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Unformatted text preview: Problem 251 Determine the magnitude of force F so that the resultant FRof the three forces is as small
as possible. What is the minimum magnitude of FR? Units Used: v
kN = 1000 N
Given:
——————X
F 1 = 5 kN 
I
F2 = 4 kN ‘3 i
E M x
a t. 1
t9 = 30 deg F? _ x‘:
 
1.
Solution: Scalar Notation: Suming the force components
algebrically, we have ‘ + _—> FRx=ZFx; FRx = F] —Fsin(t9) +T FRy= 2F); FR), = Fcos(6) — F2 The magnitude of the resultant force FR is FR = ,/ 1:sz + FR; = (F; _ Fsin(6))2 + (“03(0) _ F2)2 56 FR2 = F12 + F22 + F2 — 2FF1 sin(€) _ 2F; Fcos(l9) dF‘
ZFR—R = 2F — 2F] sin(0)  2172 cos(t9)
dF
dF
IfF is aminimum, then ((1—15 = ] F = F] sin(¢9) + F2 005(0) FR = (F1 _ mm)? + (“05(0) _ F2)2 Problem 258 Express each force in Cartesian vector form. Units Used:
kN = 103 N
Given:
F] = 5 kN
F2 = 2 kN
91 = 60 deg
62 = 60 deg
63 = 45 deg‘
Solution:
cos(62)
Flv = F1 cos(03)
cos(61)
0
sz = F2 1 X iiiii Problem 261 The stock S mounted on' the lathe is subjected to a
force F, which is caused by the die D. Determine the coordinate direction angle ﬂ and express the force as
a Cartesian vector.  Given:
F = 60 N
a = 60 deg y = 30 deg 65 Solution:
cos2(a) + cosz(ﬂ) + cos2(y) = 1 ﬂ = acos(1 — cos(a)2 — cos(y)2) ,6 = 90deg
Loos(a) ~30 n = F mo) F 0 N
—cos(y) 5 _‘ ~52 Problem 268 Determine the magnitude and coordinate direction angles of the resultant force. , z
Given: F1=350N a=60deg F2 = 250N ,6 = 60 deg 0:3 7:45 deg mi
61 = 4 t9 = 30 deg ﬂ
3 1’
Solution: 1: Fa
cos(a) 175
Flv = F] cos(ﬂ) Flv = 175 N
—cos(7) —247.5
F F d F F C
2h = 2 — 2y = 2 —
~ \/c2+d2 J3”?
F2hCOS(6') 173.2
FZV : —F2hsin(0) F2v = —100 N
F2y 150 [FR = F1v+F2v i won.) ...
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 Fall '06
 Kane

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