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# Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch12

(Parte **1** de 4)

Chapter 12

1. (a) The center of mass is given by

com

(b) Similarly, we have

com

(c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have i i i i

i i i xm g x m g x m g x m g x m g x m g x m gx m g m g m g m g m g m gmg

(d) Similarly, ycog = [0 + (2.0)(m)(7.80) + (4.0)(m)(7.60) + (4.0)(m)(7.40) + (2.0)(m) (7.60) + 0]/(8.00m + 7.80m + 7.60m + 7.40m + 7.60m + 7.80m) = 1.97 m.

2. The situation is somewhat similar to that depicted for problem 10 (see the figure that accompanies that problem). By analyzing the forces at the “kink” where F is exerted, we find (since the acceleration is zero) 2T sin θ = F, where θ is the angle (taken positive) between each segment of the string and its “relaxed” position (when the two segments are collinear). Setting T = F therefore yields θ = 30º. Since α = 180º – 2θ is the angle between the two segments, then we find α = 120º.

3. The object exerts a downward force of magnitude F = 3160 N at the midpoint of the rope, causing a “kink” similar to that shown for problem 10 (see the figure that accompanies that problem). By analyzing the forces at the “kink” where F is exerted, we find (since the acceleration is zero) 2T sinθ = F, where θ is the angle (taken positive) between each segment of the string and its “relaxed” position (when the two segments are colinear). In this problem, we have

CHAPTER 12

Therefore, T = F/(2sinθ ) = 7.92 × 103 N.

4. From τ=×rF, we note that persons 1 through 4 exert torques pointing out of the page (relative to the fulcrum), and persons 5 through 8 exert torques pointing into the page.

(a) Among persons 1 through 4, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 2.

(b) Among persons 5 through 8, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 7.

5. Three forces act on the sphere: the tension force T of the rope

(acting along the rope), the force of the wall NF (acting horizontally away from the wall), and the force of gravity mg (acting downward). Since the sphere is in equilibrium they sum to zero. Let θ be the angle between the rope and the vertical. Then Newton’s second law gives

vertical component : | T cos θ – mg = 0 |

horizontal component: | FN – T sin θ = 0. |

(a) We solve the first equation for the tension: T = mg/ cos θ. We substitute cosθ=+LLr/22to obtain

9.4 N0.080 m mg L rT L

(b) We solve the second equation for the normal force: sinNFTθ=. Using sinθ=+rLr/2, we obtain

(0.080 m)N Tr mg L r r mgrF

L L r L r

6. Our notation is as follows: M = 1360 kg is the mass of the automobile; L = 3.05 m is the horizontal distance between the axles; (3.051.78) m1.27 m=−=lis the horizontal distance from the rear axle to the center of mass; F1 is the force exerted on each front wheel; and, F2 is the force exerted on each back wheel.

(a) Taking torques about the rear axle, we find

(b) Equilibrium of forces leads to1222,FFMg+=from which we obtainF2338910=×.N .

7. We take the force of the left pedestal to be F1 at x = 0, where the x axis is along the diving board. We take the force of the right pedestal to be F2 and denote its position as x = d. W is the weight of the diver, located at x = L. The following two equations result from setting the sum of forces equal to zero (with upwards positive), and the sum of torques (about x2) equal to zero:

+ − = (a) The second equation gives

L dF W d which should be rounded off to 311.210 NF=−×. Thus, 31||1.210 N.F=× (b) Since F1 is negative, indicating that this force is downward.

which should be rounded off to 321.710 NF=×. Thus, 32||1.710 N.F=× (d) The result is positive, indicating that this force is upward.

(e) The force of the diving board on the left pedestal is upward (opposite to the force of the pedestal on the diving board), so this pedestal is being stretched.

(f) The force of the diving board on the right pedestal is downward, so this pedestal is being compressed.

8. Let l115=.mand2(5.01.5) m3.5 m=−=l. We denote tension in the cable closer to the window as F1 and that in the other cable as F2. The force of gravity on the scaffold itself (of magnitude msg) is at its midpoint, l325=.mfrom either end.

(a) Taking torques about the end of the plank farthest from the window washer, we find

CHAPTER 12 w sm g m gF + = =

l l l l

(b) Equilibrium of forces leads to which (using our result from part (a)) yieldsF225310=×.N.

9. The forces on the ladder are shown in the diagram on the right. F1 is the force of the window, horizontal because the window is frictionless.

F2 and F3 are components of the force of the ground on the ladder. M is the mass of the window cleaner and m is the mass of the ladder.

The force of gravity on the man acts at a point 3.0 m up the ladder and the force of gravity on the ladder acts at the center of the ladder. Let θ be the angle between the ladder and the ground. We use

22cos/ or sin/ dLLdLθθ==− to find θ = 60º. Here L is the length of the ladder (5.0 m) and d is the distance from the wall to the foot of the ladder (2.5 m).

(a) Since the ladder is in equilibrium the sum of the torques about its foot (or any other point) vanishes. Let lbe the distance from the foot of the ladder to the position of the window cleaner. Then,

M mL gF L θθ

This force is outward, away from the wall. The force of the ladder on the window has the same magnitude but is in the opposite direction: it is approximately 280 N, inward.

(b) The sum of the horizontal forces and the sum of the vertical forces also vanish:

The first of these equations gives FF3122810==×.Nand the second gives

The magnitude of the force of the ground on the ladder is given by the square root of the sum of the squares of its components:

2 2 2 228 10 83 10 8 10( . ( | N) N) N.2 2 |

(c) The angle φ between the force and the horizontal is given by tan φ = F3/F2 = 830/280 = 2.94, so φ = 71º. The force points to the left and upward, 71º above the horizontal. We note that this force is not directed along the ladder.

10. The angle of each half of the rope, measured from the dashed line, is

Analyzing forces at the “kink” (where F is exerted) we find

2sin 2sin1.9 FT θ

1. The x axis is along the meter stick, with the origin at the zero position on the scale. The forces acting on it are shown on the diagram below. The nickels are at x = x1 = 0.120 m, and m is their total mass. The knife edge is at x = x2 = 0.455 m and exerts force F. The mass of the meter stick is M, and the

force of gravity acts at the center of the stick, x = x3 = 0.500 m. Since the meter stick is in equilibrium, the sum of the torques about x2 must vanish:

0.455m 0.120m (10.0g)=74.4 g. 0.500m 0.455m x xM m x x

12. (a) Analyzing vertical forces where string 1 and string 2 meet, we find

cos cos35

CHAPTER 12

(b) Looking at the horizontal forces at that point leads to

(c) We denote the components of T3 as Tx (rightward) and Ty (upward). Analyzing horizontal forces where string 2 and string 3 meet, we find Tx = T2 = 28 N. From the vertical forces there, we conclude Ty = wB =50 N. Therefore,

2 23 57N.x yT T T= + = (d) The angle of string 3 (measured from vertical) is

13. (a) Analyzing the horizontal forces (which add to zero) we find Fh = F3 = 5.0 N.

(b) Equilibrium of vertical forces leads to Fv = F1 + F2 = 30 N. (c) Computing torques about point O, we obtain

30 NvFdFbFad⇒

14. The forces exerted horizontally by the obstruction and vertically (upward) by the floor are applied at the bottom front corner C of the crate, as it verges on tipping. The center of the crate, which is where we locate the gravity force of magnitude mg = 500 N, is a horizontal distance l=0375.mfrom C. The applied force of magnitude F = 350 N is a vertical distance h from C. Taking torques about C, we obtain

350N mgh F

15. Setting up equilibrium of torques leads to a simple “level principle” ratio:

16. With pivot at the left end, Eq. 12-9 leads to – ms g – Mgx + TR L = 0 where ms is the scaffold’s mass (50 kg) and M is the total mass of the paint cans (75 kg). The variable x indicates the center of mass of the paint can collection (as measured from the left end), and TR is the tension in the right cable (722 N). Thus we obtain x = 0.702 m.

17. The (vertical) forces at points A, B and P are FA, FB and FP, respectively. We note that FP = W and is upward. Equilibrium of forces and torques (about point B) lead to

(a) From the second equation, we find

(b) The direction is upward since FA > 0. (c) Using this result in the first equation above, we obtain

(d) FB points downward, as indicated by the minus sign.

18. Our system consists of the lower arm holding a bowling ball. As shown in the free-body diagram, the forces on the lower arm consist of T from the biceps muscle, F from the bone of the upper arm, and the gravitational forces, mg and Mg . Since the system is in static equilibrium, the net force acting on the system is zero:

In addition, the net torque about O must also vanish:

(a) From the torque equation, we find the force on the lower arms by the biceps muscle to be

CHAPTER 12

19. (a) With the pivot at the hinge, Eq. 12-9 gives TLcosθ – mg = 0 . This leads to 78.θ=° Then the geometric relation tanθ = L/D gives D = 0.64 m.

(b) A higher (steeper) slope for the cable results in a smaller tension. Thus, making D greater than the value of part (a) should prevent rupture.

20. With pivot at the left end of the lower scaffold, Eq. 12-9 leads to

– m2 g – mgd + TR L2 = 0 where m2 is the lower scaffold’s mass (30 kg) and L2 is the lower scaffold’s length (2.0 m). The mass of the package (m = 20 kg) is a distance d = 0.50 m from the pivot, and TR is the tension in the rope connecting the right end of the lower scaffold to the larger scaffold above it. This equation yields TR = 196 N. Then Eq. 12-8 determines TL (the tension in the cable connecting the right end of the lower scaffold to the larger scaffold above it): TL = 294 N. Next, we analyze the larger scaffold (of length L1 = L2 + 2d and mass m1, given in the problem statement) placing our pivot at its left end and using Eq. 12-9:

– m1 g – TL d – TR (L1 – d) + T L1 = 0 .

This yields T = 457 N.

21. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force on the wheel, and the only forces acting are the force F applied horizontally at the axle, the force of gravity mg acting vertically at the center of the wheel, and the force of the step corner, shown as the two components fh and fv. If the minimum force is applied the wheel does not accelerate, so both the total force and the total torque acting on it are zero.

We calculate the torque around the step corner. The second diagram indicates that the distance from the line of F to the corner is r – h, where r is the radius of the wheel and h is the height of the step.

The distance from the line of mg to the corner is rrhrhh2222+−=−bg. Thus,

F r h mg rh h− − − =b g 2 02 . The solution for F is

rh hF mg r h

= 2. As shown in the free-body diagram, the forces on the climber consist of T from the rope, normal forceNF on her feet, upward static frictional force sf and downward gravitational force mg . Since the climber is in static equilibrium, the net force acting on her is zero. Applying Newton’s second law to the vertical and horizontal directions, we have

net, net,

0 sin y s

F F T F T f mgφφ = = −

In addition, the net torque about O (contact point between her feet and the wall) must also vanish:

From the torque equation, we obtain sin/sin(180).Tmgθθφ=°−− Substituting the expression into the force equations, and noting that ssNfFµ=, we find the coefficient of static friction to be cos sin cos /sin(180 ) sin sin sin /sin(180 )

1 sin cos /sin(180 ) sin sin /sin(180 ) s s f mg T mg mg

F T mg φ θ φ θ φµ φ θ φ θ φ

With 40θ=° and 30φ=°, the result is

sin sin /sin(180 ) sin 40 sin30 /sin(180 40 30 )s θ φ θ φµ θ φ θ φ

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23. (a) All forces are vertical and all distances are measured along an axis inclined at θ =

30º. Thus, any trigonometric factor cancels out and the application of torques about the contact point (referred to in the problem) leads to

tripcep

(b) The direction is upward since tricep0F> (c) Equilibrium of forces (with upwards positive) leads to

and thus to3

(d) The minus sign implies thathumerFpoints downward.

24. As shown in the free-body diagram, the forces on the climber consist of the normal forces 1NF on his hands from the ground and

2NF on his feet from the wall, static frictional force sf and downward gravitational force mg. Since the climber is in static equilibrium, the net force acting on him is zero. Applying Newton’s second law to the vertical and horizontal directions, we have

x N s

y N

F F f F F mg = = −

In addition, the net torque about O (contact point between his feet and the wall) must also vanish:

net 20 cos sinNO mgd F Lτ θ θ= = −∑ .

The torque equation gives 2cos/sincot/NFmgdLmgdLθθθ==. On the other hand, from the force equation we have 2NsFf= and 1.NFmg= These expressions can be combined to yield

On the other hand, the frictional force can also be written as 1ssNfFµ=, where sµ is the coefficient of static friction between his feet and the ground. From the above equation and the values given in the problem statement, we find sµ to be d a d

25. The beam is in equilibrium: the sum of the forces and the sum of the torques acting on it each vanish. As shown in the figure, the beam makes an angle of 60º with the vertical and the wire makes an angle of 30º with the vertical.

(a) We calculate the torques around the hinge. Their sum is

TL sin 30º – W(L/2) sin 60º = 0.

Here W is the force of gravity acting at the center of the beam, and T is the tension force of the wire. We solve for the tension:

(b) Let Fh be the horizontal component of the force exerted by the hinge and take it to be positive if the force is outward from the wall. Then, the vanishing of the horizontal component of the net force on the beam yields Fh – T sin 30º = 0 or

(c) Let Fv be the vertical component of the force exerted by the hinge and take it to be positive if it is upward. Then, the vanishing of the vertical component of the net force on the beam yields Fv + T cos 30º – W = 0 or

26. (a) The problem asks for the person’s pull (his force exerted on the rock) but since we are examining forces and torques on the person, we solve for the reaction force 1NF (exerted leftward on the hands by the rock). At that point, there is also an upward force of static friction on his hands f1 which we will take to be at its maximum value 11NFµ. We note that equilibrium of horizontal forces requires 12NNFF=(the force exerted leftward on his feet); on this feet there is also an upward static friction force of magnitude µ2FN2. Equilibrium of vertical forces gives mgf f mg F µ µ − ⇒ ×

(b) Computing torques about the point where his feet come in contact with the rock, we find

CHAPTER 12 mg d w F w mg d w f w F h h

(c) Both intuitively and mathematically (since both coefficients are in the denominator) we see from part (a) that 1NF would increase in such a case.

(d) As for part (b), it helps to plug part (a) into part (b) and simplify:

h d w d= + +2 1a fµ µ from which it becomes apparent that h should decrease if the coefficients decrease.

(Parte **1** de 4)