Physics 101 - Homework # 8 Solutions

• Problem 11-3
• Problem 11-9
• Problem 11-18
• Problem 11-24
• Problem 11-31
• Problem 11-39
• Problem 12-6
• Problem 12-10
• Problem 12-19
• Problem 12-26
• Problem 12-53
• Problem 12-56

Chapter 11

Problem 3)

We choose the coordinate system shown, with positive torques chosen to be clockwise. Let's write = I about the point A from the force diagram for the board:

_A = MgD - F_N2L = 0

which gives

F_N2 = Mg(D/L) = (24 kg)(9.8 m/s²)(0.9 m)/(2.2 m)

= 96.2 N

We write   F_y = ma_y   from the force diagram for the board and worker:

F_N1 + F_N2 - Mg = 0

which gives

F_N1 = Mg - F_N2

= (24 kg)(9.8 m/s²) - 96 N = 139 N

The forces on the workmen are the reactions to these normal forces:

139 N down     and   96.2 N down

Problem 9)

We choose the coordinate system shown, with positive torques clockwise. As the plank is moved out from the roof the effective normal forces acts at a point closer to the edge. When the normal force reaches the edge, the plank is on the verge of tipping. We write   = I   about the edge A from the force diagram for the plank, the load and the concrete:

_A = M₁gx - M₂g(L-x) =0

which gives

x = M₂L/(M₁ + M₂) = (15 kg)(2.4 m)/(30 kg + 15 kg)

= 0.80 m

Problem 18)

In the coordinate system shown, the positive torques are chosen counterclockwise. The angle between the ladder and the ground is found from

sin = h/L = (8 m)/(8.5 m)

which gives

= 70.2°

We write = I about the point A from the force diagram for the ladder and man:

_A = mg(½L)cos + Mgdcos - F_NtopLsin = 0

which gives

F_Ntop = (½mL + Md)gcos/(L sin)

= [½(26 kg)(8.5 m) + (75 kg)(7 m)](9.8 m/s²)(cos70.2°)/(8.5 m)(sin70.2°)

= 2.63 x 10² N

We write   F_x = ma_x   from the force diagram for the ladder and man:

F_Ntop - f_bottom = 0

which gives

f_bottom = F_Ntop = 2.63 x 10² N

From   F_y = ma_y   the force diagram for the ladder and man gives

F_Nbottom - (m + M)g = 0

yielding

F_Nbottom = 9.90 x 10² N

The force on the ladder from the man is the reaction to the force on the man by the ladder, which must have the magnitude of the man's weight. Thus the forces on the ladder are

F_Nbottom = 9.90 x 10² N upwards

f_bottom = 2.63 x 10² N toward the wall

F_Ntop = 2.63 x 10² N away from the wall

F_man = 7.35 x 10² N down

W_ladder = 2.55 x 10² N down

Problem 24)

We choose the coordinate system shown, with the positive torques chosen to be clockwise. If we take torques about point A, then the forces F_wallH and F_wallV will cause no torque, so the equation will be simple. From the force diagram for the flagpole, we write = I about the point A as:

_A = Mg(L/2) - (Tsin)L = 0

which gives

T = (1/2)Mg/(sin)

= (1/2)(6 kg)(9.8 m/s²)/(sin 25°)

T = 70 N

We write F_x = ma_x using the force diagram to get

F_wallH - Tcos = 0

which gives

F_wallH = (70 N)cos25° = 63 N

Now, writing F_y = ma_y we have

F_wallV + Tsin - Mg = 0

which gives

F_wallV = (6 kg)(9.8 m/s²) - (70 N)sin25° = 29 N

Combining these two components we get

F_wall = (63 N)i + (29 N)j

Problem 31)

We choose the coordinate system shown, with positive torques clockwise. Let's write   = I   about the point B from the force diagram for the beam:

_B = mg(½Lcos) + Mg(L-d)sin - Tcos(Lsin) - Tsin(Lcos) = 0

where we have broken the tension T into its x-component (-Tcos) and its y-component (-Tsin), and seen that the lever arm for the x-component is Lsin and that for the y-component is Lcos. Solving this for M gives

M = {[TL(cossin -sincos)/g] - ½mLcos}/[(L-d)sin]

Since the maximum M corresponds to the maximum T, we have

M_max = {[(10 x 10³N)(3 m)(cos30°sin45° - sin30°cos45°)/(9.8 m/s²)] - ½(100 kg)(3 m)cos45°}/[(3 m - 0.25 m)sin45°]

= 353 kg

Problem 39)

We choose the coordinate system shown. Choose counterclockwise for the direction of positive torques.

a) Write = I about the point A, using the force diagram for the beam:

_A = Mg(L/2)cos - TLsin = 0

which gives

T = (1/2)Mgcot

b) When the cable snaps, the tension is then zero, so we have (remembering the relation for the moment of inertia of a thin beam for rotations about one end)

_A = I

Mg(L/2)cos = (1/3)ML²

which yields

= (3gcos)/2L

c) The angular acceleration is not constant here. One can either integrate the angular acceleration as a function of , or, it is easier to use energy considerations. Since the force at the pivot does no work on the beam, only gravity does work, and we have

W = K + U

0 = [(1/2)I_A²-0] + Mg[0 - (L/2)sin]

Now, substituting in the moment of inertia I_A = (1/3)ML², we get

= [(3gsin)/L]^1/2

Chapter 12

Problem 6)

The gravitational force between the proton and the electron is

F = GMm/r²

= (6.67 x 10^-11 Nm²/kg²)(1.67 x 10^-27 kg)(9.1 x 10^-31 kg)/(0.6 x 10^-10 m)² = 2.8 x 10^-47 N

we will see in Physics 102 that this is miniscule compared to the electrical forces between the proton and electron.

Problem 10)

The radius of the moon is R_moon = 1.74x10⁶ m (see problem 12-8), so the radius of the satellite's orbit is R = R_moon + 3.0 x 10⁵ m = 2.04 x 10⁶ m. We can find the period using Kepler's third law, given that the mass of the moon is 7.35 x 10²² kg:

T² = 4²R³/GM

= 4²(2.04 x 10⁶ m)³/ (6.67 x 10^-11 Nm²/kg²)(7.35 x 10²² kg)

which gives

T = 8.26 x 10³ s = 2.30 hours

Problem 19)

Because the escape speed is the speed at the surface necessary to get far away with no final speed, we use energy conservation, where we have chosen the zero for potential energy to be at infinity:

K_i + U_i = K_f + U_f

(1/2)mv_esc² - (GM_sm/R_s) = 0 + 0

v_esc = (2GM/R)^1/2

where M and R are the mass and radius, respectively, of the object from which we are escaping.

a) use M = M_moon= 7.35 x 10²² kg, R = R_moon = 1.74 x 10⁶ m; we have

v_esc = [2(6.67 x 10^-11 Nm²/kg²)(7.35 x 10²² kg)/(1.74 x 10⁶ m)^1/2

= 2.37 x 10³ m/s

b) use M = M_mars= 6.42 x 10²³ kg, R = R_mars = 3.40 x 10⁶ m; we have

v_esc = [2(6.67 x 10^-11 Nm²/kg²)(6.42 x 10²³ kg)/(3.40 x 10⁶ m)^1/2

= 5.02 x 10³ m/s

c) use M = M_jupiter= 1.90 x 10²⁷ kg, R = R_jupiter = 7.14 x 10⁷ m; we have

v_esc = [2(6.67 x 10^-11 Nm²/kg²)(1.90 x 10²⁷ kg)/(7.14 x 10⁷ m)^1/2

= 5.96 x 10⁴ m/s

Problem 26)

We use conservation of energy, with the reference level for potential energy at infinity:

K_i + U_i = K_f + U_f

½mv₀² - (GM_mm/R_m) = 0 + 0

which gives

½mv₀² = (GM_mm/R_m) = (6.67 x 10^-11 Nm²/kg²)(7.35 x 10²² kg)(1200 kg)/(1.74 x 10⁶ m)

= 3.38 x 10⁹ J

Problem 53)

We want the magniture of the centripetal acceleration to have the magnitude of terrestial gravity:

a = r² = g

9.8 m/s² = (60 m)²

which gives

= 0.40 rad/s

Problem 56)

a) g = GM/R² = (6.67 x 10^-11 Nm²/kg²)(4 x 10³⁰ kg)/(10 x 10³ m)²

= 2.67 x 10¹² m/s²

(a little bit more than 9.8 m/s²!)

b) The gravitational force is   F = GMm/r²

Because a change in r of 1 cm is much smaller than the radius of 10 km, we approximate the changes by differentials:

F/r = dF/dr = -2GMm/r³ = - 2mg/r

Thus

F = - 2mgr/r = -2(1 x 10^-3 kg)(2.67 x 10¹² m/s²)(1 x 10^-2)/(10 x 10³)

= - 5.34 x 10³ N

as mentioned in lecture, these tidal forces are large - unless that dumbbell is very strong, it will be torn apart by this tidal force!

An alternative (uglier) solution is to directly calculate the force at 10,000.00 m and subtract from it the force at 10,000.01 m - this involves taking the difference of two large numbers, and one needs to be careful that your calculator stores enough significant figures...