11. the Forces on the Ladder Are Shown in the Diagram Below. F1 Is the Force of the Window

Chap 12

11. The forces on the ladder are shown in the diagram below. 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  be the angle between the ladder and the ground. We use to find  = 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 be the distance from the foot of the ladder to the position of the window cleaner. Then, , and

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 and 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:

tan  = F3/F2 = 830/280 = 2.94,

so  = 71º. The force points to the left and upward, 71º above the horizontal. We note that this force is not directed along the ladder.

21. The beam is in equilibrium: the sum of the forces and the sum of the torques acting on it each vanish. As we see 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

31. The diagram below shows the forces acting on the plank. Since the roller is frictionless the force it exerts is normal to the plank and makes the angle  with the vertical. Its magnitude is designated F. W is the force of gravity; this force acts at the center of the plank, a distance L/2 from the point where the plank touches the floor. is the normal force of the floor and f is the force of friction. The distance from the foot of the plank to the wall is denoted by d. This quantity is not given directly but it can be computed using d = h/tan.

The equations of equilibrium are:

The point of contact between the plank and the roller was used as the origin for writing the torque equation.

When  = 70º the plank just begins to slip and f = sFN, where s is the coefficient of static friction. We want to use the equations of equilibrium to compute FN and f for  = 70º, then use s = f/FN to compute the coefficient of friction.

The second equation gives F = (W – FN)/cos  and this is substituted into the first to obtain

f = (W – FN) sin /cos  = (W – FN) tan .

This is substituted into the third equation and the result is solved for FN:

where we have use d = h/tan and multiplied both numerator and denominator by tan . We use the trigonometric identity 1+ tan2 = 1/cos2 and multiply both numerator and denominator by cos2 to obtain

Now we use this expression for FN in f = (W – FN) tan  to find the friction:

We substitute these expressions for f and FN into s = f/FN and obtain

Evaluating this expression for  = 70º, we obtain

35. The diagrams below show the forces on the two sides of the ladder, separated. FA and FE are the forces of the floor on the two feet, T is the tension force of the tie rod, W is the force of the man (equal to his weight), Fh is the horizontal component of the force exerted by one side of the ladder on the other, and Fv is the vertical component of that force. Note that the forces exerted by the floor are normal to the floor since the floor is frictionless. Also note that the force of the left side on the right and the force of the right side on the left are equal in magnitude and opposite in direction.

Since the ladder is in equilibrium, the vertical components of the forces on the left side of the ladder must sum to zero: Fv + FA – W = 0. The horizontal components must sum to zero: T – Fh = 0. The torques must also sum to zero. We take the origin to be at the hinge and let L be the length of a ladder side. Then

FAL cos  – W(L/4) cos  – T(L/2) sin  = 0.

Here we recognize that the man is one–fourth the length of the ladder side from the top and the tie rod is at the midpoint of the side.

The analogous equations for the right side are FE – Fv = 0, Fh – T = 0, and FEL cos  – T(L/2) sin  = 0.

There are 5 different equations:

The unknown quantities are FA, FE, Fv, Fh, and T.

(a) First we solve for T by systematically eliminating the other unknowns. The first equation gives FA = W – Fv and the fourth gives Fv = FE. We use these to substitute into the remaining three equations to obtain

The last of these gives FE = Tsin/2cos = (T/2) tan. We substitute this expression into the second equation and solve for T. The result is

To find tan, we consider the right triangle formed by the upper half of one side of the ladder, half the tie rod, and the vertical line from the hinge to the tie rod. The lower side of the triangle has a length of 0.381 m, the hypotenuse has a length of 1.22 m, and the vertical side has a length of . This means

tan  = (1.16m)/(0.381m) = 3.04.

Thus,

(b) We now solve for FA. Since Fv = FE and FE = T sin2cos, Fv = 3W/8. We substitute this into Fv + FA – W = 0 and solve for FA. We find

36. (a) The Young’s modulus is given by

(b) Since the linear range of the curve extends to about 2.9 × 108 N/m2, this is approximately the yield strength for the material.

39. (a) Let FA and FB be the forces exerted by the wires on the log and let m be the mass of the log. Since the log is in equilibrium FA + FB – mg = 0. Information given about the stretching of the wires allows us to find a relationship between FA and FB. If wire A originally had a length LA and stretches by , then , where A is the cross–sectional area of the wire and E is Young’s modulus for steel (200 × 109 N/m2). Similarly, . If is the amount by which B was originally longer than A then, since they have the same length after the log is attached, . This means

We solve for FB:

We substitute into FA + FB – mg = 0 and obtain

The cross–sectional area of a wire is . Both LA and LB may be taken to be 2.50 m without loss of significance. Thus

(b) From the condition FA + FB – mg = 0, we obtain

(c) The net torque must also vanish. We place the origin on the surface of the log at a point directly above the center of mass. The force of gravity does not exert a torque about this point. Then, the torque equation becomes FAdA – FBdB = 0, which leads to

40. (a) Since the brick is now horizontal and the cylinders were initially the same length , then both have been compressed an equal amount. Thus,

which leads to

When we combine this ratio with the equation FA + FB = W, we find FA/W = 4/5 = 0.80.

(b) This also leads to the result FB/W = 1/5 = 0.20.

Chap 13

6. The gravitational forces on m5 from the two 5.00g masses m1 and m4 cancel each other. Contributions to the net force on m5 come from the remaining two masses:

The force is directed along the diagonal between m2 and m3, towards m2. In unit-vector notation, we have

15. The acceleration due to gravity is given by ag = GM/r2, where M is the mass of Earth and r is the distance from Earth’s center. We substitute r = R + h, where R is the radius of Earth and h is the altitude, to obtain ag = GM /(R + h)2. We solve for h and obtain . According to Appendix C, R = 6.37  106 m and M = 5.98  1024 kg, so

21. Using the fact that the volume of a sphere is 4R3/3, we find the density of the sphere:

When the particle of mass m (upon which the sphere, or parts of it, are exerting a gravitational force) is at radius r (measured from the center of the sphere), then whatever mass M is at a radius less than r must contribute to the magnitude of that force (GMm/r2).

(a) At r = 1.5 m, all of Mtotal is at a smaller radius and thus all contributes to the force:

(b) At r = 0.50 m, the portion of the sphere at radius smaller than that is

Thus, the force on m has magnitude GMm/r2 = m (3.3  107 N/kg).

31. (a) The work done by you in moving the sphere of mass mB equals the change in the potential energy of the three-sphere system. The initial potential energy is

and the final potential energy is

The work done is

(b) The work done by the force of gravity is (UfUi) = 5.0  1013 J.

32. Energy conservation for this situation may be expressed as follows:

where M = 5.0  1023 kg, r1 = R = 3.0  106 m and m = 10 kg.

(a) If K1 = 5.0  107 J and r2 = 4.0  106 m, then the above equation leads to

(b) In this case, we require K2 = 0 and r2 = 8.0  106 m, and solve for K1:

33. (a) We use the principle of conservation of energy. Initially the particle is at the surface of the asteroid and has potential energy Ui = GMm/R, where M is the mass of the asteroid, R is its radius, and m is the mass of the particle being fired upward. The initial kinetic energy is . The particle just escapes if its kinetic energy is zero when it is infinitely far from the asteroid. The final potential and kinetic energies are both zero. Conservation of energy yields GMm/R + ½mv2 = 0. We replace GM/R with agR, where ag is the acceleration due to gravity at the surface. Then, the energy equation becomes agR + ½v2 = 0. We solve for v:

(b) Initially the particle is at the surface; the potential energy is Ui = GMm/R and the kinetic energy is Ki = ½mv2. Suppose the particle is a distance h above the surface when it momentarily comes to rest. The final potential energy is Uf = GMm/(R + h) and the final kinetic energy is Kf = 0. Conservation of energy yields

We replace GM with agR2 and cancel m in the energy equation to obtain

The solution for h is

(c) Initially the particle is a distance h above the surface and is at rest. Its potential energy is Ui = GMm/(R + h) and its initial kinetic energy is Ki = 0. Just before it hits the asteroid its potential energy is Uf = GMm/R. Write for the final kinetic energy. Conservation of energy yields

We substitute agR2 for GM and cancel m, obtaining

The solution for v is

45. (a) The greatest distance between the satellite and Earth’s center (the apogee distance) is Ra = (6.37  106 m + 360  103 m) = 6.73  106 m. The least distance (perigee distance) is Rp = (6.37  106 m + 180  103 m) = 6.55  106 m. Here 6.37  106 m is the radius of Earth. From Fig. 13-14, we see that the semi-major axis is

(b) The apogee and perigee distances are related to the eccentricity e by Ra = a(1 + e) and Rp = a(1 e). Add to obtain Ra + Rp = 2a and a = (Ra + Rp)/2. Subtract to obtain RaRp = 2ae. Thus,

51. In our system, we have m1 = m2 = M (the mass of our Sun, 1.99  1030 kg). Withr = 2r1 in this system (so r1 is one-half the Earth-to-Sun distance r), and v = r/T for the speed, we have

With r = 1.5  1011 m, we obtain T = 2.2  107 s. We can express this in terms of Earth-years, by setting up a ratio:

56. Although altitudes are given, it is the orbital radii which enter the equations. Thus, rA = (6370 + 6370) km = 12740 km, and rB = (19110 + 6370) km = 25480 km

(a) The ratio of potential energies is

(b) Using Eq. 13-38, the ratio of kinetic energies is

(c) From Eq. 13-40, it is clear that the satellite with the largest value of r has the smallest value of |E| (since r is in the denominator). And since the values of E are negative, then the smallest value of |E| corresponds to the largest energy E. Thus, satellite B has the largest energy.

(d) The difference is

Being careful to convert the r values to meters, we obtain E = 1.1  108 J. The mass M of Earth is found in Appendix C.

60. (a) The pellets will have the same speed v but opposite direction of motion, so the relative speed between the pellets and satellite is 2v. Replacing v with 2v in Eq. 13-38 is equivalent to multiplying it by a factor of 4. Thus,

(b) We set up the ratio of kinetic energies:

Chap 14

4. The magnitude F of the force required to pull the lid off is F = (po – pi)A, where po is the pressure outside the box, pi is the pressure inside, and A is the area of the lid. Recalling that 1N/m2 = 1 Pa, we obtain

18. (a) The force on face A of area AAdue to the water pressure alone is

Adding the contribution from the atmospheric pressure,

F0= (1.0  105 Pa)(5.0 m)2 = 2.5  106 N,

we have

(b) The force on face Bdue to water pressure alone is

Adding the contribution from the atmospheric pressure,

F0= (1.0  105 Pa)(5.0 m)2 = 2.5  106 N,

we have

19. (a) At depth y the gauge pressure of the water is p = gy, where  is the density of the water. We consider a horizontal strip of width W at depth y, with (vertical) thickness dy, across the dam. Its area is dA = W dy and the force it exerts on the dam is dF = p dA = gyW dy. The total force of the water on the dam is

(b) Again we consider the strip of water at depth y. Its moment arm for the torque it exerts about O is D – y so the torque it exerts is d = dF(D – y) = gyW (D – y)dy and the total torque of the water is

23. Eq. 14-13 combined with Eq. 5-8 and Eq. 7-21 (in absolute value) gives

mg = kx .

With A2 = 18A1 (and the other values given in the problem) we find m = 8.50 kg.

25. (a) The anchor is completely submerged in water of density w. Its effective weight is Weff = W – wgV, where W is its actual weight (mg). Thus,

(b) The mass of the anchor is m = V, where  is the density of iron (found in Table

14-1). Its weight in air is

32. (a) An object of the same density as the surrounding liquid (in which case the “object” could just be a packet of the liquid itself) is not going to accelerate up or down (and thus won’t gain any kinetic energy). Thus, the point corresponding to zero K in the graph must correspond to the case where the density of the object equals liquid. Therefore, ball = 1.5 g/cm3 (or 1500 kg/m3).

(b) Consider the liquid = 0 point (where Kgained = 1.6 J). In this case, the ball is falling through perfect vacuum, so that v2 = 2gh (see Eq. 2-16) which means that K = mv2 = 1.6 J can be used to solve for the mass. We obtain mball = 4.082 kg. The volume of the ball is then given by mball/ball = 2.72  103 m3.

35. The volume Vcav of the cavities is the difference between the volume Vcast of the casting as a whole and the volume Viron contained: Vcav = Vcast – Viron. The volume of the iron is given by Viron = W/giron, where W is the weight of the casting and iron is the density of iron. The effective weight in water (of density w) is Weff = W – gwVcast. Thus, Vcast = (W – Weff)/gw and

38. (a) Since the lead is not displacing any water (of density w), the lead’s volume is not contributing to the buoyant force Fb. If the immersed volume of wood is Vi, then

which, when floating, equals the weights of the wood and lead:

Thus,

(b) In this case, the volume Vlead = mlead/lead also contributes to Fb. Consequently,

which leads to

47. (a) The equation of continuity leads to

which gives v2 = 3.9 m/s.

(b) With h = 7.6 m and p1 = 1.7  105 Pa, Bernoulli’s equation reduces to

55. (a) Since Sample Problem 14-8 deals with a similar situation, we use the final equation (labeled “Answer”) from it:

The stream of water emerges horizontally (0 = 0° in the notation of Chapter 4), and setting y – y0 = –(H – h) in Eq. 4-22, we obtain the “time-of-flight”

Using this in Eq. 4-21, where x0 = 0 by choice of coordinate origin, we find

(b) The result of part (a) (which, when squared, reads x2 = 4h(H – h)) is a quadratic equation for h once x and H are specified. Two solutions for h are therefore mathematically possible, but are they both physically possible? For instance, are both solutions positive and less than H? We employ the quadratic formula:

which permits us to see that both roots are physically possible, so long as xH. Labeling the larger root h1 (where the plus sign is chosen) and the smaller root as h2 (where the minus sign is chosen), then we note that their sum is simply

Thus, one root is related to the other (generically labeled h' and h) by h' = H – h. Its numerical value is

(c) We wish to maximize the function f = x2 = 4h(H – h). We differentiate with respect to h and set equal to zero to obtain

or h = (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water will travel the maximum horizontal distance.

59. (a) The continuity equation yields Av = aV, and Bernoulli’s equation yields , where p = p1 – p2. The first equation gives V = (A/a)v. We use this to substitute for V in the second equation, and obtain . We solve for v. The result is

(b) We substitute values to obtain

Consequently, the flow rate is

Chap 15

5. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s.

(b) The frequency is the reciprocal of the period: f = 1/T = 1/(0.500 s) = 2.00 Hz.

(d) The angular frequency is related to the spring constant k and the mass m by . We solve for k: k = m2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m.

(e) Let xm be the amplitude. The maximum speed is vm = xm = (12.6 rad/s)(0.350 m) = 4.40 m/s.

(f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by Fm = kxm = (79.0 N/m)(0.350 m) = 27.6 N.

15. The maximum force that can be exerted by the surface must be less than sFN or else the block will not follow the surface in its motion. Here, µs is the coefficient of static friction and FN is the normal force exerted by the surface on the block. Since the block does not accelerate vertically, we know that FN = mg, where m is the mass of the block. If the block follows the table and moves in simple harmonic motion, the magnitude of the maximum force exerted on it is given by F = mam = m2xm = m(2f)2xm, where am is the magnitude of the maximum acceleration,  is the angular frequency, and f is the frequency. The relationship  = 2f was used to obtain the last form. We substitute F = m(2f)2xm and FN = mg into F µsFN to obtain m(2f)2xm µsmg. The largest amplitude for which the block does not slip is