Problem 267 | Resultant of Non-Concurrent
Problem 266 | Resultant of Non-Concurrent Force System
Force System
The Howe roof truss shown in Fig. P-267 carries the given
Problem 266
loads. The wind loads are perpendicular to the inclined
Determine the resultant of the three forces acting on the
members. Determine the magnitude of the resultant, its
dam shown in Fig. P-266 and locate its intersection with
inclination with the horizontal, and where it intersects AB.
the base AB. For good design, this intersection should
occur within the middle third of the base. Does it? Rx=ΣFx
Rx=
(1120+2240+1120)
(15√)+2000
Rx=4003.52 N to the right
Rx=ΣFx Rx=10000−6000sin60∘
Rx=4803.85 lb to the right
Ry=ΣFy Ry=24000+6000cos60∘
Ry=27000 lb downward
Ry=ΣFy Ry=(1120+2240+1120)(2/√5)+3000+2000+1000
√R=Rx2+Ry2 √R=4803.852+270002
Ry=10007.03 N downward
R=27424.02 lb
√R=Rx2+Ry2 √R=4003.522+10007.032
R=10778.16 N
tanθx=(Ry/Rx) tanθx=(27000/4803.85)
tanθx=RyRx tanθx=10007.034003.52 θx=68.2∘
θx=79.91∘
Righting moment
RM=24000(18−7)+6000(4) RM=288000 lb⋅ft
Overturning moment
OM=10000(6) OM=60000 lb⋅ft
Moment at the toe (downstream side - point B)
MA=ΣFd
MB=RM−OM MB=288000−60000
MA=2240(3.354)+1120(3.354)(2)+2000(1.5)+3000(3)+2000(
MB=228000 lb⋅ft counterclockwise
6)+1000(9) MA=48026.37 N⋅m clockwise
Ryx=MA
10007.03x=48026.37
x=4.8 m to the right of A
Thus, R = 10 778.16 N downward to the right at θx =
68.2° passing 4.8 m to the right of A.
Problem 268 | Resultant of Non-Concurrent Problem 269 | Resultant of Non-Concurrent
Force System Force System
Problem 268 Repeat Prob. 268 is the resultant is 390 lb directed down to the right at
The resultant of four forces, of which three are shown in a slope of 5 to 12 passing through point A. Also determine the x and y
Fig. P-268, is a couple of 480 lb·ft clockwise in sense. If intercepts of the missing force F.
each square is 1 ft on a side, determine the fourth force
completely.
Let F4 = the fourth force and for
couple resultant, R is zero.
Rx=0
110+150(3/5)+F4x=0
F4x=−200 lb
F4x=200 lb to the left
Ry=0
150(4/5)−120+F4y=0
F4y=0
Thus, F4=200 lb to the left
Assuming F4 is above point O and clockwise rotation to
be positive
MO=C
110(4)+120(2)−F4d=480
110(4)+120(2)−200d=480
d=1 ft
d is positive, thus, the assumption is correct that F 4 is
above point O.
Therefore, the fourth force is 200 lb acting horizontally
to the left at 1 ft above point O. answer
Problem 270 | Resultant of Non-
Concurrent Force System
The three forces shown in Fig. P-270 are required to cause a
horizontal resultant acting through point A. If F = 316 lb,
determine the values of P and T. Hint: Apply MR = ΣMB to
determine R, then MR = ΣMC to find P, and finally MR = ΣMD or
Ry = ΣY to compute T.
For horizontal resultant,
Ry = 0 and Rx = R
MR=ΣMB
R(1)=Fx(2)+Fy(1)
R(1)=316(1/√10)(2)
+316(3/√10)(1)
R=499.64 lb to the right at A
ΣMC=MR
Fx(4)−Fy(2)+Py(4) =R(3)
316(1/√10)(4)−316(3/√10)(2)+P(2/√5)(4)=499.64(3)
P=474.82 lb answer
ΣMD=MR
Fx(4)+Fy(2)−Ty(4)=R(3)
316(1/√10)(4)+316(3/√10)(2)−T(2/√13)(4)=499.64(3)
T=−225.18 lb answer
