Download 9. Find the mass m of the counterweight needed to balance a truck

9. Find the mass m of the counterweight needed to balance a truck with mass M = 1
500 kg on an incline of θ = 45º (Fig. P12.9). Assume both pulleys are friction-less
and massless.
Figure P12.9
SOL. The second condition for equilibrium at the pulley is
Στ = 0 = mg(3r)−Tr
and from equilibrium at the truck, we obtain
2T  Mg sin 45.0  0
Mg sin 45.0
2
(1500 kg) g sin 45.0

2
 530 g N
T
solving for the mass of the counterweight from [1] and substituting gives
m
T 530 g

 177 kg
3g
3g
ANS. FIG. P12.9
12. A vaulter holds a 29.4 N pole in equilibrium by exerting an upward force
U
with her leading hand and a downward force D with her trailing hand as shown
in Figure P12.12. Point C is the center gravity of the pole. What are the
magnitudes of (a) U and (b) D ?
Figure P12.12
SOL.
(a) To find U, measure distances and forces from point A. Then, balancing
torques,
(0.750 m)U = (29.4 N)(2.25) U = 88.2 N
(b)
To find D, measure distances and forces from point B. Then,
balancing torques,
(0.750 m)D = (1.50 m)(29.4 N) D = 58.8 N
Also, notice that U = D+ Fg, so ΣFy = 0.
17. Figure P12.17 shows a claw hammer being used to pull a nail out of a horizontal
board. The mass of the hammer is 1.00 kg. A force of 150 N is exerted
horizontally as shown, and the nail does not yet move relative to the board. Find
(a) the force exerted by the hammer claws on the nail and (b) the force exerted by
the surface on the point of contact with the hammer head. Assume the force the
hammer exerts on the nail is parallel to the nail.
Figure P12.17
SOL. (a)
In Figure P12.17, let the “Single point of contact” be point P, the force the
nail exerts on the hammer claws be R, the mass of the hammer
(1.00 kg) be M, and the normal force exerted on the hammer at
point P be n, while the horizontal static friction exerted by the
surface on the hammer at P be f.
Taking moments about P,
(R sin 30.0°)0 + (R cos 30.0°)(5.00 cm) + Mg(0)
− (150 N)(30.0 cm) = 0
R = 1 039.2 N = 1.04 kN
The force exerted by the hammer on the nail is equal in magnitude
and opposite in direction:
1.04 kN at 60° upward and to the right
(b)
From the first condition for equilibrium,
ΣFx = f – R sin30.0° + 150 N = 0→ f = 370 N
ΣFy = n −Mg – R cos 30.0° = 0
→n = (1.00 kg)(9.80 m/s2)+(1 040 N)cos 30.0° = 910 N


Fsurface  370ˆi +910ˆj N
35. A 2.00-m-long cylindrical steel wire with a cross-sectional diameter of 4.00 mm
is placed over a light, frictionless pulley. An object of mass m1 = 5.00 kg is hung
from one end of the wire and an object of mass m2 = 3.00 kg from the other end as
shown in Figure P12.35. The objects are released and allowed to move freely.
Compared with its length before the objects were attached, by how much has the
wire stretched while the objects are in motion?
SOL. Let the 3.00-kg mass be mass #1, with the 5.00-kg mass, mass # 2. Applying
Newton’s second law to each mass gives
m1a = T − m1g
and
[1]
m2a = m2g – T
[2]
where T is the tension in the wire.
Solving equation [1] for the acceleration gives a 
T
g
m1
m2
T  m2 g  m2 g  T
m2
and substituting this into equation [2] yields
Solving for the tension T gives
2m m g 2  3.00 kg  5.00 kg   9.80 m/s
T 1 2 
m2  m1
8.00 kg
2
  36.8 N
From the definition of Young’s modulus,  
L 
the wire is:
FLi
, the elongation of
A  L 
 36.8 N  2.00 m 
TLi

A  2.00 1011 N/m 2    2.00 103 m 2
 0.029 2 mm
39. In exercise physiology studies, it is sometimes important to determine the
location of a person’s center of mass. This determination can be done with the
arrangement shown in Figure P12.39. A light plank rests on two scales, which
read Fg1 = 380 N and Fg2 = 320 N. A distance of 1.65 m separates the scales. How
far from the woman’s feet is her center of mass?
Figure P12.39
SOL.
ΣFy = 0: + 380 N − Fg + 320 N = 0
Fg = 700 N
Take torques about her feet:
Στ = 0:
−380 N(1.65 m) + (700 N)x + (320 N)0 = 0
x = 0.896 m
ANS. FIG. P12.39
41. The arm in Figure P12.41 weighs 41.5 N. The gravitational force on the arm acts
through point A. Determine the magnitudes of the tension force F t in the deltoid
muscle and the force F s exerted by the shoulder on the humerus (upper-arm
bone) to hold the arm in the position shown.
Figure P12.41
SOL.
We reproduce the forces in ANS. FIG. P12.41.
ANS. FIG. P12.41
Requiring that Στ = 0, using the shoulder joint at point O as a pivot, gives
Στ = Ft (sin12.0°)(0.080 0 m)−(41.5 N)(0.290 m) = 0
or
Ft = 724 N
Then ΣFy = 0 ⇒ − Fsy +(724 N)sin12.0° − 41.5 N = 0
yielding Fsy = 109 N
ΣFx = 0 then gives Fsx − (724 N)cos 12.0° = 0, or Fsx = 708 N
Therefore, Fs  Fsx2  Fsy2 
 708 N   109 N 
2
2
 716 N
42. When a person stands on tiptoe on one foot (a strenuous position), the position of
the foot is as shown in Figure P12.42a. The total gravitational force F g on the
body is supported by the normal force n exerted by the floor on the toes of one
foot. A mechanical mode l of the situation is shown in Figure P12.42b, where T
is the force exerted on the foot by the Achilles tendon and R is the force exerted
on the foot by the tibia. Find the values of T, R, and θ when Fg = 700 N.
Figure P12.42
SOL. In the free-body diagram of the foot given at the right, note that the force R
(exerted on the foot by the tibia) has been replaced by its horizontal and
vertical components. Employing both conditions of equilibrium (using
point O as the pivot point) gives the following three equations:
ANS. FIG. P12.42
ΣFx = 0  Rsin15.0° −T sinθ = 0
or
R
T sin 
sin15.0
[1]
Fy  0  700 N  R cos15.0  T cos   0
[2]
ΣτO = 0  −(700 N)[(18.0 cm)cos θ]+T(25.0 cm− 18.0 cm) = 0
or
T = (1 800 N)cos θ
[3]
Substituting equation [3] into equation [1] gives
 1800 N 
R
 sin  cos 
 sin15.0 
[4]
Substituting equations [3] and [4] into equation [2] yields
 1800 N 
2

 sin  cos   1800 N  cos   700 N
 tan15.0 
which reduces to: sinθ cosθ = (tan 15.0°)cos2θ + 0.104 2
Squaring this result and using the identity sin2θ = 1 − cos2θ gives
[tan2 (15.0°) + 1]cos4θ
+ [(2 tan15.0°)(0.104 2) – 1]cos2θ + (0.104 2)2 = 0
In this last result, let u = cos2θ and evaluate the constants to obtain the
quadratic equation:
(1.071 8)u2 − (0.944 2)u + 0.010 9 = 0
The quadratic formula yields the solutions u = 0.869 3 and u = 0.0117.
Thus,   cos1


0.8693  21.2
or   cos1


0.0117  83.8
We ignore the second solution since it is physically impossible for the
human foot to stand with the sole inclined at 83.8° to the floor. We are the
left with θ = 21.2°.
Equation [3] then yieldsT = (1 800 N) cos 21.2° = 1.68 kN
1.68 10 N  sin 21.2
R
3
and equation [1] gives
sin15.0
 2.34 kN
52. The large quadriceps muscle in the upper leg terminates at its lower end in a
tendon attached to the upper end of the tibia (Fig. P12.52a, page 386). The forces
on the lower leg when the leg is extended are modeled as in Figure P12.52b,
where T is the force in the tendon, F g ,leg is t he gravitational force acting on the
lower leg, and F g ,foot is the gravitational force acting on the foot. Find T when the
tendon is at an angle of  = 25.0º with the tibia, assuming Fg,leg = 30.0 N, Fg,foot =
12.5 N, and the leg is extended at an angle θ = 40.0º with respect to the vertical.
Also assume the center of gravity of the tibia is at its geometric center and the
tendon attaches to the lower leg at a position one-fifth of the way down the leg.
Figure P12.52
SOL. First, we resolve all forces into components parallel to and perpendicular to the
tibia, as shown. Note that θ = 40.0° and
wy = (30.0 N)sin 40.0° = 19.3 N
Fy = (12.5 N)sin 40.0° = 8.03 N
and
Ty = T sin 25.0°
Using Στ = 0 for an axis perpendicular to the page and through the upper
end of the tibia gives
ANS. FIG. P12.52
T sin 25.0  d5  19.3 N  d2  8.03 N  d  0
or T = 209 N
59. Two racquetballs, each having a mass of 170 g, are placed in a glass jar as shown
in Figure P12.59. Their centers lie on a straight line that makes a 45° angle with
the horizontal. (a) Assume the walls are frictionless and determine P1, P2, andP3.
(b) Determine the magnitude of the force exerted by the left ball on the right ball.
Figure P12.59
SOL. (a)
Take both balls together. Their weight is 2mg = 3.33 N and their CG is at
their contact point.
ΣFx = 0:+ P3 − P1 = 0→P3 = P1
ΣFy = 0:+ P2 − 2mg = 0→P2 = 2mg = 3.33 N
For torque about the contact point (CP) between the balls:
Στ CP = 0: P1 (Rcos 45.0°)− P2 (Rcos 45.0°)+ P3 (Rcos 45.0°)
−mg(R cos 45.0°)+mg(R cos 45.0°) = 0
→P1 − P2 + P3 = 0→P1 + P3 = P2
Substituting P3 = P1, we find
2P1 = P2 = 2mg→ P1 = mg
Therefore, P1 = P3 = 1.67 N
ANS. FIG. P12.59(a)
(b)
Take the upper ball. The lines of action of its weight, of P1, and of
the normal force n exerted by the lower ball all go through its
center, so for rotational equilibrium there can be no frictional force.
ΣFx = 0: n cos 45.0° − P1 = 0
n
1.67 N
 2.36 N
cos 45.0
ΣFy = 0: nsin 45.0° − 1.67 N = 0 gives the same result.
ANS. FIG. P12.59(b)