0
2
13
)=F(
0.3883P = 35
4
41
) + 35
F = 0.7104 (90.14)
F = 64.03 kN
P=90.14 kN
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Plate No. 1 Part III:
18. Determine the reactions R1 and R2 of the beam in Fig. P-333 loaded with a concentrated load of
1600 lb and a load varying from zero to an intensity of 400 lb per ft.
Solution:
ΣMR4 = 0
1
12R3 = 4 [ (12) (400)]
2
R3 = 800 lb
ΣMR2 = 0
16R1 = 13 (1600) + 12R3
16R1 = 13( 1600) + 12 (800)
R1 = 1900 lb
ΣMR3 = 0
1
12R4 = 8 [2 (12) (400)]
R4 = 1600 lb
ΣMR1 = 0
16R2 = 3 (1600) + 4R3+16R4
16R2 = 3 (1600) + 4 (800) + 16 (1600)
R2 = 2100 lb
19. Determine the reactions for the beam loaded as shown in Fig. P-334.
Solution:
ΣMR2 = 0
7.5R1 = 6 (12) + 4.5 [ 3 (6)] + 1[
R1 = 23.4 kN
1
(3) (15)]
2
ΣMR1 = 0
7.5R2 = 1.5 (12) + 3[ 3 (6) ] + 6.5 [
R2 = 29.1 kN
1
2
(3) (15)]
20. The roof truss in Fig. P-335 is supported by a roller at A and a hinge at B. Find the value
reactions.
Solution:
Replace the 3-20 kN forces and 2-10 kN forces by a single 80
kN force
ΣMB = 0
15RA = 10 (60) + 7.5 (80) + 5 (50)
RA = 96.67 kN
ΣMA = 0
15RB = 5 (60) + 7.5 (80) + 10 (50)
RB = 93.33 kN
21. The upper beam in Fig. P-337 is supported at D and a roller at C which separates the upper
and lower beams. Determine the values of the reactions at A, B, C, and D. Neglect the weight of the
beams.
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Solution:
ΣMA= 0
ΣMC = 0
10RB = 4 (400) + 14 (160)
10RD + 4 (60) = 6 (190)
ΣMD= 0
RB = 384 kN
RD = 90 kN
ΣMB = 0
10RA + 4 (160) = 6 (400)
10RC = 14 (60) + 4 (190)
RA = 176 kN
RC = 60 kN
22. The two 12-ft beams shown in Fig. 3-16 are to be moved horizontally with respect to each
other and load P shifted to a new position on CD so that all three reactions are equal. How far apart will
R2 and R3 then be? How far will P be from D?
Solution:
From FBD of beam CD
ΣFV = 0
RC + R3 = P
RC + 0.5RC = 960
RC = 640 lb
ΣMC = 0
12R3 = 960x
12 (320) = 960x
x = 4 ft
R3 = 0.5 (640)
R3 = 320 lb
Thus, P is 8 ft to the left of D.
From the figure above, Rc is at the midspan of AB to produce equal
reactions R1 and R2. Thus, R2 and R3 are 6 ft apart.
From FBD of beam AB
R1 = 0.5 (640)
R1 = 320 lb
R2 = 0.5 (640)
R2 = 320 lb
23. The weight W of a traveling crane is 20 tons acting as shown in Fig. P-343. To prevent the
crane from tipping to the right when carrying a load P of 20 tons, a counterweight Q is used. Determine
the value and position of Q so that the crane will remain in equilibrium both when the maximum load P
is applied and when the load P is removed.
Solution:
When load P is removed
ΣMA = 0
Qx = 20(5+1)
Qx = 120
→ Equation (1)
When load P is applied
ΣMB = 0
Q( x + 5 ) = 20 (1) + 20 (10)
Qx + 5Q = 220
Plate No. 1 Part IV:
From Equation (1), Qx = 120, thus,
120 + 5Q = 220
5Q = 100
Q = 20 tons
24. A boom AB is supported in a horizontal position by a hinge A and a cable which runs from C
over a small pulley at D as shown in Fig. P-346. Compute the tension T in the cable and the horizontal
and vertical components of the reaction at A. Neglect the size of the pulley at D
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Solution:
4(
ΣMA = 0
2
5
T ) = 2 (200) + 6 (100)
T = 279.51 lb
AV +
2
5
AV +
ΣFV = 0
2
5
T = 200 + 100
(279.51) = 300
AV = 50 lb
ΣFH = 0
AH =
AH =
1
5
1
5
T
(279.51)
AH = 125 lb
25. Repeat the last problem if the cable pulls the boom AB into a position at which it is inclined at
30° above the horizontal. The loads remain vertical.
Solution:
sin60o =
�
4
X = 4sin60o
6
tanθ =
tanθ =
�
tanθ =
6
4sin60°
3
Because θ = 60°, T is perpendicular to AB.
MA= 0
4T = 200 (2cos30o) + 100 (6cos30o)
T = 216.51 lb
θ = 60o
ΣFH = 0
AH = Tcosθ
AH = 216.51cos60o
AH = 108.26 lb
ΣFV = 0
AV + Tsinθ = 200 + 100
AV + 216.51sin60o = 200 + 100
AV = 112.50 lb
26. The frame shown in Fig. P-348 is supported in pivots at A and B. Each member weighs 5 kN/m.
Compute the horizontal reaction at A and the horizontal and vertical components of the reaction at B
Solution:
Length of DF
LDF2 = 42 + 32
LDF2 = 25
LDF = 5 m
Weights of members
WAB = 6 (5) = 30 kN
WCE = 6 (5) = 30 kN
WDF = 5 (5) = 25 kN
ΣFV = 0
BV = WAB + WCE + WDF + 200
ΣMB = 0
6AH = 3WCE + 2WDF + 6 (200)
6AH = 3 (30) + 2 (25) + 6 (200)
AH = 223.33 kN
ΣFH = 0
BH = AH
BH = 223.33 kN
BV= 30 + 30 + 25 + 200
BV= 285 kN
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27. The truss shown in Fig. P-349 is supported on roller at A and hinge at B. Solve for the
components of the reactions
Solution:
ΣMB = 0
24AV + 16 (240) = 36 (400) + 12 (600)
AV = 740 lb
ΣMA = 0
24BV + 12 (400) = 16 (240) + 12 (600)
BV = 260 lb
ΣFH = 0
BH = 240 lb
28. The beam shown in Fig. P-351 is supported by a hinge at A and a roller on a 1 to 2 slope at B.
Determine the resultant reactions at A and B.
Solution:
ΣMA = 0
2
4( RB) = 3 (40)
5
RB = 33.54 kN
ΣMB = 0
4AV = 1(40)
AV = 10 kN
AH =
ΣFH = 0
1
RB =
5
AH = 15 kN
1
5
(33.54)
RA =
�2� + �2� =
RA = 18.03 kN
tanθAx =
��
��
θAx = 33.69o
=
152 + 102
10
15
Thus, RA = 18.03 kN up to the right at
33.69o from horizontal.
29. A pulley 4 ft in diameter and supporting a load 200 lb is mounted at B on a horizontal beam as
shown in Fig. P-352. The beam is supported by a hinge at A and rollers at C. Neglecting the weight of
the beam, determine the reactions at A and C.
Solution:
From FBD of pulley
T = 200 lb
ΣFV = 0
BV + Tsin30o = 200
BV + 200sin30o = 200
BV = 100 lb
ΣFH = 0
BH = Tcos30o
BH = 200cos30o
BH = 173.21 lb
From FBD of beam
ΣMA = 0
8RC = 4BV
8RC = 4 (100)
RC = 50 lb
ΣMC = 0
8AV = 4BV
8AV = 4 (100)
AV = 50 lb
ΣFH = 0
AH = BH
AH = 173.21 lb
RA =
RA = 173.212 + 502
RA = 180.28 lb
tanθAx =
tanθAx =
��
��
50
173.21
θAx = 16.1o
Thus, RA = 180.28 lb up to the right
at 16.1o from horizontal.
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30. The forces acting on a 1-m length of a dam are shown in Fig. P-353. The upward ground
reaction varies uniformly from an intensity of p1 kN/m to p2 kN/m at B. Determine p1 and p2 and also
the horizontal resistance to sliding.
Solution:
Horizontal resistance to sliding
ΣFH = 0
Rx + Fcos30o = 1000
Rx + 600cos30o = 1000
ΣFV = 0
Rx = 480.38 kN
Ry = W + Fsin30o
Ry = 2400 + 600sin30
Ry = 2700 kN
o
Righting moment
MR = 11 (2400) + 4 (600)
MR = 28800 kN · m
Overturning moment
MO = 6 (1000)
MO = 6000 kN · m
ΣMB = 0
� Ry = MR − MO
� (2700) = 28800 − 6000
� = 8.44 m to the left of B
Eccentricity
1
e = 2B − x
1
e = 2 (18) − 8.44
e = 0.56 m
Foundation pressure
R
6�
p = y (1 ± )
p1 =
�
�
2700
1 −
6 (0.59)
2700
1 +
6 (0.59)
18
p1 = 121 kN/m
p2 =
18
p2 = 180 kN/m
18
18
31. Compute the total reactions at A and B on the truss shown in Fig. P-354
Solution:
ΣMA = 0
2
1
24RB + 3 (20) + 3 ( 5) (22.4) = 18 ( 5) (22.4) + 18 (30) + 12 (20) + 6 (10)
RB = 46.27 kN
ΣMB = 0
2
1
24AV = 3 (20) + 3 ( 5) (22.4) + 6 ( 5) (22.4) + 6 (30) + 12 (20) + 18 (10)
AV = 33.76 kN
ΣFH = 0
AH = 20 +
1
5
AH = 30.02 kN
(22.4)
RA =
�� 2 + �� 2
RA = 30.022 + 33.762
RA = 45.18 kN
tanθAx =
tanθAx =
��
��
33.76
30.02
θAx = 48.36o
Thus, RA=45.18 kN up to the right at 48.36o from horizontal.
32. Determine the reactions at A and B on the Fink truss shown in Fig. P-355. Members CD and
FG are respectively perpendicular to AE and BE at their midpoints.
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Solution:
tanθ =
4.5
9
θ = 26.57o
cosθ = 4.5
AC =
��
4.5
cosθ
=
AC = 5.03 m
cosθ = ��
��
��
AD =
cosθ
=
AD = 5.62 m
FB = AD
4.5
cos26.57°
FB = 5.62 m
ΣFH = 0
BH + 40sinθ = RAsin30o
5.03
cos26.57°
BH + 40sin26.57o
=53.59sin30o
BH = 8.90 kN
ΣMB = 0
18 (RAcos30o) = 12 (18−4.5) + (40cosθ) (FB) + 40 (FB) + 20 (18−AD)
18( RAcos30o) = 12 (13.5) + (40cos26.57o) (5.62) + 40(5.62) + 20 (18−5.62)
15.59RA = 835.46
RA = 53.59 kN
ΣMA = 0
18BV = 12 (4.5) + (40cosθ) (18−FB) + 40 (18−FB) + 20 (AD)
18BV = 12 (4.5) + (40cos26.57o) (18−5.62) + 40 (18−5.62) + 20(5.62)
18BV = 1104.50
BV = 61.36 kN
33. The cantilever truss shown in Fig. P-356 is supported by a hinge at A and a strut BC. Determine
the reactions at A and B.
Solution:
From right triangles ACD and ACB.
��
��
cos30o =
=
6
��
AB=6 m
Notice also that triangle ABD is an equilateral triangle of sides 6 m.
ΣMA = 0
6 (RBcos30o) = 3(10) + 6 (10) + 9 (10)
RB = 20 3 kN
RB = 34.64 kN
ΣFH = 0
AH + 4 (10sin30o) = RBcos30o
AH+ 4 (10sin30o) = 20 3cos30o
AH = 10 kN
RA =
�� 2 + �� 2
RA = 102 + 17.322
RA = 20 kN
tanθAx =
tanθAx =
θAx =60o
��
��
17.32
10
ΣFV = 0
AV + RBsin30o = 4 (10cos30o)
AV + 20 3sin30o = 4(10cos30o)
AV = 10 3 kN
AV = 17.32 kN
Thus, RA= 20 kN up to the right at 60o from horizontal.
34. The uniform rod in Fig. P-357 weighs 420 lb and has its center of gravity at G. Determine the
tension in the cable and the reactions at the smooth surfaces at A and B.
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Solution:
Distance AB
AB =
ΣFH = 0
82 + 82
T = RBcos45o
AB = 8 2 m
T = 254.56cos45o
T = 180 lb
ΣMA = 0
2T + 6 (420) = 8 2RB
2T + 2520 = 8 2RB
T + 1260 = 4 2RB
o
RBcos45 + 1260 = 4 2RB
4.9497RB = 1260
ΣFV = 0
RA + RBsin45o = 420
RA + 254.56sin45o = 420
RA = 240 lb
RB = 254.56 lb
35. A bar AE is in equilibrium under the action of the five forces shown in Fig. P-358. Determine P,
R, and T.
Solution:
ΣFV = 0
4
5
T + R = 60
R = 60 −
4
5
ΣMA = 0
T
10T + 4 (4) R = 4 (60) + 3 (3) (40)
10T + 16R = 600
5T + 8R = 300
5T + 8 (60 7
4
5
T) = 300
−5 T + 480 = 300
7
5
T = 180
T = 128.57 kN up to the left
R = 60 −
4
5
(128.57)
R = −42.86 kN
R = 42.86 kN downward
ΣFH = 0
P+
P+
3
5
3
5
T = 40
(128.57) = 40
P = −37.14 kN
P = 37.14 kN to the right
36. A 4-m bar of negligible weight rests in a horizontal position on the smooth planes shown in Fig.
P-359. Compute the distance x at which load T = 10 kN should be placed from point B to keep the bar
horizontal.
Solution:
From the Force Polygon
��
���45°
=
20 + 10
���105°
RA = 21.96 kN
From the Free Body Diagram
ΣMB = 0
4 (RAcos30o) = 20 (3) + 10x
4(21.96cos30o)=20(3) + 10x
10x = 16.072
x = 1.61 m
MECH 121 –Static of Rigid Bodies | Instructor: Engr. Jennifer C. Paglingayen5
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37. Referring to Problem 359, what value of T acting at x = 1 m from B will keep the bar
horizontal.
Solution:
From the Force Polygon
��
���45°
=
20 + �
���105°
RA = 0.732 (20 + T)
RA = 14.641 + 0.732T
From the Free Body Diagram
ΣMB=0
4 (RAcos30o) = 3 (20) + 1(T)
4 (14.641 + 0.732T) cos30o = 60 + T
50.7179 + 2.5357T = 60 + T
1.5357T = 9.2821
T = 6.04 kN
MECH 121 –Static of Rigid Bodies | Instructor: Engr. Jennifer C. Paglingayen5
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