1、第8頁共9頁理論力學試卷（A）XXX工業大學20XX20XX學年第XX學期期末考試試卷學院 班級 姓名 學號理論力學課程試卷 B （附參考答案）Instructor: Yijian(Time: 2 Hours) Course Code:題號123456總得分題分201515151520得分1. Choose the correct answers with proper justification (102=20, 20%)義12345678910DDCABBACBC(1) For vector addition you have to use law.A) Newton ' s Se

2、cond B) the arithmetic C) Pascal ' sD) the parallelogram(2) A particle moves along a horizontal path with its velocity varying with time as shown in theright figure. The average acceleration of the particleA) 0.4 m/s2 -B) 0.4 m/s2 一C) 1.6 m/s2 -D) 1.6 m/s2 一(3) If the position of a particle is d

3、efined by r = (1.5t2 + 1) i + (4t - 1) j (m), its speed at t=1 sis .A) 2 m/sB) 3 m/sC) 5 m/s D) 7 m/s(4) As shown in the right figure, the speed of block B is .A) 1 m/sB) 2 m/sC) 4 m/sD) None of the above.(5) If the path function of a particle is s = 10 sin 2 Q the acceleration, a,is . (Noting that

4、0 =w t, is6the angular velocity at time t).A) 20 a sin2 0B) 20 a cos2 9-40 c/sin2 0C) 20 a cos2 0D) -40 a sin2 0(6) As shown in the right figure, the velocity of plane A with respect to plane B is A) (400 i + 520 j) km/hrB) (1220 i - 300 j) km/hrC) (-181 i - 300 j) km/hrD) (-1220 i + 300 j ) km/hr(7

5、) If the pendulum is released fromvertical position is A) 3.8 m/sB) 6.9 m/sC) 14.7 m/sD) 21 m/s(8) If a slender bar rotates about end A, its angular momentum with respect to A isA) (1/12) m l2 3B) (1/6) m l2 coC) (1/3) m l2 coD) m l2 9(9) The tangential acceleration of an object .A) represents the r

6、ate of change of the velocity vector' s direction.B) represents the rate of change in the magnitude of the velocity.C) is a function of the radius of curvature.D) Both B and C.(10) Point A on the rod has a velocity of 8 m/s to the right. Where is the instantaneous center (IC) for the rod? .A) Po

7、int A. B) Point B. C) Point C. D) Point D.2. (15%)Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axisSolution:Using the vector analysis, we have the x- and y-components of forces 600N and 800N asFix600 N cos30o 300 ,3 N 519.6NRy600 N si

8、n30o300 N1y(2%)F2x 0F2y800NHence, the force vectors are obtained asFiF1xi Eyj 519.6i 300j(4%)F2 F2xi F2yj 800jAccording to the vector addition rule, the resultant force is expressed in the vector form(5%)FrFi F2Fix F2x iF1yF2y jFRxi FRyj519.6i 500jSo, the magnitude and direction of the resultant for

9、ce are obtained as519.6 2500 2 721.10NFrx arccos Fr519.6 arccos721.143.90°(4%)3. (15%)The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles % 0 of the resultant force. Take x=20 m, y=15 m.Solu

10、tion:Going from D to A, one must travel -24 k m, then 20 i m and finally 15 j m. Thus, the unitvector for rope DA is (2%)u DArA 20i 15j 24k m rA, 202 15224 2 m20 i 15 .34.661 34.66J&k 34.6620Fa Fauda400N i34.661524j k34.6634.66230.81i 173.11j 276.98k NGoing from D to B, one must travel -24 k m,

11、then -6 i m and finally 4 j m. Thus, the unitvector for rope DB is (2%)u DBrB 6i 4j 24k mrB . 6 2 4224 2 m6424i j k25.0625.0625.06Fb Fbudb800N6 . i 25.064 .25.06 J2425.06191.54i 127.69j 766.16k NGoing from D to C one must travel -24 k m, then -18 j m and finally 16 im. Thus, the unitvector for rope

12、DC is (2%)u DCrc16i 18j24k m8 .一i9 .jkJ -c17rc16218224 m1717FCFC u DC8912600N i j k 171717282.35i 317.65j 423.53k NHence, the resultant force is (5%)FR FA FB FC321.62i 16.85j 1466.67k NThe magnitude and the coordinate direction angles of FR are (4%)FR - FRxFRyfRz.,321.622216.85FRx321.62一一 oarccos -a

13、rccos 一77.63FR1501.7FRy90.64oarccos arccos -90.64°FR1501.7FRz167.6oarccos arccos 167.6FR1501.721466.671501.7N4. (15%)Determine the horizontal and vertical components of reaction at the pin A and the reaction on the beam atC.Solution:(1) Free-body diagram (5%)We draw a free-body diagram for the

14、overhanging beam as below:(2) Equations of equilibrium (5%)Consider the counterclockwise moment of the force positive. According to the free-body diagram, we have x- and y-component equations and moment about point A of equilibrium as followso -Fx0,FaxFc cos450Fy0,FAyFcsin45o4kN0M A 0,FC sin 45o 1.5

15、m 4kN 3m 0After solving, we have (5%)FAx 11.31kN, FAy8kN, Fc4kNThe negetive sign indicates that the direction sence is oppisite to that shown in the free-body diagram.5. (15%)A projectile is launched from point A with the initial conditions shown in the figure. Determine the slant distance s which l

16、ocates the point B of impact. Calculate the time of flight t.Solution:Establish a fixed x,y coordinate system (in the solution here, the origin of the coordinate system is placed at A). Apply the kinematic relations in x and y-directions.Motion in x-direction: (4%)XBxA vOxtABxa vo cos40otABxB 0120m/

17、s cos40otABMotion in y-direction: (5%)yByA v0ytAB 2gtAB Va120m/s sin40otAB 1gtAB199yB 0120m/ s sin 40 tAB - 9.81kg/ s2 tABNoting that (2%)Vb xb 800m tan20o, s yB /sin 20oAfter solving, we obtain the slant distance s and the flight time t as follows (4%) s 279.8m, t tAB 3.04s6. (20%)The rod AB has a

18、mass of 10 kg. Piston B is attached to a spring of constant k = 800 N/m. The spring is un-stretched when = 0°. Neglect the mass of the pistons. Determine the angular velocity of rod AB at = 0 ° if the rod is released from rest when = 30 :Solution:Initial position (2%)Final position (2%)(1)

19、 Potential Energy: (5%)Let ' put the datum in line with the rod when = 0 : Then, the gravitational potential energyand the elastic potential energy will be zero at position 2. => V 2 = 0Gravitational potential energy at position 1: - (10)( 9.81) ? (0.4 sin 30)Elastic potential energy at position 1: ? (800) (0