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Unformatted text preview: MEEN 221 Summer 2009  Worksheet 1 Dr. A. Palaggolo (1) HWSl Due at Beginning of Next Class
Chl (1, 8, 21, 48), Ch 2 (1, 13, 30, 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) Ch2 problems can be solved by the Law of Sines or the Component Sum approach.
Either method is OK. I SYSTEMS 0F UNITS Re ‘. Section 1.4 Table 16 S.I. —En' lish Conversions Base Units Multiples of Units ‘ Derived Units
Quantiﬂ Unit Smbol Mega (M) 10 6 MomentN Length Meter M Kilo (K) 10 3 Pressure W: Pascal
Mass Kilogram Kg Centi (C) 10‘2 Energy N. M= Joule
Time Second 3 Milli (M) 10'3 Acceleration M/s2 Current Ampere A Micro(p.) 106
Temp. Kelvin °K UNITS CONVERSION EXAMPLE
(1) Convertqn acceleration of 1 cm/s2 to its equivalent in Km/yr2 . Use “unit table”.
____—
_—___ (2) Pb.) 133 e. Riley
Express a speed of 20 nm/hr 11 nautical mile = 6076 ft.) in units of
kilometers per minute. Use “units table” approach. _————
_———_ (3) Pb. 1—45 in Riley
1 hp = 33000 ﬁJb/ min and 1 Watt = 1 N.M/S
Prove :1 hp = 745.7 Watts 33000 ﬂ 1b —_
—__ Min G) 1112 NEWTON ’S LAW OF GRAVITATION (Mass 1 Weight : Gravitation Constant g 2 CD.)
See Table 1—] [or Solar System Masses and Distances Mass does not va here in the solar . stem . Wei ht and does va Cavendish balance' “Netting val. “Jaw em you. Newton? The force between any two masses M1 and M2 is MGM”? (I) r
where G = 6.673 x 10 ’“ M3/ (Kg.Sz) (2)
r = distance between the mass centers (1) Pb. 15 in Riley
The planet Venus has a diameter of 7700 mi. and a mass of 3.34 x 1023
slugs. Determine the gravitational acceleration at the surface of the planet. MMv G
R3 F = Weight'(w) = mgv = M (III) ' Dimensional Homogeneity
' Ref. Riley 15.1
Every valid equation must be dimensionally homogeneous: that is, all
iitive terms on both sides of the equation must have the same dimensions. The equation 0(m/s) : 00 (m/s) + g(m / s2)t(s)
is dimensionally homogeneous since each of the terms 0,00 ',gt have the
dimensions length /tirne. ' '(1) Riley 155. In the dimensionallyhomogeneous equation p Mc =—+
GA 1 0' is stress in N/M2 , A is area in M2 , M is a moment of a force and c is a
length. Determine the dimensions of P and I. (IV) Newton’s 3 Laws of Motion 1St A particle originally at rest or moving in a straight line with constant
velocity will remain in this state provided that the particle is not
subjected to an unbalanced force. . ——> 2nd A particle acted on by an unbalanced force F experiences an
: acceleration a that has the same direction as the force and a magnitude
' that is directly proportional to the force. ——> ——>
F=ma 3rd For every force acting on a particle, the particle, exerts an equal,
opposite and collinear reactive force. (1) Riley 147. Newton’s Law of Gravitation can be expressed in equation
form by MIM2 r2 F=G where F is a force, Ml and M2 are masses and r is a distance. Determine
the dimension of G. C3) ngidlbodyis the samefor, all points of application of the force along its
line of action.” F has same effect on motion if
applied at A, B or C. A : (2) ConcurrentForces R('gg. 372: “A ' force system is said to be c0ncun'ent if the action lines of all forces intersect at a comnion point.” (3) 1.Force Vector Analysis. with Laws at Sines and Coesines: ﬂ: F1~F
Smdéd' Force , __ ' 93 ” I ResultantR= FiFz ’ , ' . (1) I Law of Sines: F /sma F 2/sin ,6: R/sin?
I Law of cosines: R 2F2 + F22 + 2FF2 mm (2) ‘
Note sinﬂ 2 F2 SKEW p: [mm{EM ' . (3) C41 a (4) Force Vector Analysis by Component Sums .
Let R be the vector sum (resultant) of a set of n force vectors 13,. : R=Rzéx +Ryéy =23: =§Fkéx +Fiyéy 71 av E then ' Rx =2}; and Ry :25, g .
i=1 ' i=1 lRlzJsz +R2, 6=tan“1(Ry /R ) Rx The angle 6 is measured from the positive x direction in the ccw direction. 6“) Chime/1; ﬂab/9
2’; ax+ér= C
aad’ dX+eY:/ 7X91 X: (ecééIO/ﬂ
)1.— C4/‘ CdJ/A Q5 Exam le 22‘ R 238 Two cables are used to support a stop light as shown. The
resultant R of the cable forcezFu and Fv has a. magnitude of
1350. N and its line of action is vertical. Determine the
magnitudes of forces Fu and Fv. Example 3 : (R228) Find the Magnitude of the ResultantR and the Angle 6 @
Between the + x axis)" and the direction of R _ .—l< Example 4 : 244 = A gusset plate is used to transfer forces from three
bars to a beam as shown in Fig. P244. The magnitude of
the resultant R of the three forces is 5000 .N. If force F1 has
a magnitude of 1000 N, determine the magnitudes of forces
F2 and F3. ...
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 Spring '08
 McVay

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