This preview shows pages 1–14. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: V MEEN 221 Summer 2002  Worksheet 2 Dr. A. Palazgolo (l) HWS2 Due at Beginning of Next Class
Ch2 (58, 84), Ch 3 (3, 8, 10, 23, 37)
(2) Attendance Mandatory for 10:00 am— 12:45 pm , unless excused by instructor.
Quizzes may be given at any time during this period.
(3) Subjects: Concurrent Force System, Static Equilibrium, Free Body Diagram Rectangular Comgonen'ts at a Force: Let F pass through point q with coordinates (a, b, 0). Then the unit
vector (3,, along the line of action of F is given by: ' A A A A 'a': b 1‘ C . . I’C‘
eF =Zt+21+gk=emz+6“,y+eEz _ Q Casgze —‘
Fx 4 where a’ =ch2 +b2 +c2 . The vector F may be represented by _F =F éi (where F is the magnitude ofAF) A ‘ (OS 6% 5 eFa :
. =FX i.+F, j+.Fz k=Fein +Feﬂjr+Feﬂk _ The orientation angles of F are given by: .K 6, =cos“'(eFx), 61, =‘cos"(e,‘;,)_, 62 =cos"(em) .Resultants by Rectangular Comgonents Let 13‘} be represented in rectangular form as: F}. =ng +Fﬁj+1~gk, j=1,2,...n Then the resultant of the concurrent force system shown above is given by:
— 0—. A A A
R=2Fj =in +Ryj+Rzk
i=1
“(hem .Rx :iFAjn Ry =53Frj’ R2 zinj j=l F1 ' j=l Note: _ ” [Ma/R: +123 +_R§ . 9 R _. Rt _ R ‘ Izco’s 6R? : cos 2,008 wiﬁté Ceneurrent Fortes Suppose that the rigid body in Fig. 1(a) is represented as a collection of
an inﬁnite number of particles subjected to external forces 13],}?2 F;
and interaction forces i = force on article i (1116 to article '= — 7 b Newton's 3rd Law
f, p 13 J J y ‘ (wherefﬂ is the force on particle j due to particle i ) (l)
‘ Newton's second law for each particle becomes
jﬁéeﬁ+gﬁj I
(2) 0 _. a)
0 %=R2 +jz=lf2j where R is the resultant external force on particle i. All ai's are zero. '
since the body is stationary. Sum all equations in (2) to obtain 0=§E+iin (» i=1 j=1 ' The double summation in (3) is zero by .eq. 1 and the ﬁrst sum equals: ' the of all external forces aetin on the body. Therefore (3 ‘ a roves . (Static EQuilibriwn) R (2—61 1:
a) Determine the x, y and z scalar components of the
force shown below, and _. 1, (—H)+\a,'7) 1C1:
/\ 1’ A.» I __ \ :ibOOlbr F, 3 F J
./ Riley (2 — 66 )
TWO forces are applied to an eyebolt as shown .
(a) Determine the components of F2 (b) Write; F2 in Cartesian vector form .I (c) Determine the magniturle of the component of F2 along FI '17.: Fi=3Q.k‘U
F2.’—50KN (a; ‘3, wmeﬁrerr '. (S; “3) 1)/V\9:‘r err 32;;
Pl. Q96 283*'Determine the magnitude R of the resultant and the angles
9 , 9 , and 9 between the x y z . line of action of the resultant
and the positive x—, y, angz—
coordinate axes for the ﬁhree forces shown in Fig. PZBS; ' > restart
>‘#sauusBLE , (3 3312) Harnlng, premature and of Input . ’ [> ca. := 3.1415931180 : v ‘
> eqns :={TE300*9.81=0 , '1'C*coé (60fga) 'TD=0 , TC*Sin (60*ca} ~TE=0 ,
[ TB*cos (30*ca.)T11*cos(40*ca)TC*cds(60*ca)=0 , '
, TB*sin (30*ca) +‘1‘A*sin (40*ca) TC*sini(60 *ca.) =0} :
[> varsz= {TE,TC,TD,TA,TB}_:
. >. solve(eqmsi,vars) , i i .  , .
i '= 2943., TD = 1699141392, T C n 3398283459, TA = 1808.l%8952, T B = 3561436524} EXAMPLE 1 R(3—12) Statement: A body with a mass of 300 kg is supported by the ﬂexible cable
system shown in Fig. 1. Determine the tensions in cables A, B, C, D,‘ E. Exam 1e ~ 2% ~ w.—
Determine the resultant of the 4‘ Forces and its line of action with
respect to the axis of the plane. Fi‘ez 9 swig 9\ «(fireman 84
_..CHAPTER 3 STATICS OF PARTICLES »
TABLE 6—1 TWODIMENSIONAL REACTIONS AT SUPPORTS AND CONNECTIONS
1. Gravitational attraction The gravitational attraction of the Earth on a body (see Fig. 61) is the
weight W of the body. The line of action of the force W passes through the
center of gravity of the body and is directed toward the center of the Earth, Cylinder supported by
smooth surfaces Fig. 61 2. Flexible cord, rope, chain, or cable A ﬂexible cord, rope, chain, or cable (see Fig. 6—2) always exerts a tensile
force R on the body. The line of action of the force R is known; it is
tangent to the cord, rope, chain, or cable at the point of attachment. Free—body diagram Figure 31 Freebody diagmm for a
cylinder supported by two smooth
surfaces. A rigid link (see Fig. 63) can exert either a tensile or a compressive force
R on the body. The line of action of the force R is known; it must be
directed along the axis of the link (see Section 63.1 for proof). Fig. 63 A ball, roller, or rocker (see Fig. 64) can exert a compressive force R on
the body. The line of action of the force F. is perpendicular to the surface
supporting the ball, roller, or rocker. Block held on a smooth inclined
surfnce with a ﬂexible cable Freebod) diagram Figure 32 Freebody diagram for a
block held on a smooth inclined sur
face with a ﬂexible cable. (b) Constructing a FreeBody Diagram i i Step1. Dedde which body or combination of bodies is to be 1
shown on the freebody diagram. Step 2. Prepare a drawing or sketch of the outline of this isolated I
or free body. Step 3. Carefully trace around the boundary of the free body and ,
identify all the forces exerted by contacting or attracting I
bodies that were removed during the isolation process. , Step 4. Choose the set of coordinate axes to be used in solving [
the problem and indicate their directions on the ﬂeebody ;
diagram. Place any dimensions required for solution of I
the problem on the diagram. l @0) [> #EXAMPLE 2 (R. c3—40) d: 5.60.04“ > W1:=5*9.81 : Theta := arccos(2/3*b/d 1) : éy/ <§$ven)
[ Thetan:=subs(d:100,Theta): ' ‘l . w . v
a [> Df:= WA/sin('1‘hetan): ' MAPLE 'COﬂE ‘ [> Az=WA*cos(Thetan)/sin(Thetan):
> plot({n£,1inesty1e=1,p.,11nesty1e=2},b=200.‘.290,1£it1e="n and A
Forces vs. b , with d = 100nm",l_abels=["b in m","rorce in
N"],thickhess=2,axgs=box)a‘ ‘ :I' , ’ DandAForcesya b , With d,=100mm’ cheinN \‘¢?‘ ,2;
‘47 40 200 2%] 24) ‘4“ ﬂ ,_ > # EXAMPLE. 3 c3—39 FROM Riley
> ThetaI. :é arcsin(ld/10_) :_
' ThetaR:=a.rctan (d/ (30—10mm (arcsin' (d/lO) ) ) ) :
> de1:= sin('1'heta.L +' ThetaR) . V v .
>_'1'L:= 1000*cos(ThetaR)/de1: P:= 1000*cos (ThetaL) Idel: MP; 9.6 CODE‘ vs. d ",1abe13=["d in ft. " ,"For_ces in
lbs . "] ,thickness=2 ,a3es=box) ;
> : P and TL Forces vs. d 6000 ‘ P44300011»: = meaeéi Forces in lbs. 2000 [ I:  ‘ > plot‘({TL,1inestj1eé1,P,linestyle=2},d=1..10,title¥"P and n. Edrces
1000 a: 3 C3 3 ‘83 m: 59G R0?
F;,‘o\ CQBXQ +2nJth. ...
View
Full
Document
This note was uploaded on 07/16/2009 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.
 Spring '08
 McVay

Click to edit the document details