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# ws 12 - MEEN 221 Summer 2009 Worksheet 1 Dr A Palaggolo(1...

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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 1-6 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 Mill-i (M) 10'3 Acceleration M/s2 Current Ampere A Micro(p.) 10-6 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. 1-5 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 1-5.1 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 1-55. In the dimensionally-homogeneous 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 1-47. 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 same-for, 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= F-i-Fz ’ , ' . (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 2-38 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 : (R2-28) Find the Magnitude of the ResultantR and the Angle 6 @ Between the + x axis)" and the direction of R _ .—l< Example 4 : 2-44 = 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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