ISV SM Ch06 할리데이 10판 6장 솔루션 PDF

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할리데이 10판 6장 솔루션...

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Chapter 6 1. The free-body diagram for the block is shown to the right with the various forces acting on the block placed over the rough inclined surface. As the block has a tendency to slip down the inclined plane, the frictional force acts upward as shown in free body diagram. Because the block is on the verge of sliding, that frictional force is at its maximum magnitude (f = ms FN ). For equilibrium of block the sum of all forces on r r block must be zero, that is, å F = 0 . This will be possible if the net forces along horizontal and vertical directions are separately zero. Equating the net force along horizontal direction (parallel to inclined surface) to zero, å Fx = 0 , we get F + ms FN - mg sin q = 0 Similarly, equating the net forces along vertical direction (perpendicular to inclined surface) to zero, å F y = 0 , we get FN - mg cosq = 0

Combining the two equations gives F + ms mg cos q = mg sin q

Þ

F = mg (sin q - m s cosq )

which leads to F = mg (sin q - m s cosq ) = (6.0 kg)(9.8 m/s 2 )[sin 60° - (0.60) cos 60°] = 33.28 N » 33 N

Hence, a force of magnitude F = 33 N is required if the block is on the verge of sliding down the ramp.

43

44

CHAPTER 6

2. Applying Newton’s second law to the horizontal motion, we have F - mk m g = ma, where we have used Eq. 6-2, assuming that FN = mg (which is equivalent to assuming that the vertical force from the broom is negligible). Eq. 2-16 relates the distance traveled and the final speed to the acceleration: v2 = 2aDx. This gives a = 1.3 m/s2. Returning to the force equation, we find (with F = 25 N and m = 3.5 kg) that mk = 0.60.

45 3. (a) The free body diagram for the figure given in problem is drawn below indicating the various forces.

r

r

r

From the figure shown above and using å F = ma for each mass, where a is the acceleration, we have (1) 10 g - T 2 = 10a T2 - T1 - f k = 3.0a (2) T1 - f k = 2.0a (3) where the kinetic frictional force fk can be expressed as fk = μk N =[0.3(2.0g)] Adding Eqs. (1)–(3), we get 10 g - 2 fk = 15 a 10 ´ 9.8 - 2 ´ 0.3(2.0 g ) 98 - 11.76 Þa = = = 5.75 m/s2 15 15 Thus, the acceleration of both the masses is a = 5.75 m/s2. (b) Substituting the value of a in Eq. (3), we get T1 = 2.0 a + f k = 2.0a + mk (2.0g ) = 2.0(5.75) + 0.3 ´ 2 ´ 9.8 = 11.5 + 5.88 = 17.38 N » 17 N (c) Substituting the value of a in Eq. (1), we get T2 = 10 g - 10 a = 10(9.8) - 10(5.75) = 98 - 57.5 = 40.5 N » 40 N. Hence, the tension force in the strings is 17 N and 40 N.

46

CHAPTER 6

4. (a) When mass m is at its minimum value, that block is on the verge of sliding up its plane and the block with mass M is on the verge of sliding down its plane, and the static frictional forces on both blocks are at their maximum values. Since the system is at rest, the net force on both blocks must be zero. The free body diagram for the two blocks is shown below.

From the above free body diagram, for the block M to be in equilibrium the net force on it should be zero. Equating net force along vertical direction (normal to the surface of inclined) on block M, we get å Fy = 0 N - Mg cos 45° = 0 N = Mg cos 45° =

Mg 2

(1)

Similarly, equating net force along horizontal direction (parallel to surface of incline) to zero we get

å Fx = 0 T + ms N - Mg sin 45 ° = 0 T + ms N = Mg sin 45 ° =

Mg 2

(2)

From Eqs. (1) and (2), we get Mg æ Mg ö - ms ç ÷ 2 è 2ø Mg T= (1 - ms ) 2 T=

(3)

For the block m, under equilibrium the net force on it will be zero. Therefore, on equating net vertical force (along direction perpendicular to incline surface) to zero, we get

47

å Fy = 0 N ¢ - mg cos 45° = 0 mg 2

N ¢ = mg cos 45° =

(4)

Similarly, equating the horizontal force along direction parallel to incline surface to zero, we get

å Fx = 0

T - ms N¢ - mgsin 45 ° = 0 T = ms N¢ + mg sin 45° T = ms N¢ +

mg

(5)

2

From Eqs. (4) and (5), we get T =ms T ==

mg

+

mg

2

mg 2

2

(1 + ms )

(6)

Equations (3) and (6) give Mg

(1 - ms ) =

mg

(1 + ms ) 2 2 2(1 - 0.28) = m(1 + 0.28)

(7)

1.44 = 1.28m m=

1.44 = 1.125 kg » 1.12 kg 1.28

(b) When then mass m has its maximum value, that block is on the verge of sliding down its plane and the other block is on the verge sliding up its plane, the static frictional forces on both blocks are at their maximum values. Therefore, the directions of the frictional forces on the blocks are now reversed. Thus, we can rewrite Eq. (7) as Mg 2 2 2

(1 + ms ) =

(1 + 0.28) =

mg 2 m

(1 - ms )

(1 - 0.28) 2 2.56= 0.72m m = 3.56 kg

Hence, the minimum and maximum value of mass m for which system remains at rest are m =1.12 kg and m = 3.56 kg.

48

CHAPTER 6

5. In addition to the forces already shown in Fig. 6-17, a free-body diagram would r r include an upward normal force FN exerted by the floor on the block, a downward mg r representing the gravitational pull exerted by Earth, and an assumed-leftward f for the kinetic or static friction. We choose +x rightwards and +y upwards. We apply Newton’s second law to these axes: F - f = ma P + FN - mg = 0 where F = 6.0 N and m = 2.5 kg is the mass of the block. (a) In this case, P = 8.0 N leads to FN = (2.5 kg)(9.8 m/s2) – 8.0 N = 16.5 N. Using Eq. 6-1, this implies f s,max = msF N = 6.6 N , which is larger than the 6.0 N rightward force – so the block (which was initially at rest) does not move. Putting a = 0 into the first of our equations above yields a static friction force of f = P = 6.0 N. (b) In this case, P = 10 N, the normal force is FN = (2.5 kg)(9.8 m/s2) – 10 N = 14.5 N. Using Eq. 6-1, this implies f s,max = m sF N = 5.8 N , which is less than the 6.0 N rightward force – so the block does move. Hence, we are dealing not with static but with kinetic friction, which Eq. 6-2 reveals to be f k = m k FN = 3.6 N . (c) In this last case, P = 12 N leads to FN = 12.5 N and thus to f s, max = m sF N = 5.0 N , which (as expected) is less than the 6.0 N rightward force – so the block moves. The kinetic friction force, then, is f k = mk FN = 3.1 N .

49 r 6. The free-body diagram for the player is shown to the right. FN is r the normal force of the ground on the player, mg is the force of r gravity, and f is the force of friction. The force of friction is related to the normal force by f = mkFN.

We use Newton’s second law applied to the vertical axis to find the normal force. The vertical component of the acceleration is zero, so we obtain FN – mg = 0; thus, FN = mg. Consequently, 485 N ms = = 0.60 2 (83 kg)(9.8 m/s )

50

CHAPTER 6

7. THINK A force is being applied to accelerate a crate in the presence of friction. We apply Newton’s second law to solve for the acceleration. diagram for the crate is shown to EXPRESS The free-body r the right. We denote F as the horizontal force of the person r exerted on the crate (in the +x direction), f k is the force of kinetic friction (in the –x direction), FN is the vertical normal force exerted by the floor (in the +y direction), and r mg is the force of gravity. The magnitude of the force of friction is given by (Eq. 6-2): fk = mkFN . Applying Newton’s second law to the x and y axes, we obtain F - f k = ma FN - mg = 0 respectively. ANALYZE (a) The second equation above yields the normal force FN = mg, so that the friction is f k = mk FN = mk mg = ( 0.30)( 55 kg) (9.8 m/s2 ) = 1.6 ´ 102 N . (b) The first equation becomes

F - m k mg = ma

which (with F = 260 N) we solve to find F a= - m k g = 1.8 m/s2 . m LEARN For the crate to accelerate, the condition F > fk = mk mg must be met. As can be seen from the equation above, the greater the value of mk , the smaller the acceleration with the same applied force.

51 8. To maintain the stone’s motion, a horizontal force (in the +x direction) is needed that cancels the retarding effect due to kinetic friction. Applying Newton’s second to the x and y axes, we obtain F - f k = ma FN - mg = 0 respectively. The second equation yields the normal force FN = mg, so that (using Eq. 6-2) the kinetic friction becomes fk = mk mg. Thus, the first equation becomes F - m kmg = ma = 0 where we have set a = 0 to be consistent with the idea that the horizontal velocity of the 2 stone should remain constant. With m = 20 kg and mk = 0.80, we find F = 1.6 ´ 10 N.

52

CHAPTER 6

9. We choose +x horizontally rightwards and +y upwards and observe that the 15 N force has components Fx = F cos q and Fy = – F sin q . (a) We apply Newton’s second law to the y axis: FN - F sin q - mg = 0 Þ FN = (15 N) sin 40° + (3.5 kg) (9.8 m/s 2 ) = 44 N.

With mk = 0.25, Eq. 6-2 leads to fk = 11 N. (b) We apply Newton’s second law to the x axis: F cos q - f k = ma Þ a =

(15 N ) cos 40° - 11 N = 0.14 m/s2 . 3.5 kg

Since the result is positive-valued, then the block is accelerating in the +x (rightward) direction.

53 10. The free body diagram for the block of weight W is shown below

Since the block is on the verge of moving the sum of forces acting horizontally and vertically must be zero at that instant, that is, å Fx = 0 and å Fy = 0. Also, the value of the static frictional force must be at its maximum. From the free body diagram shown below, equating the sum of vertical forces on the block to zero, we get

åFy = 0

W W sin 30° - W - = 0 2 2 W W N + sin 30° = W + 2 2 N+

N=

3W W 5W = 2 4 4

Similarly, equating the sum of horizontal forces on the block to zero, we get

åFx = 0 f-

Þm =

W 2 W 2 W 2 W 2

cos30° = 0 cos30 ° = f cos30 ° = m N æ 5W ö cos30 ° = m ç ÷ è 4 ø

(W / 2) ´ 0.866 2 ´ 0.866 = = 0.35. 5W / 4 5

(1)

54

CHAPTER 6

11. THINK Since the crate is being pulled by a rope at an angle with the horizontal, we need to analyze the force components in both the x and y-directions.

r The free-body diagram for the crate is shown on the right. T is the r tension force of the rope on the crate, FN is the normal force of the r r floor on the crate, mg is the force of gravity, and f is the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up. We assume the crate is motionless. The equations for the x and the y components of the force according to Newton’s second law are: T cos q – f = 0 T sin q + F N - mg = 0 where q = 15° is the angle between the rope and the horizontal. The first equation gives f = T cos q and the second gives FN = mg – T sin q. If the crate is to remain at rest, f must be less than m s FN, or T cos q < m s (mg – T sinq ). When the tension force is sufficient to just start the crate moving, we must have T cos q = ms (mg – T sin q). ANALYZE (a) We solve for the tension: ( 0.65 ) (68 kg ) ( 9.8 m/s 2 ) m smg = = 382 N. T= cos q + ms sin q cos 15° + 0.65 sin 15° (b) The second law equations for the moving crate are T cos q – f = ma FN + T sin q – mg = 0. Now f =mkFN, and the second equation gives FN = mg – Tsin q , which yields f = m k (mg - T sinq ) . This expression is substituted for f in the first equation to obtain T cos q – m k (mg – T sin q ) = ma, so the acceleration is T ( cos q + mk sin q) a= - mk g m (304 N)(cos15 ° + 0.35 sin 15°) 2 2 = - (0.35) (9.8 m/s ) = 1.3 m/s . 68 kg LEARN Let’s check the limit where q = 0 . In this case, we recover the familiar expressions: T = ms mg and a = (T - mk mg ) / m .

55 12. There is no acceleration, so the (upward) static friction forces (there are four of them, one for each thumb and one for each set of opposing fingers) equals the magnitude of the (downward) pull of gravity. Using Eq. 6-1, we have 4m sFN = mg = (79 kg)(9.8 m/s 2) 2 which, with ms = 0.70, yields FN = 2.8 ´ 10 N.

56

CHAPTER 6

13. We denote the magnitude of 110 N force exerted by the worker on the crate as F. The magnitude of the static frictional force can vary between zero and f s, max = m s FN . (a) In this case, application of Newton’s second law in the vertical direction yields FN = mg . Thus, f s, max = m s FN = m s mg = (0.37 )( 35kg ) (9.8m / s 2 ) = 1.3´ 102 N

which is greater than F. (b) The block, which is initially at rest, stays at rest since F < fs, max. Thus, it does not move. (c) By applying Newton’s second law to the horizontal direction, that the magnitude of the frictional force exerted on the crate is fs = 1.1 ´102 N . (d) Denoting the upward force exerted by the second worker as F2, then application of Newton’s second law in the vertical direction yields FN = mg – F2, which leads to f s,m ax = m s FN = m s ( mg - F2 ) .

In order to move the crate, F must satisfy the condition F > fs,max = m s (mg - F2) , or 2 110N > (0.37 ) éë (35kg)(9.8m/s ) - F2 ùû .

The minimum value of F2 that satisfies this inequality is a value slightly bigger than 45.7 N , so we express our answer as F2, min = 46 N. (e) In this final case, moving the crate requires a greater horizontal push from the worker than static friction (as computed in part (a)) can resist. Thus, Newton’s law in the horizontal direction leads to F + F2 > f s, max

Þ

110 N+ F2 > 126.9 N

which leads (after appropriate rounding) to F2, min = 17 N.

57 14. (a) Using the result obtained in Sample Problem – “Friction, applied force at an angle,” the maximum angle for which static friction applies is q max = tan -1 m s =tan -1 0.63 » 32 ° . This is greater than the dip angle in the problem, so the block does not slide. (b) Applying Newton’s second law, we have F + mg sin q - f s, max = ma = 0 FN - mg cos q = 0. Along with Eq. 6-1 (fs, max = msFN) we have enough information to solve for F. With q = 24° and m = 1.5 ´ 107 kg, we find F = mg ( ms cos q - sin q )= 2.5´ 107 N .

58

CHAPTER 6

15. The free body diagram for the block is shown below

Since the block is at rest, it is in equilibrium and hence the net force on the block is zero. Using the free body diagram for the block, equating the net force along the vertical direction to zero, we get

åF

y

=0

N -mg cos q = 0 N = mg cosq

(1)

Similarly, equating the net force along horizontal direction equal to zero, we get

åF

x

=0

f - mg sin q = 0 f = mg sin q

(2)

The net contact force is Fnet =

N2 + f 2

= (mg cos q )2 + (mg sin q )2 = m 2 g 2(cos 2 q + sin 2 q ) = mg

Hence, with m = 5.0 kg, the force exerted by the ramp on the block is mg = 49 N.

59 16. The force of friction at different contact surfaces is as shown in the following figures.

Here, the frictional force between small block and the upper of big block is f2 = m 2 (mg )

And the force of friction between the big block and the ground is f1 = m1(11 mg)

Given that m 2 > 11m 1 . Therefore, f 2 > f 1. The retardation of the upper block is f2 = m2 g a1 = m The acceleration of the lower block is f - f1 ( m2 - 11 m1 ) g a2 = 2 = 10 m 10 The relative retardation of the upper block is é æ m - 11m1 öù ar = a1 + a2 = ê m2 + ç 2 ÷ú g è 10 øû ë æ11m 2 - 11m1 ö =ç ÷g 10 è ø 11 = (m 2 - m1 )g 10 (a) For minimum velocity (v min) of the upper block which enables it to fall off the lower one, using third kinematics relation, we get 2 0 = vmin - 2 a rl

vmin =

2ar l =

22 (m - m1 ) gl 10 2

(b) Using first kinematics relation, we get 0 = vmin - ar t Þt=

vmin = ar

20 l 11(m 2 - m 1 )g

60

CHAPTER 6

17. If the block is sliding then we compute the kinetic friction from Eq. 6-2; if it is not sliding, then we determine the extent of static friction from applying Newton’s law, with zero acceleration, to the x axis (which is parallel to the incline surface). The question of whether or not it is sliding is therefore crucial, and depends on the maximum static friction force, as calculated from Eq. 6-1. The forces are resolved in the incline plane coordinate system in Figure 6-5 in the textbook. The acceleration, if there is any, is along the x axis, and we are taking uphill as +x. The net force along the y axis, then, is certainly zero, which provides the following relationship:

r

åF

y

= 0 Þ FN = W cos q

where W = mg = 45 N is the weight of the block, and q = 15° is the incline angle. Thus, FN = 43.5 N, which implies that the maximum static friction force should be fs,max = (0.50) (43.5 N) = 21.7 N. r (a) For P = (- 5.0 N)iˆ, Newton’s second law, applied to the x axis becomes

f - | P | - mg sin q = ma . r Here we are assuming f is pointing uphill, as shown in Figure 6-5, and if it turns out that it points downhill (which is a possibility), then the result for fs will be negative. If f = fs then a = 0, we obtain

fs = | P | + mg sinq = 5.0 N + (43.5 N)sin15° =17 N, r or fs = (17 N)iˆ. This is clearly allowed since fs is less than fs, max. r r (b) For P = ( - 8.0 N)iˆ, we obtain (from the same equation) f s = (20 N)iˆ, which is still allowed since it is less than fs, max.

r (c) But for P = (- 15 N)iˆ, we obtain (from the same equation) fs = 27 N, which is not allowed since it is larger than fs, max. Thus, we conclude that it is the kinetic friction instead of the static friction that is relevant in this case. The result is r f k = m k FN ˆ i = (0.34)(43.5 N) ˆ i = (15 N) ˆ i.

61 18. (a) We apply Newton’s second law to the “downhill” direction: mg sinq – f = ma, where, using Eq. 6-11,

f = fk = mkFN = mk mg cos q .

Thus, with mk = 0.600, we have a = g(sinq - mk cosq ) = –3.71 m/s2 which means, since we have chosen the positive direction in the direction of motion (down the slope) then the acceleration vector points “uphill”; it is decelerating. With v0 = 18.0 m/s and Dx = d = 30.0 m, Eq. 2-16 leads to

v = v20 + 2ad = (18 m/s)2 + 2(-3.71 m/s2 )(30 m) = 10.0 m/s. (b) In this case, we find a = g (sinq - mk cosq ) = (9.8 m/s2 )[ sin12° - (0.10) cos12° ] = 1.08 m/s2

and the speed (when impact occurs) is

v = v02 + 2ad = (18 m/s)2 + 2(+ 1.08 m/s2 )(30 m) = 19.7 m/s.

62

CHAPTER 6

r 19. (a) The free-body diagram for the block is shown below. F is r the applied force, FN is the normal force of the wall on the block, r r f is the force of friction, and mg is the force of gravity. To determine if the block falls, we find the magnitude f of the force of friction required to hold it without accelerating and also find the normal force of the wall on the block. We compare f and msFN. If f < msFN, the block does not slide on the wall but if f > msFN, the block does slide. The horizontal component of Newton’s second law is F –FN = 0, so FN = F = 12 N and msFN = (0.60)(12 N) = 7.2 N. The vertical component is f – mg = 0, so f = mg = 5.0 N. Since f < msFN the block does not slide. (b) Since the block does not move f = 5.0 N and FN = 12 N. The force of the wall on the block is r j Fw = - FN iˆ+ f ˆj = - (12N ) iˆ+ (5.0N ) ˆ where the axes are as shown on Fig. 6-26 of the text.

63 20. Treating the two boxes as a single system of total mass mC + mW =1.0 + 3.0 = 4.0 kg, subject to a total (leftward) friction of magnitude 2.0 N + 3.5 N = 5.5 N, we apply...