2011_8_9e1bcf17

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班級:光電一乙 學號:4980106 姓名:高柏翔 98 學年度第 1 學期物理作業(chap4) 1. Find the tension in each cable supporting the 600-N cat burglar in Figure 1. a x= a 37 cos = ) ( 5 4 N a a y =a 37 sin = ) ( 5 3 N a b x =600 270 cos =0(N) b y =600 270 sin =-600(N) Figure 1 c x =c 180 cos =-c(N) c y =c 180 sin =0(N) 0 ) ( 0 5 4 c a c b a F x x x X ------(1) 0 0 ) 600 ( 5 3 a c b a F y y y y ----(2) 解聯立 (2): 600 5 3 a a=1000(N)代入(1) c=800 Ans:a=1000(N) b=600(N) c=800(N) 2. Two people are pulling a boat through the water as in Figure 2. Each exerts a force of 600 N directed at a 30.0° angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force F exerted by the water on the boat. c x =c ) ( 180 cos N c ) ( 0 180 sin N c c y ) ( 6 . 519 30 cos 600 N a x ) ( 300 30 sin 600 N a y Figure 2 ) ( 6 . 519 330 cos 600 N b x ) ( 300 330 sin 600 N b y 0 6 . 519 6 . 519 x y x x X c c b a F ) ( 2 . 1039 N c x ) ( 0N c y 0 0 300 300 y y y Y c b a F ) ( 2 . 1039 0 ) 2 . 1039 ( 2 N c

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Page 1: 2011_8_9e1bcf17

班級:光電一乙 學號:4980106 姓名:高柏翔

98 學年度第 1 學期物理作業(chap4)

1. Find the tension in each cable supporting the 600-N cat burglar in Figure 1.

ax=a 37cos = )(54

Na

ay=a 37sin = )(53

Na

bx=600 270cos =0(N)by=600 270sin =-600(N)

Figure 1 cx=c 180cos =-c(N)cy=c 180sin =0(N)

0)(054

cacbaF xxxX ------(1)

00)600(53

acbaF yyyy ----(2)

解聯立 由(2): 60053

a a=1000(N)代入(1) c=800

Ans:a=1000(N)b=600(N)c=800(N)

2. Two people are pulling a boat through the water as in Figure 2. Each exerts a force of 600 N directed at a30.0° angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the

resistive force F

exerted by the water on the boat.

cx=c )(180cos Nc

)(0180sin Ncc y

)(6.51930cos600 Nax

)(30030sin600 Na y

Figure 2 )(6.519330cos600 Nbx

)(300330sin600 Nby

06.5196.519 xyxxX ccbaF )(2.1039 Ncx )(0 Nc y

00300300 yyyY cbaF )(2.10390)2.1039( 2 Nc

Page 2: 2011_8_9e1bcf17

3. An object with mass m1 = 5.00 kg rests on a frictionless horizontal table and is connected to a cablethat passes over a pulley and is then fastened to a hanging object with mass m2 = 10.0 kg, as shown inFigure 3. Find the acceleration of each object and the tension in the cable.

akgTmaF xX )00.5( (1)

Next consider the block with move vertically.The force on it are thetension T and weight 98.0N.

akgTNmaF yy )0.10(0.98 (2)

Figure 3Note that both block must have the same magnitude of acceleration.Equation(1) and (2) can be solved simultaneously to give.

253.6s

ma and T=32.7N

4. In Figure P4.30, m1 = 10 kg and m2 = 4.0 kg. The coefficient of static friction between m1 and thehorizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released fromrest, what will its acceleration be? (b) If the system is set in motion with m2 moving downward, whatwill be the acceleration of the system?

(a) The static friction force attempting to prevent motion may reach a maximum value of

Ns

mkggmnf sss 49)80.9)(10)(50.0()( 211max

This exceeds the force attempting to move the system,the weight of m2.Hence,the system remains at restAnd the acceleration is a=0

(b) gmnf kkk 11

2

2

21

12 70.0100.4

)80.9()10(30.00.4

sm

kgkgs

mkgkg

mmgmgm

mF

a k

total

net

Page 3: 2011_8_9e1bcf17

5. Two boxes of fruit on a frictionless horizontal surface are connected by a light string as in Figure4,where m1 = 10 kg and m2 = 20 kg. A force of 50 N is applied to the 20-kg box. (a) Determine theacceleration of each box and the tension in the string. (b) Repeat the problem for the case where thecoefficient of kinetic friction between each box and the surface is 0.10.

Figure 4

)( 111 gmnf kk mf k (2

ammffNFX )(50 2121

gmm

Na k

21

50(1)

amfTFX 11 ,or )(1 gamT k (2)

(a) 0k 代入(1)

221

7.13050

050

sm

kgN

mmN

a

代入(2)

Ns

mkgT 17)07.1)(10( 2

(b) 10.0k 代入(1)

22 69.0)80.9)(10.0(3050

sm

sm

kgN

a 代入(2)

Ns

ms

mkgT 17)80.9)(10.0(69.0)10( 22

m1T

a

m1g

Figure 2