Solutions: Chapter 18

Transcription

1 Solutions: Chapter (a) The force of an side of the cube is F pa ( nrt/v ) A ( nrt )/L, since the ratio of area to volume is A/ V / L. For T 00C 93 K, nrt (3 mol)(834 J/mol K)(93 K) 4 F 3660 N L 000 m (b) For T 0000 C 373 K, nrt (3 mol)(8 34 J/ mol K)(373 K) 4 F 460 N L 000m Ke: F pa and pv nrt V/A L 8.9. (a) pv nrt gives pv (.03 0 Pa)( m ) 4 n.48 0 mol. The mass of this amount of gas is RT (8.3 J/mol K)(9 K) m nm (.48 0 mol)(8.0 0 kg/mol) kg. (b) Ke: 6 m kg kg/m. V m pv nrt ρ = m/v Use vrms to calculate T: vrms 3RT M so 3 Mvrms ( kg/mol)(8 m/s) T 37.0 K. 3R 3(8.34 J/mol K) nrt The ideal gas law gives: p. V 3 m kg 3 n mol. M kg/mol Solving for p gives 3 ( mol)(8.34 J/mol K)(37.0 K) 3 p.69 0 Pa m 3RT M Ke: v rms pv = nrt,

2 8.37. (a) m v av kt ( ) ( 38 0 J/molecule K)(300 K) 6 0 J (b) We need the mass m of one molecule: 3 M 30 0 kg/mol 6 m 34 0 kg/molecule N molecules/mol A Then av mv ( ) 6 0 J (from part (a)) gives (60 J) (60 J) av 6 ( v ) 34 0 m /s m 34 0 kg (c) (d) v rms 4 rms ( v ) 34 0 m /s 484 m/s 6 3 p mv rms (34 0 kg)(484 m/s) 7 0 kg m/s (e) Time between collisions with one wall is 00 m 00 m s t v 484 m/s rms In a collision v changes direction, so dp F so dt (f) (g) F 3 av 4 rms p 4 0 kg m/s N t 430 s 3 3 p mv (7 0 kg m/s) 4 0 kg m/s 9 7 pressure F/A 4 0 N/(00 m) 4 0 Pa (due to one molecule) pressure atm 03 0 Pa Number of molecules needed is Pa/( 40 Pa/molecule) 87 0 molecules (h) pv NkT (Eq. 8.8), so 3 pv (03 0 Pa)(00 m) N 40 molecules kt 3 (38 0 J/molecule K)(300 K) (i) From the factor of 3 in av v 3 av ( v ) ( ) 8.4. (a) V Q nc T n R T. Q (. mol)( )(8.34 J/mol K)(0.0 K) 600 J. 3 3 (b) Q nc T n( R) T. Q (. mol)( )(8.34 J/mol K)(0.0 K) 60 J. V Ke: dq=ncvdt

3 8.43. (a) c C 076 J/mol K c 74 J/kg K. w M kg/mol cn N 3 6; c w is over five time larger. (b) To warm the water, Q 49 0 J m 6 kg. c T (74 J/kg K)(0 0 K) N Q mcw T ( 00 kg)(490 J/mol K)(0 0 K) 49 0 J. For air, 4 mrt (6 kg)(834 J/mol K)(93 K) 3 V 48 m. Mp 3 (804 0 kg/mol)( 030 Pa) Ke: C Mc, where C is the molar heat capacit and c is the specific heat capacit. m pv nrt RT M (a) c C /M (49 J/mol K)/(8 00 kg/mol) 380 J/kg K V V C 6 R 3R 3(834 J/mol K) 49 J/mol K The specific heat capacit is V (b) For water vapor the specific heat capacit is 3 C Mc (8 00 kg/mol)(000 J/kg K) 360 J/mol K 3 c 000 J/kg K The molar heat capacit is Ke: R contribution to C V for each degree of freedom. The molar heat capacit C is related to the specific heat capacit c b C Mc

4 Chapter IDENTIFY: Eample 9. shows that for an isothermal process W nrt ln( p/ p). Solve for p. SET UP: For a compression (V decreases) W is negative, so W 468 J. T 9. K. W p EXECUTE: (a) ln. nrt p p e p W/ nrt. W 468 J nrt (0.30 mol)(8.34 J/mol K)(9. K) / 0.63 p p e W nrt (.76 atm) e 0.94 atm. (b) In the process the pressure increases and the volume decreases. The pv-diagram is sketched in Figure 9.. EVALUATE: W is the work done b the gas, so when the surroundings do work on the gas, W is negative. The gas was compressed at constant temperature, so its pressure must have increased, which means that p < p, which is what we found. Figure IDENTIFY: The gas is undergoing an isobaric compression, so its temperature and internal energ must be decreasing. SET UP: The pv diagram shows that in the process the volume decreases while the pressure is constant. 3 3 L 0 m and atm.03 0 Pa. V nr EXECUTE: (a) pv nrt. n, R and p are constant so constant. T p V T a a V T b b. T T /4 (0.00 L) 0. L. b a Vb Va T a T a (b) For a constant pressure process, W p V (.0 atm)(0. L 0.00 L) and m.030 Pa W ( 0.6 Latm) 7.0 J. W is negative since the volume decreases. L atm Since W is negative, work is done on the gas. (c) For an ideal gas, U because the temperature decreases. nct so U decreases when T decreases. The internal energ of the gas decreases

5 (d) For a constant pressure process, Q nc T. T decreases so T is negative and Q is therefore negative. Negative Q means heat leaves the gas. p EVALUATE: W nr T and Q nc T. C R, so more energ leaves as heat than is added b work done on the gas, and the internal energ of the gas decreases. p p 9..IDENTIFY: Part ab is isochoric, but bc is not an of the familiar processes. SET UP: pv nrt determines the Kelvin temperature of the gas. The work done in the process is the area under the curve in the pv diagram. Q is positive since heat goes into the gas. 3 3 L 0 m. U Q W. EXECUTE: (a) The lowest T occurs when pv has its smallest value. This is at point a, and 3 3 pv a a (0.0 atm)(.03 0 Pa/atm)(.0 L)(.0 0 m /L) Ta 78 K. nr (0.07 mol)(8.3 J/mol K) (b) a to b: V 0 so W = 0. atm.03 0 Pa. b to c: The work done b the gas is positive since the volume increases. The magnitude of the work is the area under the curve so W (0.0 atm 0.30 atm)(6.0 L.0 L) and 3 3 W (.6 Latm)( 0 m /L)(.03 0 Pa/atm) 6 J. (c) For abc, W 6 J. U Q W J 6 J 3 J. EVALUATE: J of heat energ went into the gas. 3 J of energ staed in the gas as increased internal energ and 6 J left the gas as work done b the gas on its surroundings. 9.9.IDENTIFY: For a constant pressure process, W p V, Q ncpt and U ncv T U Q W and Cp CV R For an ideal gas, pv nrt SET UP: From Table 9., C 846 J/mol K EXECUTE: (a) The pv diagram is given in Figure 9.9. (b) V W pv pv nr( T T ) (00 mol)(8 34 J/mol K)(00 0 K) 08 J (c) The work is done on the piston. (d) Since Eq. (9.3) holds for an process, U nc T (00 mol)(8 46 J/mol K)(00 0 K) 7 J V 9..IDENTIFY: For an ideal gas, U C T, and at constant pressure, pv nrt V SET UP: CV 3 R for a monatomic gas. EXECUTE: U n R T p V (400 0 Pa)(800 0 m 00 0 m ) 360 J

6 EVALUATE: 3 W nrt U 40 J 3 Q nc T n R T U 600 J 600 J of heat energ p flows into the gas. 40 J leaves as epansion work and 360 J remains in the gas as an increase in internal energ. 9.3.IDENTIFY: For an adiabatic process of an ideal gas, W ( p ) V pv and p V p V. SET UP: 40 for an ideal diatomic gas. atm 03 0 Pa and 3 3 L 0 m EXECUTE: Q U W 0 for an adiabatic process, so U W ( p ) V pv p 0 Pa 4 p p ( V / V ) ( 0 Pa)(3) 68 0 Pa ([ 68 0 Pa][0 0 3 m 3 ] [ 0 Pa][ m 3 ]) J W The internal 040 energ increases because work is done on the gas ( U 0) and Q 0 The temperature increases because the internal energ has increased. EVALUATE: In an adiabatic compression W 0 since V 0 Q 0 so U W U 0 and the temperature increases (a) IDENTIFY and SET UP: In the epansion the pressure decreases and the volume increases. The pv-diagram is sketched in Figure Figure 9.33 (b) Adiabatic means Q 0 Then U Q W gives W U ncv T ncv ( T T ) (Eq. 9.). C 47 J/mol K (Table 9.) V EXECUTE: W (040 mol)( 47 J/mol K)(00 C 00C) 4 J W positive for V 0 (epansion) (c) U W 4 J

7 EVALUATE: There is no heat energ input. The energ for doing the epansion work comes from the internal energ of the gas, which therefore decreases. For an ideal gas, when T decreases, U decreases.

8 CHAPTER IDENTIFY: This ccle involves adiabatic (ab), isobaric (bc), and isochoric (ca) processes. SET UP: ca is at constant volume, ab has Q 0, and bc is at constant pressure. For a constant pressure pv process W p V and Q ncp T. pv nrt gives nt, so Q pv. R R C p If.40 the gas is diatomic and 7 C R. For a constant volume process W 0 and Q nc T. p C V Vp pv nrt gives nt, so Q Vp. For a diatomic ideal gas CV R. R R atm.03 0 Pa. EXECUTE: (a) V b m, p. atm and b V a Vb m pava pbv b. pa pb. atm.3 atm. V 3 3 a.0 0 m (b) Heat enters the gas in process ca, since T increases..0 0 m. For an adiabatic process.4 V C V ( m 3 )(.3 atm. atm)(.03 0 Pa/atm) 470 J. Q Vp R Q 470 J. H (c) Heat leaves the gas in process bc, since T increases. C p Q pv R 7 (. atm)(.03 0 Pa/atm)( m 3 ) 373 J. (d) W QH QC 470 J ( 373 J) 747 J. QC 373 J. (e) W 747 J e %. Q 470 J H EVALUATE: We did not use the number of moles of the gas. W Q 0.4. IDENTIFY: H QC Q C 0, Q H 0 W Q e For a Carnot ccle, C TC QH QH TH SET UP: T C 300 K, T H 0 K 3 Q J. EXECUTE: (a) C QC QH T H H T K 3 (6.40 J) J. 0 K (b) W QH QC J J.73 0 J 3 W.730 J (c) e ,. Q J H EVALUATE: We can verif that e TC/ TH also gives e 43,

9 0.3.IDENTIFY: constant. SET UP: For water, Q S T for each object, where T must be in kelvins. The temperature of each object remains L J/kg. EXECUTE: (a) The heat flow into the ice is f Q ml f (0.30 kg)( J/kg).7 0 J. The heat flow occurs at T 73 K, so Q.7 0 J S 49 J/K Q is positive and S is positive. T 73 K (b) Q.7 0 J flows out of the heat source, at T 98 K Q.7 0 J S 393 J/K. T 98 K Q is negative and S is negative. (c) Stot 49 J/K ( 393 J/K) 36 J/K. EVALUATE: For the total isolated sstem, S 0 and the process is irreversible. 0.. IDENTIFY: Both the ice and the room are at a constant temperature, so Q S For the melting T phase transition, Q ml Conservation of energ requires that the quantit of heat that goes into the ice is the amount of heat that comes out of the room. SET UP: For ice, of an object, Q 0 f f 3 L J/kg When heat flows into an object, Q 0, and when heat flows out EXECUTE: (a) Irreversible because heat will not spontaneousl flow out of kg of water into a warm room to freeze the water. (b) ice room 3 3 mlf mlf (0 kg)(334 0 J/kg) (0 kg)(334 0 J/kg) S Sice Sroom T T 73 K 93 K S 0 J/K EVALUATE: This result is consistent with the answer in (a) because S 0 for irreversible processes IDENTIFY: No heat is transferred, but the entrop of the He increases because it occupies a larger volume and hence is more disordered. To calculate the entrop change, we need to find a reversible process that connects the same initial and final states. SET UP: The reversible process that connects the same initial and final states is an isothermal epansion at T 93 K, from V 0.0 L to V 3.0 L. For an isothermal epansion of an ideal gas U 0 and Q W nrt ln( V/ V).

10 EXECUTE: (a) Q (3.0 mol)(8.3 J/mol K)(93 K)ln(3.0 L/0.0 L) 9767 J. (b) The isolated sstem has S 0 so the process is irreversible. Q 9767 J S 33.3 J/K. T 93 K EVALUATE: The reverse process, where all the gas in 3.0 L goes through the hole and into the tank does not ever occur. 0.. IDENTIFY: Use U Q W and the appropriate epressions for Q, W and U for each tpe of process. pv nrt relates T to p and V values. e W Q H, where QH is the heat that enters the gas during the ccle. SET UP: For a monatomic ideal gas, 3 C R and C R p V (a) ab: The temperature changes b the same factor as the volume, and so Cp 3 Q ncpt pa( Va Vb ) ()(300 0 Pa)(0300m ) 0 J R The work pv is the same ecept for the factor of, so W J U Q W 30 J bc: The temperature now changes in proportion to the pressure change, and 3 3 Q ( p p ) V (.)( 00 0 Pa)(0800 m ) 40 0 J, and the work is zero c b b ( V 0) U Q W 40 0 J ca: The easiest wa to do this is to find the work done first; W will be the negative of area in the p-v plane bounded b the line representing the process ca and the verticals from points a and c. The area of this trapezoid is (300 0 Pa 00 0 Pa)(0800 m 000 m ) J and so the work is J U must be be U W 04 0 J (b) See above; 0 0 J (since U 0 for the ccle, anticipating part (b)), and so Q must Q W J, U 0 (c) The heat added, during process ab and ca, is is W e 0, Q 70 0 H EVALUATE: For an ccle, U 0 and QW. 0 J 04 0 J 70 0 J and the efficienc

11 Chapter 4 dv 4.4. IDENTIFY: The force and acceleration are related b Newton s second law. a, so a is dt the slope of the graph of SET UP: The graph of v versus t. v versus t consists of straight-line segments. For t 0 to t 00 s, a 4 00 m/s. For t 00 s to 6.00 s, a 0. For t 6 00 s to 0.0 s, F ma, with m 7 kg. F is the net force. a 00 m/s. EXECUTE: (a) The maimum net force occurs when the acceleration has its maimum value. F ma (7 kg)(400 m/s ) 0 N. This maimum occurs in the interval t 0 to t 00 s. (b) The net force is zero when the acceleration is zero. This is between.00 s and 6.00 s. (c) Between 6.00 s and 0.0 s, a 00 m/s, so (7 kg)( 00 m/s ) 7 N. EVALUATE: The net force is largest when the velocit is changing most rapidl IDENTIFY: The sstem is accelerating so we use Newton s second law. SET UP: The acceleration of the entire sstem is due to the 00-N force, but the acceleration of bo B is due to the force that bo A eerts on it. F ma applies to the two-bo sstem and to each bo individuall. F EXECUTE: For the two-bo sstem: 00 N 4 0 m/s. a Then for bo B, where F A is the force kg eerted on B b A, F A m a (0 kg)(40 m/s ) 0 N. B EVALUATE: The force on B is less than the force on A IDENTIFY: The surface of block B can eert both a friction force and a normal force on block A. The friction force is directed so as to oppose relative motion between blocks B and A. Gravit eerts a downward force w on block A. SET UP: The pull is a force on B not on A. EXECUTE: (a) If the table is frictionless there is a net horizontal force on the combined object of the two blocks, and block B accelerates in the direction of the pull. The friction force that B eerts on A is to the right, to tr to prevent A from slipping relative to B as B accelerates to the right. The free-bod diagram is sketched in Figure 4.8a. f is the friction force that B eerts on A and n is the normal force that B eerts on A. (b) The pull and the friction force eerted on B b the table cancel and the net force on the sstem of two blocks is zero. The blocks move with the same constant speed and B eerts no friction force on A. The free-bod diagram is sketched in Figure 4.8b.

12 EVALUATE: If in part (b) the pull force is decreased, block B will slow down, with an acceleration directed to the left. In this case the friction force on A would be to the left, to prevent relative motion between the two blocks b giving A an acceleration equal to that of B. Figure IDENTIFY: Note that in this problem the mass of the rope is given, and that it is not negligible compared to the other masses. Appl F ma to each object to relate the forces to the acceleration. (a) SET UP: The free-bod diagrams for each block and for the rope are given in Figure 4.4a. Figure 4.4a T t is the tension at the top of the rope and T b is the tension at the bottom of the rope. EXECUTE: (b) Treat the rope and the two blocks together as a single object, with mass m 600 kg 400 kg 00 kg 0 kg. Take upward, since the acceleration is upward. The free-bod diagram is given in Figure 4.4b. Figure 4.4b F ma F mg ma F mg a m 00 N (0 kg)(980 m/s ) a 33 m/s 0 kg

13 (c) Consider the forces on the top block ( m 6 00 kg), since the tension at the top of the rope ( T t ) will be one of these forces. F ma F mg Tt ma Tt F m( g a) T 00 N (600 kg)(9 80 m/s 33 m/s ) 0 N Figure 4.4c Alternativel, can consider the forces on the combined object rope plus bottom block ( m 9 00 kg): F ma Tt mg ma Tt m( g a) 900 kg(980 m/s 33 m/s ) 0 N, which checks Figure 4.4d (d) One wa to do this is to consider the forces on the top half of the rope ( m 00 kg). Let T m be the tension at the midpoint of the rope. F ma t T T mg ma m Tm Tt m( g a) 0 N 00 kg(980 m/s 33 m/s ) 93 3 N Figure 4.4e To check this answer we can alternativel consider the forces on the bottom half of the rope plus the lower block taken together as a combined object ( m 00 kg 00 kg 7 00 kg): F ma Tm mg ma Tm m( g a) 700 kg(980 m/s 33 m/s ) 93 3 N, which checks Figure 4.4f

14 EVALUATE: The tension in the rope is not constant but increases from the bottom of the rope to the top. The tension at the top of the rope must accelerate the rope as well the.00-kg block. The tension at the top of the rope is less than F; there must be a net upward force on the 6.00-kg block.

15 Chapter.4. IDENTIFY: Appl Newton s second law to the three sleds taken together as a composite object and to each individual sled. All three sleds have the same horizontal acceleration a. SET UP: The free-bod diagram for the three sleds taken as a composite object is given in Figure.4a and for each individual sled in Figure.4b d. Let m 60 0 kg. tot be to the right, in the direction of the acceleration. EXECUTE: (a) F ma for the three sleds as a composite object gives P mtota and P a m 600 kg tot N 08 m/s. (b) F ma applied to the 0.0 kg sled gives P TA m0a and TA P m0a N (0 0 kg)(08 m/s ) 04 N. F ma applied to the 30.0 kg sled gives TB m30 a (300 kg)(08 m/s ) 6 4 N. EVALUATE: If we appl F ma to the 0.0 kg sled and calculate a from T A and T B found in part (b), we get T T m0 a. A B calculated in part (a). TATB a m 00 kg 0 04 N 6 4 N 08 m/s, which agrees with the value we Figure.4

16 .34. IDENTIFY: Constant speed means zero acceleration for each block. If the block is moving, the friction force the tabletop eerts on it is kinetic friction. Appl F ma to each block. SET UP: The free-bod diagrams and choice of coordinates for each block are given b Figure.34. m 4.9 kg and m. kg. A B EXECUTE: (a) F ma with a 0 applied to block B gives m g T 0 and T.0 N. F ma with a 0 applied to block A gives T fk 0 and fk.0 N. n m g 4.0 N and fk.0 N k 0.6. n 4.0 N A (b) Now let A be block A plus the cat, so m 9.8 kg. n 90.0 N and A fk k n (0.6)(90.0 N) 0.0 N. F ma for A gives T fk m a. F ma for block B gives mbg T mba. a for A equals and a A a for B, so adding the two equations gives m g fk ( m m ) a B A A A B A B mbg fk.0 N 0.0 N.3 m/s. The acceleration is upward and block B slows down. m m 9.8 kg. kg A B EVALUATE: The equation m g fk ( m m ) a has a simple interpretation. If both blocks are B A B considered together then there are two eternal forces: that acts oppositel. The net force of B k mbg that acts to move the sstem one wa and m g f must accelerate a total mass of m m. A B f k Figure IDENTIFY: The acceleration of the car at the top and bottom is toward the center of the circle, and Newton s second law applies to it. SET UP: Two forces are acting on the car, gravit and the normal force. At point B (the top), both forces are toward the center of the circle, so Newton s second law gives mg n ma. At point A (the bottom), gravit is downward but the normal force is upward, so n mg ma. EXECUTE: Solving the equation at B for the acceleration gives A B (0 800 kg)(9 8 m/s ) 6 00 N 7 3 m/s. mg n B a m 0800 kg Solving the equation at A for the normal force gives n m( g a) (0800 kg)(98 m/s 73 m/s ) 7 N. A

17 EVALUATE: The normal force at the bottom is greater than at the top because it must balance the weight in addition to accelerate the car toward the center of its track..97. IDENTIFY: Appl F ma to the block. The cart and the block have the same acceleration. The normal force eerted b the cart on the block is perpendicular to the front of the cart, so is horizontal and to the right. The friction force on the block is directed so as to hold the block up against the downward pull of gravit. We want to calculate the minimum a required, so take static friction to have its maimum value, f s s n. SET UP: The free-bod diagram for the block is given in Figure.97. EXECUTE: F ma n ma f n ma s s s Figure.97 F ma f s mg 0 s ma mg a g / s EVALUATE: An observer on the cart sees the block pinned there, with no reason for a horizontal force on it because the block is at rest relative to the cart. Therefore, such an observer concludes that n 0 and thus fs 0, and he doesn t understand what holds the block up against the downward force of gravit. The reason for this difficult is that F ma does not appl in a coordinate frame attached to the cart. This reference frame is accelerated, and hence not inertial. The smaller block pinned against the front of the cart. s is, the larger a must be to keep the.9. IDENTIFY: Appl F ma to the circular motion of the bead. Also use Eq. (.6) to relate a rad to the period of rotation T. SET UP: The bead and hoop are sketched in Figure.9a.

18 The bead moves in a circle of radius Rr sin The normal force eerted on the bead b the hoop is radiall inward. Figure.9a The free-bod diagram for the bead is sketched in Figure.9b. EXECUTE: Figure.9b F ma ncos mg 0 n mg /cos F ma nsin ma rad Combine these two equations to eliminate n: mg sin ma cos sin cos a g rad / rad rad a v R and v R/ T, so R rsin, so a 4 rsin T rad arad 4 R/ T, where T is the time for one revolution. Use this in the above equation: sin 4 sin cos r T g This equation is satisfied b sin 0, so 0, or b 4, cos r T g which gives T g cos. 4 r (a) 4.00 rev/s implies T (/4 00) s 00 s Then (00 s) (980 m/s ) cos 4 (000 m) and 8. (b) This would mean 90. But cos90 0, so this requires T 0. So approaches 90 as the hoop rotates ver fast, but 90 is not possible.

19 (c).00 rev/s implies T 00 s The cos T g 4 r equation then sas (00 s) (980 m/s ) cos 48, 4 (000 m) which is not possible. The onl wa to have the the bottom of the hoop. F ma equations satisfied is for sin 0 This means 0; the bead sits at EVALUATE: 90 as T 0 (hoop moves faster). The largest value T can have is given b T g/(4 r ) so T r/ g 063 s. This corresponds to a rotation rate of (/0 63) rev/s 8 rev/s. For a rotation rate less than.8 rev/s, 0 is the onl solution and the bead sits at the bottom of the hoop. Part (c) is an eample of this.

20 Chapter IDENTIFY: All forces are constant and each block moves in a straight line, so W Fscos. The onl direction the sstem can move at constant speed is for the.0 N block to descend and the 0.0 N block to move to the right. SET UP: Since the.0 N block moves at constant speed, a 0 for it and the tension T in the string is T 0 N. Since the 0.0 N block moves to the right at constant speed the friction force f k on it is to the left and fk T 0 N. EXECUTE: (a) (i) 0 and W ( 0 N)(070 m)cos J. (ii) 80 and W ( 0 N)(070 m)cos J. (b) (i) 90 and W 0. (ii) 0 and W ( 0 N)(070 m)cos J. (iii) 80 and W ( 0 N)(070 m)cos J. (iv) 90 and W 0. (c) Wtot 0 for each block. EVALUATE: For each block there are two forces that do work, and for each block the two forces do work of equal magnitude and opposite sign. When the force and displacement are in opposite directions, the work done is negative IDENTIFY: From the work-energ relation, W W grav K rock. SET UP: As the rock rises, the gravitational force, F mg, does work on the rock. Since this force acts in the direction opposite to the motion and displacement, s, the work is negative. Let h be the vertical distance the rock travels. EXECUTE: (a) Appling Wgrav K K we obtain mgh mv mv. Dividing b m and solving for, v v gh. Substituting h0 m and v 0 m/s, v v (0 m/s) (980 m/s )( 0 m) 30 3 m/s (b) Solve the same work-energ relation for h. At the maimum height v 0. mgh mv mv and v (303 m/s) (00 m/s) v h 468 m. g (980 m/s ) EVALUATE: Note that the weight of 0 N was never used in the calculations because both gravitational potential and kinetic energ are proportional to mass, m. Thus an object, that attains.0 m/s at a height of.0 m, must have an initial velocit of 30.3 m/s. As the rock moves upward gravit does negative work and this reduces the kinetic energ of the rock.

21 6.34. IDENTIFY: The magnitude of the work can be found b finding the area under the graph. SET UP: The area under each triangle is / base height. F 0, so the work done is positive when increases during the displacement. EXECUTE: (a) / (8 m)(0 N) 40 J. (b) / (4m)(0 N) 0 J. (c) / ( m)(0 N) 60 J. EVALUATE: The sum of the answers to parts (a) and (b) equals the answer to part (c) IDENTIFY and SET UP: Appl Eq. (6.6) to the glider. Work is done b the spring and b gravit. Take point to be where the glider is released. In part (a) point is where the glider has traveled.80 m and K 0 There are two points shown in Figure 6.4a. In part (b) point is where the glider has traveled 0.80 m. EXECUTE: (a) Wtot K K 0. Solve for, the amount the spring is initiall compressed. Figure 6.4a Wtot Wspr W w 0 So Wspr Ww (The spring does positive work on the glider since the spring force is directed up the incline, the same as the direction of the displacement.) The directions of the displacement and of the gravit force are shown in Figure 6.4b. Figure 6.4b W ( wcos ) s ( mg cos30 0 ) s w W (00900 kg)(980 m/s )(cos30 0 )( 80 m) - 00 J w (The component of w parallel to the incline is directed down the incline, opposite to the displacement, so gravit does negative work.) Wspr W w J W spr k so Wspr (00 J) 0 06 m k 640 N/m (b) The spring was compressed onl 0.06 m so at this point in the motion the glider is no longer in contact with the spring. Points and are shown in Figure 6.4c.

22 Wtot K K K K Wtot K 0 Figure 6.4c Wtot Wspr Ww From part (a), Wspr 00 J and W ( mg cos30 0 ) s (00900 kg)(9 80 m/s )(cos30 0 )(080 m) 0 44 J w Then K Wspr W w + 00 J 044 J + 07 J EVALUATE: The kinetic energ in part (b) is positive, as it must be. In part (a), 0 since the spring force is no longer applied past this point. In computing the work done b gravit we use the full 0.80 m the glider moves IDENTIFY: Appl Wtot K K to the blocks. SET UP: If X is the distance the spring is compressed, the work done b the spring is kx. At maimum compression, the spring (and hence the block) is not moving, so the block has no kinetic energ and 0. EXECUTE: (a) The work done b the block is equal to its initial kinetic energ, and the maimum compression is found from m 00 kg and X v0 (6 00 m/s) m k 00 N/m kx mv 0 k 00 N/m (b) Solving for v 0 in terms of a known X, v0 X (0 0 m) 0 m/s m 00 kg EVALUATE: The negative work done b the spring removes the kinetic energ of the block. \

23 Chapter IDENTIFY: The normal force does no work, so onl gravit does work and Eq. (7.4) applies. SET UP: K 0. The crate s initial point is at a vertical height of d sin above the bottom of the ramp. EXECUTE: (a) 0, dsin. K Ugrav, K Ugrav, gives U grav, K. mgd sin mv and v gd sin. (b) 0, dsin. K Ugrav, K Ugrav, gives 0 K Ugrav,. 0 mv ( mgd sin ) and v gd sin, the same as in part (a). (c) The normal force is perpendicular to the displacement and does no work. EVALUATE: When we use upward. Ugrav mg we can take an point as 0 but we must take to be F k 7.3. IDENTIFY: Onl the spring does work and Eq. (7.) applies. a, where F is the force m m the spring eerts on the mass. SET UP: Let point be the initial position of the mass against the compressed spring, so K 0 and U J. Let point be where the mass leaves the spring, so Uel, 0. EXECUTE: (a) K Uel, K Uel, gives U el, K. el, mv U and U el, ( J) v 303 m/s. m 0 kg K is largest when U el is least and this is when the mass leaves the spring. The mass achieves its maimum speed of 3.03 m/s as it leaves the spring and then slides along the surface with constant speed. (b) The acceleration is greatest when the force on the mass is the greatest, and this is when the spring has its maimum compression. U U k el ( J) so 0099 m. The minus sign k 00 N/m el indicates compression. F k ma and k (00 N/m)( m) a 99 m/s. m 0 kg EVALUATE: If the end of the spring is displaced to the left when the spring is compressed, then (b) is to the right, and vice versa. a in part

24 7.9. IDENTIFY: Since the force is constant, use W Fscos. SET UP: For both displacements, the direction of the friction force is opposite to the displacement and 80. EXECUTE: (a) When the book moves to the left, the friction force is to the right, and the work is ( N)(30 m) 36 J. (b) The friction force is now to the left, and the work is again 3 6 J. (c) 7 J. (d) The net work done b friction for the round trip is not zero, and friction is not a conservative force. EVALUATE: The direction of the friction force depends on the motion of the object. For the gravit force, which is conservative, the force does not depend on the motion of the object IDENTIFY: Appl Eq. (7.6). SET UP: The sign of F indicates its direction. EXECUTE: du F 4 (48 J/ m ). d 4 3 F ( m) (4.8 J/m )( 0.80 m).46 N. The force is in the -direction. EVALUATE: F 0 when 0 and F 0 when 0, so the force is alwas directed towards the origin IDENTIFY: Appl Eq. (7.6). SET UP: du d is the slope of the U versus graph. EXECUTE: (a) Considering onl forces in the -direction, du F and so the force is zero when the d slope of the U vs graph is zero, at points b and d. (b) Point b is at a potential minimum; to move it awa from b would require an input of energ, so this point is stable. (c) Moving awa from point d involves a decrease of potential energ, hence an increase in kinetic energ, and the marble tends to move further awa, and so d is an unstable point. EVALUATE: At point b, F is negative when the marble is displaced slightl to the right and F is positive when the marble is displaced slightl to the left, the force is a restoring force, and the equilibrium is stable. At point d, a small displacement in either direction produces a force directed awa from d and the equilibrium is unstable.

25 7.49. IDENTIFY: Appl Eq. (7.7) to the motion of the stone. SET UP: K U Wother K U Let point be point A and point be point B. Take 0 at point B. EXECUTE: mg mv mv, with h 0 0 m and v 0 0 m/s v v gh m/s EVALUATE: The loss of gravitational potential energ equals the gain of kinetic energ. (b) IDENTIFY: Appl Eq. (7.8) to the motion of the stone from point B to where it comes to rest against the spring. SET UP: Use K U W other K U, with point at B and point where the spring has its maimum compression. EXECUTE: U U K 0; K mv with v m/s other f el k W W W - mgs k, with s00 m The work-energ relation gives K W other 0. mv mgs k k 0 Putting in the numerical values gives The positive root to this equation is 6 4 m. EVALUATE: Part of the initial mechanical (kinetic) energ is removed b friction work and the rest goes into the potential energ stored in the spring. (c) IDENTIFY and SET UP: Consider the forces. EXECUTE: When the spring is compressed Fel k 38 N. The maimum possible static friction force is ma f mg (080)( 0 kg)(980 m/s ) 8 N. s s 6 4 m the force it eerts on the stone is EVALUATE: The spring force is less than the maimum possible static friction force so the stone remains at rest IDENTIFY: Use Eq. (7.6) to relate F and U( ). The equilibrium is stable where U( ) is a local minimum and the equilibrium is unstable where U( ) is a local maimum. SET UP: / du d is the slope of the graph of U versus. K E U, so K is a maimum when U is a minimum. The maimum is where E U. EXECUTE: (a) The slope of the U vs. curve is negative at point A, so F is positive (Eq. (7.6)).

26 (b) The slope of the curve at point B is positive, so the force is negative. (c) The kinetic energ is a maimum when the potential energ is a minimum, and that figures to be at around 0.7 m. (d) The curve at point C looks prett close to flat, so the force is zero. (e) The object had zero kinetic energ at point A, and in order to reach a point with more potential energ than U( A ), the kinetic energ would need to be negative. Kinetic energ is never negative, so the object can never be at an point where the potential energ is larger than U( A ). On the graph, that looks to be at about. m. (f) The point of minimum potential (found in part (c)) is a stable point, as is the relative minimum near.9 m. (g) The onl potential maimum, and hence the onl point of unstable equilibrium, is at point C. EVALUATE: If E is less than U at point C, the particle is trapped in one or the other of the potential "wells" and cannot move from one allowed region of to the other.

27 Chapter IDENTIFY: The force is constant during the.0 ms interval that it acts, so J F t. J p p m( v v ). SET UP: Let J m( v v ). be to the right, so v 00 m/s. Onl the component of J is nonzero, and EXECUTE: (a) The magnitude of the impulse is direction of the impulse is the direction of the force. J (b) (i) v v m. J 0 N s. v 3 3 J Ft (0 0 N)( 000 s) 0 Ns. The 0 N s 00 m/s 6 m/s. The stone s velocit has 00 kg magnitude 6. m/s and is directed to the right. (ii) Now J 0 N s and v 0 N s 00 m/s 37 m/s. The stone s velocit has magnitude 3.7 m/s and is directed to the 00 kg right. EVALUATE: When the force and initial velocit are in the same direction the speed increases and when the are in opposite directions the speed decreases IDENTIFY and SET UP: Let the -direction be horizontal, along the direction the rock is thrown. There is no net horizontal force, so EXECUTE: 0 m v m v cos3 0 v A A A B B mv B B cos3 0 m/s m EVALUATE: A P is constant. Let object A be ou and object B be the rock. P is not conserved because there is a net eternal force in the vertical direction; as ou throw the rock the normal force eerted on ou b the ice is larger than the total weight of the sstem IDENTIFY: During the collision, momentum is conserved. After the collision, mechanical energ is conserved. SET UP: The collision occurs over a short time interval and the block moves ver little during the collision, so the spring force during the collision can be neglected. Use coordinates where is to the right. During the collision, momentum conservation gives P P. After the collision, mv = k. EXECUTE: Collision: There is no eternal horizontal force during the collision and P P, so (300 kg)(800 m/s) ( 0 kg) v (300 kg)( 00 m/s) and vblock, 00 m/s. block,

28 Motion after the collision: When the spring has been compressed the maimum amount, all the initial kinetic energ of the block has been converted into potential energ k that is stored in the compressed spring. Conservation of energ gives ( 0 kg)(00 m/s) (00 0 kg), so m. EVALUATE: We cannot sa that the momentum was converted to potential energ, because momentum and energ are different tpes of quantities. 8.. IDENTIFY: Appl Eq SET UP: m kg, m kg, m 0 00 kg. A A A B B C C EXECUTE: cm. ma mb mc cm B m m m C (0300 kg)(000 m) (0400 kg)(000 m) (000 kg)( 0300 m) m kg 0400 kg 000 kg m m m A A B B C C cm. ma mb mc cm (0300 kg)(0300 m) (0400 kg)( 0400 m) (000 kg)(0600 m) 006 m kg 0400 kg 000 kg EVALUATE: There is mass at both positive and negative and at positive and negative and therefore the center of mass is close to the origin

29 Chapter 9 d 9.3. IDENTIFY z z( t). Writing Eq. (.6) in terms of angular quantities gives dt t 0 t zdt. SET UP: d t dt n n nt and n t dt t n n EXECUTE: (a) A must have units of rad/s and B must have units of (b) 3 3 rad/s. z( t ) Bt (3 00 rad/s ) t. (i) For t 0, 0. (ii) For t 00 s, z 0 rad/s. z t 3 3 (c) t ( A Bt ) dt A( t t) B( t 3 t ). For t 0 and t 00 s, (7 rad/s)(00 s) ( 0 rad/s )(00 s) 9 0 rad. EVALUATE: Both z and z are positive and the angular speed is increasing. 9.. IDENTIFY: Appl the constant angular acceleration equations to the motion. The target variables are t and 0. SET UP: (a) z 0z z t z 0z EXECUTE: (b) 0? 0 rad/s ; 0 0 (starts from rest); 36 0 rad/s; t? z 360 rad/s 0 t 40 s 0 rad/s z z z 0 0zt zt 0 (0 rad/s )(40 s) 43 rad 0 43 rad( rev/ rad) 68 8 rev EVALUATE: We could use 0 ( z 0z ) t to calculate 0 ( rad/s)(4 0 s) 43 rad, which checks IDENTIFY and SET UP: Use constant acceleration equations to find and after each displacement. Then use Eqs. (9.4) and (9.) to find the components of the linear acceleration. EXECUTE: (a) at the start t 0 flwheel starts from rest so 0 0 atan r (0300 m)(0600 rad/s ) 0 80 m/s a rad r 0 z

30 a arad atan 0 80 m/s (b) 0 60 a tan r 0 80 m/s Calculate : 0 60 ( rad/80 ) 047 rad; 0 0; z 0 z z 0 ( ) z rad/s ;? z z z z( 0) (0600 rad/s )( 047 rad) rad/s and z. Then arad r (0300 m)( rad/s) m/s. a arad atan (0377 m/s ) (080 m/s ) 0 48 m/s (c) 0 0 a tan r 0 80 m/s Calculate : 0 0 ( rad/80 ) 094 rad; 0 0; z 0 z z 0 ( ) z rad/s ;? z z z z( 0) (0600 rad/s )(094 rad) 8 rad/s and z. Then arad r (0300 m)( 8 rad/s) 074 m/s. a arad a tan (074 m/s ) (080 m/s ) 077 m/s. EVALUATE: is constant so tan is constant. increases so a rad increases IDENTIFY: Appl conservation of energ to the sstem of stone plus pulle. v r relates the motion of the stone to the rotation of the pulle. SET UP: For a uniform solid disk, I MR. Let point be when the stone is at its initial position and point be when it has descended the desired distance. Let be upward and take 0 at the initial position of the stone, so 0 and h, where h is the distance the stone descends. EXECUTE: (a) K p p I. I p M p R ( 0 kg)(0 00 m) kg m. Kp (40 J) 34 rad/s. The stone has speed v R (000 m)(3 4 rad/s) 68 m/s. I 0000 kg m p

31 The stone has kinetic energ s mv K ( 0 kg)( 68 m/s) 39 J. K U K U gives 0 K U J 39 J mg( h). 989 J h 0673 m. (0 kg)(980 m/s ) (b) Ktot Kp Ks 9 89 J. K K p tot 40 J 4%. 989 J EVALUATE: The gravitational potential energ of the pulle doesn t change as it rotates. The tension in the wire does positive work on the pulle and negative work of the same magnitude on the stone, so no net work on the sstem IDENTIFY: Appl Eq. (9.0). SET UP: dm dv ( rl dr), where L is the thickness of the disk. M LR. EXECUTE: The analsis is identical to that of Eample 9.0, with the lower limit in the integral being zero and the upper limit being R. The result is I MR. EVALUATE: Our result agrees with Table 9.(f).

32 Chapter IDENTIFY: Use Fl rfsin for the magnitude of the torque and the right-hand rule for the direction. SET UP: In part (a), r 0 0 m and 37. EXECUTE: (a) (7 0 N)(00 m)sin37 6 N m. The torque is counterclockwise. (b) The torque is maimum when 90 and the force is perpendicular to the wrench. This maimum torque is (7 0 N)(00 m) 4 N m. EVALUATE: If the force is directed along the handle then the torque is zero. The torque increases as the angle between the force and the handle increases. 0.. IDENTIFY: Use z Iz to calculate. Use a constant angular acceleration kinematic equation to relate z, z and t. SET UP: For a solid uniform sphere and an ais through its center, I MR. Let the direction the sphere is spinning be the positive sense of rotation. The moment arm for the friction force is torque due to this force is negative. EXECUTE: (a) z (0000 N)(000 m) z 48 rad/s I (0 kg)(000 m) l 0.00 m and the (b) z 0z rad/s. z 0z z t gives z 0z rad/s t s. 48 rad/s EVALUATE: The fact that negative z causes z z is negative means its direction is opposite to the direction of spin. The z to decrease IDENTIFY: Appl F ma to each bo and z Izto the pulle. The magnitude a of the acceleration of each bo is related to the magnitude of the angular acceleration of the pulle b a R. SET UP: The free-bod diagrams for each object are shown in Figure 0.7a c. For the pulle, R 00 m and I MR. T and T are the tensions in the wire on either side of the pulle. m 0 kg, m 00 kg and M 00 kg. F is the force that the ale eerts on the pulle. For the pulle, let clockwise rotation be positive. EXECUTE: (a) F ma for the.0 kg bo gives T m a. m g T m a. z Iz F ma for the pulle gives for the.00 kg weight gives ( T T ) R MR. a R and T T Ma.

33 Adding these three equations gives m g ( m m M ) a and m 00 kg a g (9 80 m/s ) 7 m/s. Then m m M 0 kg 00 kg 00 kg T m a ( 0 kg)(7 m/s ) 3 6 N. mg T ma gives T m ( g a ) (00 kg)(980 m/s 7 m/s ) 3 4 N. The tension to the left of the pulle is 3.6 N and below the pulle it is 3.4 N. (b) a 7 m/s (c) For the pulle, F ma gives F T 3 6 N and F ma gives F Mg T (00 kg)(980 m/s ) 34 N 0 N. EVALUATE: The equation m g ( m m M ) a sas that the eternal force mg must accelerate all three objects. Figure IDENTIFY: Appl z Iz and constant angular acceleration equations to the motion of the wheel. SET UP: rev rad. rad/s 30 rev/min. EXECUTE: (a) z 0 z z Iz I. t rad/s (/)( 0 kg)(000 m) (00 rev/min) 30 rev/min z 0377 N m s (b) (600 rev/min)(. s) avt.0 rev 7 rad. 60 s/min (c) W (0377 N m)(7 rad) 9 J. rad/s (/)( kg)(000 m) (00 rev/min) 9 J. 30 rev/min (d) K I the same as in part (c). EVALUATE: The agreement between the results of parts (c) and (d) illustrates the work-energ theorem.

34 0.36. IDENTIFY: L I and I I disk I woman. SET UP: 00 rev/s 3 4 rad/s. disk I m R and disk woman woman. I m R EXECUTE: I ( kg 00 kg)(40 m) 680 kg m. 3 L (680 kg m )(34 rad/s) 8 0 kg m /s EVALUATE: The disk and the woman have similar values of I, even though the disk has twice the mass IDENTIFY: Appl conservation of angular momentum to the motion of the skater. SET UP: For a thin-walled hollow clinder I mr. For a slender rod rotating about an ais through its center, I ml. EXECUTE: i f L L so I i i I f f. i I 040 kg m (80 kg)( 8 m) 6 kg m. If 040 kg m (80 kg)(0 m) 090 kg m. Ii f i I f 090 kg m 6 kg m (0 40 rev/s) 4 rev/s. EVALUATE: K I L. increases and L is constant, so K increases. The increase in kinetic energ comes from the work done b the skater when he pulls in his hands IDENTIFY: As the bug moves outward, it increases the moment of inertia of the rod-bug sstem. The angular momentum of this sstem is conserved because no unbalanced eternal torques act on it. SET UP: The moment of inertia of the rod is I ML, and conservation of angular momentum gives 3 I I. EVALUATE: (a) I ML gives 3 3 3I 3(300 0 kg m ) M kg. L (000 m)

35 (b) L L, so I I. v 060 m/s 030 rad/s, r 000 m so 3 3 mbug (300 0 kg m )(0400 rad/s) (300 0 kg m (000 m) )(030 rad/s). m 3 (300 0 kg m )(0400 rad/s 030 rad/s) kg. (030 rad/s)(000 m) bug EVALUATE: This is a 3.00 mg bug, which is not unreasonable.

36 Chapter.3. IDENTIFY: Treat the rod and clamp as point masses. The center of gravit of the rod is at its midpoint, and we know the location of the center of gravit of the rod-clamp sstem. m SET UP: m cm. m m (.80 kg)(.00 m) (.40 kg).0 m..80 kg.40 kg EXECUTE: (.0 m)(.80 kg.40 kg) (.80 kg)(.00 m).3 m.40 kg EVALUATE: The clamp is to the right of the center of gravit of the sstem, so the center of gravit of the sstem lies between that of the rod and the clamp, which is reasonable..8. IDENTIFY: Appl the first and second conditions of equilibrium to the shelf. SET UP: The free-bod diagram for the shelf is given in Figure.8. Take the ais at the left-hand end of the shelf and let counterclockwise torque be positive. The center of gravit of the uniform shelf is at its center. EXECUTE: (a) z 0 gives wt (000 m) ws (0300 m) T (0400 m) 0. (0 N)(000 m) (00 N)(0300 m) T 00 N 0400 m 0 gives T T wt ws 0 and T 0 N. The tension in the left-hand wire is.0 N and the F tension in the right-hand wire is 0.0 N. EVALUATE: We can verif that z 0 is zero for an ais, for eample for an ais at the right-hand end of the shelf. Figure.8

37 .. IDENTIFY: The sstem of the person and diving board is at rest so the two conditions of equilibrium appl. (a) SET UP: The free-bod diagram for the diving board is given in Figure.. Take the origin of coordinates at the left-hand end of the board (point A). F is the force applied at the support point and F is the force at the end that is held down. Figure. EXECUTE: A 0 gives F ( 0 m) (00 N)(3 00 m) (80 N)( 0 m) 0 (00 N)(300 m) (80 N)(0 m) F 90 N 00 m (b) F ma F F 80 N 00 N 0 F F 80 N 00 N 90 N 80 N 00 N 40 N EVALUATE: We can check our answers b calculating the net torque about some point and checking that 0 for that point also. Net torque about the right-hand end of the board: z (40 N)(300 m) (80 N)( 0 m) (90 N)(00 m) 340 Nm 40 Nm 3840 Nm 0, which checks..3. IDENTIFY: Appl the first and second conditions of equilibrium to the strut. (a) SET UP: The free-bod diagram for the strut is given in Figure.3a. Take the origin of coordinates at the hinge (point A) and upward. Let F h and F v be the horizontal and vertical components of the force F eerted on the strut b the pivot. The tension in the vertical cable is the weight w of the suspended object. The weight w of the strut can be taken to act at the center of the strut. Let L be the length of the strut.

38 EXECUTE: F ma Fv w w 0 F v w Figure.3a Sum torques about point A. The pivot force has zero moment arm for this ais and so doesn t enter into the torque equation. 0 A TLsin30 0 w(( L/)cos30 0 ) w( Lcos300 ) 0 Tsin300 (3 w/)cos wcos30 0 T 60w sin30 0 Then F ma implies T Fh 0 and Fh 60 w. We now have the components of F so can find its magnitude and direction (Figure.3b). h v F F F Figure.3b F (60 w) ( 00 w) F 3 8w Fv 00w tan F 60w h 376 (b) SET UP: The free-bod diagram for the strut is given in Figure.3c.

39 Figure.3c The tension T has been replaced b its and components. The torque due to T equals the sum of the torques of its components, and the latter are easier to calculate. EXECUTE: A 0 ( T cos300 )( Lsin40 ) ( T sin300 )( Lcos40 ) w(( L/)cos4 0 ) w( Lcos40 ) 0 The length L divides out of the equation. The equation can also be simplified b noting that sin 4.0 cos4.0. Then T(cos30.0 sin30.0 ) 3 w/. 3w T 40w (cos300 sin300 ) F ma Fh Tcos300 0 Fh T cos300 (40 w)(cos30 0 ) 3 w F ma Fv w wt sin300 0 Fv w (40 w)sin w From Figure.3d, h v F F F F (3 w) (40 w) 39w Figure.3d Fv 40w tan F 3 w h 488

40 EVALUATE: In each case the force eerted b the pivot does not act along the strut. Consider the net torque about the upper end of the strut. If the pivot force acted along the strut, it would have zero torque about this point. The two forces acting at this point also have zero torque and there would be one nonzero torque, due to the weight of the strut. The net torque about this point would then not be zero, violating the second condition of equilibrium..4. IDENTIFY: Appl the first and second conditions of equilibrium to the beam. SET UP: The free-bod diagram for the beam is given in Figure.4. H v and H h are the vertical and horizontal components of the force eerted on the beam at the wall (b the hinge). Since the beam is uniform, its center of gravit is.00 m from each end. The angle has cos and sin The tension T has been replaced b its and components. EXECUTE: (a) H v, H h and T Tcos all produce zero torque. 0 gives z w(00 m) w (400 m) T sin (400 m) 0 and load (0 N)(00 m) (300 N)(400 m) T 6 N. (400 m)(0600) (b) F 0 gives Hh Tcos 0 and Hh (6 N)(0800) 00 N. F 0 gives Hv w wload T sin 0 and Hv w wload T sin 0 N 300 N (6 N)(0600) 7 N. EVALUATE: For an ais at the right-hand end of the beam, onl w and Hv produce torque. The torque due to w is counterclockwise so the torque due to must be upward, in agreement with our result from F 0. Hv must be clockwise. To produce a clockwise torque, Hv Figure.4