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Unformatted text preview: REVIEW 1 I0.132. Determine the magnitude and coordinate
direction angles of F3 so that the resultant of the three forces is zero. Eh = 21;; 0 = —180 + 300 cos30°sin40° + 1*; cosa
2 EU = 214;; 0 = 300 cos30°cos40° + F3 c035 15"“ = 21;; O = —300 sin30° + F3 cosy F =180lb
l cosza + coszﬁ + 00527 = 1 Solving :
F; = 250 1b Ans
y
at = 87.0° Ans
X
[3 = 143° Ans
F2=3001b y = 53.1 Ans
*2136. The leg is held in position by the quadriceps AB,
which is attached to the pelvis at A. Ifthe force exerted
on this muscle by the pelvis is 85 N, in the direction shown,
determine the stabilizing force component acting along
the positive y axis and the supporting force component
acting along the negative x axis
as N % Ex: 85 cos 10° = 83.7N Ans IV; = 85 sin 10° = 14.8N Ans 2139. If F1 = F; = 301b, determine the angles 9 and 4:
so that the resultant force is directed along the positive
x axis and has a'magnitude of FR = 20 lb. (30)Z : (30)2 + (20)2 — 2(30)~(20)cos9 (1} = 6 2 705° Ans 366. Ihe pipe is held in place by the vice. If the bolt
exerts a force of 50 lb on the pipe in the direction shown,
determine the forces F A and F3 that the smooth contacts
at A and B exert on the pipe. 4
F,i * acos60° — 50(3) = 0 —1«;sin60° + 50(3) = 0 1'} = 346 lb Ans F, 573 lb Ans *372. The "scale" consists of a known weight W which
is suspended at A from a cord of total length L.
Determine the weight w at B if A is at a distance y for
equilibrium. Neglect the sizes and weights of the pulleys. +TEF, =0; 2Wsin6—w=0 w = 2Wsi116 m/t—g—YI — (ET = 2W ;—,/(L — y)2 — d2 4/; Ly 9 w: 2W
W = y/(Ly)2 —d2 Ans 4162. Determine the moment of the force F about point
0. The force has coordinate direction angles of a = 60°, B = 120°, 7 = 45°. Express the result as a Cartesian
vector. F=20Ib Puma» Victor And Fore: Valor: : rm = ((60)i+(l4—0)j+(l00)k) in.
a (6i+l4j+ 10k) in. F = 20(eos 60°i+cos 120°j+ cos 45°k) lb
3 (10.0i l0.0j+ H.142“ lb Moment of Force F About Point 0 : Applying 31.4—7. we have Mo = rm xF
i 5 k
= 6 14 10 10.0 — 10.0 14.142 ={298i+15.lj2(I)k) lbin Ans 4165. Determine the magnitude of the moment of the
force FL. about the hinged axis (40 of the door. Position Vector And Force Wants: m, = [[—0.5 — (—0.5)1i + [o — (—Jm‘ + (o — mm m = {U} m t—us—t—zsm
Fc =250 N l—o.s—(—z.sn=+
[041wn.5mssn“)lP+tu—I 5mm“)? = [159.33i + 183.l5j — 59.75klN Mammt of Farce Fc About a » mm. The unit vector along the
a —— z: axis is i. Applying Eq. 4—] I, we have Ma—a = i ' (1245 X Fc) i 0 0
= 0 E 0 15933 18115 “5975 The negative sign indicate2; that Maw“ is directed toward negative x axis.
= l[1(—59.75) — (183.l5)(0)1— 0 + (l
MP,x = 59.7 N  m Ans = ~59.7 Nm *4168. The horizontal 30N force acts on the handle of
the wrench. What is the magnitude of the moment of this
force about the z axis? Position Vector And Forte Vectors:
r“ = {—0.01i +0.2j] n1
r04 = [(—0.01 — (Di + (0.2 — OH + (0.05 — 0)k} m
= [—0.01i + 0.2j + 0.05M in
F = 30(sin 45°i — cos 45°j) N
= (21.113i— 21.7.13le Moment of Fame F About 1 Axis. Tlie unit Vector along the z axis
is k. Applying Bq. 4—11. we have M: = k‘G‘M KF) 0 o 1
= 4101 0.2 0
21.213 —21.213 (1 = o — o + lE(—0.0l )(—21.2i3) — 2 .213(0.2)] = —4.03 N . m Ans
4170. If the resultant couple moment of the three
couples acting on the triangular block is to be zero,
determine the magnitudes of forces F and P. Couple moment of the [01b force : The sense of direction of this couple moment is
outward perpendicular to the inclined plane. 3 4
='+k
“ 5J 5 M =Mu=10(3)@j+§k)={183+24k} Ib in. Couple moment of P and F :
MF = 6F j Resultant moment : MR=0=EM; 0=M+Mp+Mp 0=(18j+24 k)+(6Fj)+(—6P k)
0=(18—6F)j+(246P)k Equaling j, 1: components : (ISGE')=0
(24—6P)=0 Mp=—6Pk F=31b Ans
P=4lb Ans 0r Mz=k'(l'wt XF) O 0 l
= —~0.0l 0.2 0.05
21.213 —2L2l3 0 = o  o + l[(—0.0l)(!21.213) — 21.213(0.2)] = —4.03 N v m ADS T he negative sign indicates that MI is directed along the negative
2 axis. 2 ...
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 Fall '08
 Troxel

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