Equilibrium 12
Static and Elasticity
CHAPTER OUTLINE
12.1 12.2 12.3 12.4 The Conditions for Equilibrium More on the Center of Gravity Examples of Rigid Objects in Static Equilibrium Elastic Properties of Solids
ANSWERS TO QUESTIONS
Q12.1 When you bend over, your center of gravity shifts forward. Once your CG is no longer over your feet, gravity contributes to a nonzero net torque on your body and you begin to rotate. Yes, it can. Consider an object on a spring oscillating back and forth. In the center of the motion both the sum of the torques and the sum of the forces acting on the object are (separately) zero. Again, a meteoroid flying freely through interstellar space feels essentially no forces and keeps moving with constant velocity. No--one condition for equilibrium is that
Q12.2
Q12.3
F = 0 . For this to
be true with only a single force acting on an object, that force would have to be of zero magnitude; so really no forces act on that object. Q12.4 (a) Consider pushing up with one hand on one side of a steering wheel and pulling down equally hard with the other hand on the other side. A pair of equal-magnitude oppositelydirected forces applied at different points is called a couple. An object in free fall has a non-zero net force acting on it, but a net torque of zero about its center of mass.
(b) Q12.5 Q12.6
No. If the torques are all in the same direction, then the net torque cannot be zero. (a) (b) Yes, provided that its angular momentum is constant. Yes, provided that its linear momentum is constant.
Q12.7 Q12.8
A V-shaped boomerang, a barstool, an empty coffee cup, a satellite dish, and a curving plastic slide at the edge of a swimming pool each have a center of mass that is not within the bulk of the object. Suspend the plywood from the nail, and hang the plumb bob from the nail. Trace on the plywood along the string of the plumb bob. Now suspend the plywood with the nail through a different point on the plywood, not along the first line you drew. Again hang the plumb bob from the nail and trace along the string. The center of gravity is located halfway through the thickness of the plywood under the intersection of the two lines you drew.
349
350 Q12.9 Q12.10
Static Equilibrium and Elasticity
The center of gravity must be directly over the point where the chair leg contacts the floor. That way, no torque is applied to the chair by gravity. The equilibrium is unstable. She can be correct. If the dog stands on a relatively thick scale, the dog's legs on the ground might support more of its weight than its legs on the scale. She can check for and if necessary correct for this error by having the dog stand like a bridge with two legs on the scale and two on a book of equal thickness--a physics textbook is a good choice. If their base areas are equal, the tall crate will topple first. Its center of gravity is higher off the incline than that of the shorter crate. The taller crate can be rotated only through a smaller angle before its center of gravity is no longer over its base. The free body diagram demonstrates that it is necessary to have friction on the ground to counterbalance the normal force of the wall and to keep the base of the ladder from sliding. Interestingly enough, if there is friction on the floor and on the wall, it is not possible to determine whether the ladder will slip from the equilibrium conditions alone.
Q12.11
Q12.12
FIG. Q12.12 Q12.13 When you lift a load with your back, your back muscles must supply the torque not only to rotate your upper body to a vertical position, but also to lift the load. Since the distance from the pivot--your hips--to the load--essentially your shoulders--is great, the force required to supply the lifting torque is very large. When lifting from your knees, your back muscles need only keep your back straight. The force required to do that is much smaller than when lifting with your back, as the torque required is small, because the moment arm of the load is small--the line of action of the load passes close to your hips. When you lift from your knees, your much stronger leg and hip muscles do the work. Shear deformation. The vertical columns experience simple compression due to gravity acting upon their mass. The horizontal slabs, however, suffer significant shear stress due to gravity. The bottom surface of a sagging lintel is under tension. Stone is much stronger under compression than under tension, so horizontal slabs are more likely to fail.
Q12.14 Q12.15
Chapter 12
351
SOLUTIONS TO PROBLEMS
Section 12.1 P12.1 The Conditions for Equilibrium
F 0.600 m
To hold the bat in equilibrium, the player must exert both a force and a torque on the bat to make
Fx = Fy = 0
upward force of F = 10.0 N
and
= 0
O
Fy = 0 F - 10.0 N = 0 , or the player must exert a net
To satisfy the second condition of equilibrium, the player must exert an applied torque a to make
10.0 N
= a - a0.600 mfa10.0 N f = 0 . Thus, the required torque is
a = +6.00 N m or 6.00 N m counterclockwise
FIG. P12.1
P12.2
Use distances, angles, and forces as shown. The conditions of equilibrium are:
Fy
Fy = 0 Fx = 0 = 0
Fy + R y - Fg = 0 Fx - R x = 0 Fy cos - Fg
l
Fx
FG IJ cos - F H 2K
Ry
x
sin = 0
Rx O
Fg
FIG. P12.2 P12.3 Take torques about P. m1 g mb g d m1 2 CG x nO nP P m2 m2 g
p = -n0 M 2 + d P + m1 g M 2 + d P + m b gd - m 2 gx = 0 N Q N Q
We want to find x for which n 0 = 0 . x=
L
O
L
O
bm g + m ggd + m g
1 b 1
2
m2 g
=
bm
1
+ m b d + m1 m2
g
O
2
FIG. P12.3
352
Static Equilibrium and Elasticity
Section 12.2 P12.4
More on the Center of Gravity
The hole we can count as negative mass xCG = Call the mass of each unit of pizza area. m 1 x1 - m 2 x 2 m1 - m 2
xCG = xCG = P12.5
R 2 0 - R 2
R 8 3 4
c h c- h - c h
R 2 2 R 2 R 2 2
=
R 6
4.00 cm
The coordinates of the center of gravity of piece 1 are x1 = 2.00 cm and y1 = 9.00 cm . The coordinates for piece 2 are x 2 = 8.00 cm and y 2 = 2.00 cm . The area of each piece is
18.0 cm
1
2
12.0 cm
4.00 cm
A1 = 72.0 cm 2 and A 2 = 32.0 cm 2 .
And the mass of each piece is proportional to the area. Thus,
FIG. P12.5
2
m i xi xCG = mi
and yCG =
e72.0 cm ja2.00 cmf + e32.0 cm ja8.00 cmf = =
2
72.0 cm 2 + 32.0 cm 2
3.85 cm
m i yi mi
=
e72.0 cm ja9.00 cmf + e32.0 cm ja2.00 cmf =
2 2
104 cm 2
6.85 cm .
Chapter 12
353
P12.6
Let represent the mass-per-face area. A vertical strip at position x, with width dx and height
y 1.00 m
ax - 3.00f
9
2
has mass
x - 3.00 dx dm = . 9
The total mass is
x
3.00
a
f
2
y = (x -- 3.00)2/9
x - 3 dx M = dm = 9 x =0
z
z
a f
2
x 0 dx 3.00 m
FG IJ z ex - 6 x + 9jdx H 9K F I L x 6x + 9xOP M = G JM - H 9 KN 3 2 Q
M=
3.00 0 2 3 2
FIG. P12.6 =
3.00
0
The x-coordinate of the center of gravity is
xCG =
z
xdm M
=
1 9
3.00 0
z
x x - 3 dx =
a f
2
9
3.00 0
ze
x 3 - 6 x 2 + 9 x dx =
j
1 x 4 6x3 9x 2 - + 9 4 3 2
LM N
OP Q
3.00
=
0
6.75 m = 0.750 m 9.00
P12.7
Let the fourth mass (8.00 kg) be placed at (x, y), then xCG = 0 = x=- Similarly, yCG = 0 =
a3.00fa4.00f + m axf
4
12.0 + m 4
a3.00fa4.00f + 8.00byg
12.0 + 8.00
12.0 = -1.50 m 8.00
y = -1.50 m
P12.8 In a uniform gravitational field, the center of mass and center of gravity of an object coincide. Thus, the center of gravity of the triangle is located at x = 6.67 m , y = 2.33 m (see the Example on the center of mass of a triangle in Chapter 9). The coordinates of the center of gravity of the three-object system are then: xCG = xCG = yCG = yCG =
m i xi mi m i yi mi
=
b6.00 kg ga5.50 mf + b3.00 kgga6.67 mf + b5.00 kg ga-3.50 mf a6.00 + 3.00 + 5.00f kg
14.0 kg
35.5 kg m = 2.54 m and 14.0 kg =
b6.00 kg ga7.00 mf + b3.00 kgga2.33 mf + b5.00 kgga+3.50 mf
66.5 kg m = 4.75 m 14.0 kg
354
Static Equilibrium and Elasticity
Section 12.3 P12.9
Examples of Rigid Objects in Static Equilibrium 3r
= 0 = mga3r f - Tr
2T - Mg sin 45.0 = 0 T= Mg sin 45.0 1 500 kg g sin 45.0 = 2 2 = 530 9.80 N
bg
a fa f
m=
T 530 g = = 177 kg 3g 3g
1 500 kg
m
= 45
FIG. P12.9 *P12.10 (a) For rotational equilibrium of the lowest rod about its point of support, +12.0 g g 3 cm - m1 g 4 cm (b)
= 0 .
m1 = 9.00 g
For the middle rod, + m 2 2 cm - 12.0 g + 9.0 g 5 cm = 0
b
g
m 2 = 52.5 g
(c)
For the top rod, 52.5 g + 12.0 g + 9.0 g 4 cm - m3 6 cm = 0
b
g
m3 = 49.0 g
24.0 cm 26.0 cm
P12.11
Fg standard weight
f a f F 13 I F = F G J H 12 K F F - F I 100 = F 13 - 1I 100 = GH F JK GH 12 JK
Fg 0.240 = Fg 0.260
g g g g g
Fg weight of goods sold
a
Fg
Fg
FIG. P12.11 8.33%
T H V d 30.0 196 N
*P12.12
(a)
Consider the torques about an axis perpendicular to the page and through the left end of the horizontal beam.
= +aT sin 30.0fd - a196 N fd = 0 ,
giving T = 392 N . (b) From From
FIG. P12.12
Fx = 0 ,
H - T cos 30.0 = 0 , or H = 392 N cos 30.0 = 339 N to the right . 0 .
a
f
Fy = 0 , V + T sin 30.0-200 N = 0 , or V = 196 N - a392 N f sin 30.0 =
Chapter 12
355
P12.13
(a)
Fx = f - n w = 0 Fy = n g - 800 N - 500 N = 0
Taking torques about an axis at the foot of the ladder,
nw
a800 Nfa4.00 mf sin 30.0+a500 Nfa7.50 mf sin 30.0 -n a15.0 cmf cos 30.0 = 0
w
500 N ng
Solving the torque equation, nw
a4.00 mfa800 Nf + a7.50 mfa500 Nf tan 30.0 = 268 N . =
15.0 m
f A
800 N
Next substitute this value into the Fx equation to find
FIG. P12.13
f = n w = 268 N
Solving the equation
in the positive x direction.
Fy = 0 ,
n g = 1 300 N in the positive y direction.
(b)
In this case, the torque equation
A = 0
gives:
w
a9.00 mfa800 Nf sin 30.0+a7.50 mfa500 Nf sin 30.0-a15.0 mfbn g sin 60.0 = 0
or n w = 421 N . Since f = n w = 421 N and f = fmax = n g , we find
=
fmax 421 N = = 0.324 . 1 300 N ng (1) (2)
nw m2 g
P12.14
(a)
Fx = f - n w = 0 Fy = n g - m1 g - m 2 g = 0
A = -m1 g G 2 J cos - m 2 gx cos + n w L sin = 0 H K
From the torque equation, nw =
1 2
F LI
LM 1 m g + FG x IJ m g OP cot N2 H LK Q
m1 g
Then, from equation (1): and from equation (2): (b)
L1 F xI O f = n = M m g + G J m g P cot N2 H LK Q n = bm + m g g
w 1 2 g 1 2
f A
ng
FIG. P12.14
If the ladder is on the verge of slipping when x = d , then
=
f
x=d
ng
=
e
m1 2
+
m2d L
j cot
m1 + m 2
.
356 P12.15
Static Equilibrium and Elasticity
(a)
Taking moments about P,
aR sin 30.0f0 + aR cos 30.0fa5.00 cmf - a150 Nfa30.0 cmf = 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) f = R sin 30.0-150 N = 370 N n = R cos 30.0 = 900 N Fsurface = 370 N i + 900 N j P12.16 See the free-body diagram at the right. When the plank is on the verge of tipping about point P, the normal force n1 goes to zero. Then, summing torques about point P gives
Mg 3.00 m x P n1 d n2
FIG. P12.15
a
f a
f
p = -mgd + Mgx = 0
or
F mI x = G Jd . H MK
mg 6.00 m
1.50 m
From the dimensions given on the free-body diagram, observe that d = 1.50 m Thus, when the plank is about to tip, x= P12.17
FIG. P12.16
F 30.0 kg I a1.50 mf = GH 70.0 kg JK a fb g a
0.643 m .
Torque about the front wheel is zero.
0 = 1.20 m mg - 3.00 m 2 Fr
Thus, the force at each rear wheel is
fb g
Fr = 0.200mg = 2.94 kN .
The force at each front wheel is then Ff = mg - 2 Fr = 4. 41 kN . 2 FIG. P12.17
Chapter 12
357
P12.18
Fx = Fb - Ft + 5.50 N = 0 Fy = n - mg = 0
Summing torques about point O,
(1)
5.50 N
O = Ft a1.50 mf - a5.50 mfa10.0 mf = 0
which yields Ft = 36.7 N to the left Then, from Equation (1),
10.0 m
mg
Ft 1.50 m Fb O n
Fb = 36.7 N - 5.50 N = 31.2 N to the right
FIG. P12.18 P12.19 (a) (b) P12.20 Te sin 42.0 = 20.0 N Te cos 42.0 = Tm
Te = 29.9 N Tm = 22.2 N
Ry T x Rx 20.0 y 4.00 m 5.00 m 7.00 m 19.6 kN 9.80 kN
a f vertically down a distance y = a5.00 mf sin 20.0 = 1.71 m . The
cable then makes the following angle with the horizontal:
Relative to the hinge end of the bridge, the cable is attached horizontally out a distance x = 5.00 m cos 20.0 = 4.70 m and
= tan -1
(a)
LM a12.0 + 1.71f m OP = 71.1 . N 4.70 m Q a a f f a f
Take torques about the hinge end of the bridge: R x 0 + R y 0 - 19.6 kN 4.00 m cos 20.0 -9.80 kN 7.00 m cos 20.0 = 0 which yields T = 35.5 kN
af
af
-T cos 71.1 1.71 m + T sin 71.1 4.70 m
a
f
FIG. P12.20
(b)
Fx = 0 Rx - T cos 71.1 = 0
or R x = 35.5 kN cos 71.1 = 11.5 kN right
a
f
b
g
(c)
Fy = 0 R y - 19.6 kN + T sin 71.1-9.80 kN = 0
Thus, R y = 29. 4 kN - 35.5 kN sin 71.1 = -4.19 kN = 4.19 kN down
a
f
358 *P12.21
Static Equilibrium and Elasticity
(a)
We model the horse as a particle. The drawbridge will fall out from under the horse.
Ry
0
Rx
= mg
=
1 2 1 3
cos 0 m
2
=
3g cos 0 2
3 9.80 m s 2 cos 20.0 2 8.00 m
e
a
j
f
= 1.73 rad s
2
mg FIG. P12.21(a)
(b)
1 2 I = mgh 2 1 1 1 1 - sin 0 m 2 2 = mg 2 3 2
b
g
1.56 rad s
= (c)
3g
b1 - sin g = e 8.00 m j a1 - sin 20f =
3 9.80 m s 2
0
The linear acceleration of the bridge is: 1 1 a= = 8.0 m 1.73 rad s 2 = 6.907 m s 2 2 2
Ry
a
fe
j
0
Rx
The force at the hinge + the force of gravity produce the acceleration a = 6.907 m s 2 at right angles to the bridge. R x = ma x = 2 000 kg 6.907 m s 2 cos 250 = -4.72 kN
a
mg
b
ge
j
FIG. P12.21(c)
Ry - mg = ma y
R y = m g + a y = 2 000 kg 9.80 m s 2 + 6.907 m s 2 sin 250 = 6.62 kN Thus: R = -4.72 i + 6.62 j kN . (d) Rx = 0 a=
2
e
j b
g
e
j
e
j
FG 1 IJ = b1.56 rad sg a4.0 mf = 9.67 m s H2 K
2
Ry
2
a mg FIG. P12.21(d)
Rx
R y - mg = ma
R y = 2 000 kg 9.8 m s 2 + 9.67 m s 2 = 38.9 kN Thus: R y = 38.9 j kN
b
ge
j
Chapter 12
359
P12.22
Call the required force F, with components Fx = F cos 15.0 and
Fy
400 N
Fy = - F sin 15.0 , transmitted to the
center of the wheel by the handles. Just as the wheel leaves the ground, the ground exerts no force on it.
R b 8.00 cm a distances
Fx b nx ny a forces
Fx = 0 : Fy = 0 :
F cos 15.0-n x
(1)
- F sin 15.0-400 N + n y = 0 (2)
FIG. P12.22
Take torques about its contact point with the brick. The needed distances are seen to be: b = R - 8.00 cm = 20.0 - 8.00 cm = 12.0 cm a = R - b = 16.0 cm (a)
2 2
a
f
= 0 :
F - 12.0 cm cos 15.0+ 16.0 cm sin 15.0 + 400 N 16.0 cm = 0 so (b) F= 6 400 N cm = 859 N 7.45 cm
a
- Fx b + Fy a + 400 N a = 0 , or
a
f
f
a
f
a
fa
f
Then, using Equations (1) and (2), n x = 859 N cos 15.0 = 830 N and n y = 400 N + 859 N sin 15.0 = 622 N
2 2 n = n x + n y = 1.04 kN
a
f
a
f
= tan -1
*P12.23
F n I = tan a0.749f = GH n JK
y x -1
36.9 to the left and upward
When x = x min , the rod is on the verge of slipping, so f = fs From
2.0 m n f x Fg Fg
b g
max
= sn = 0.50n .
37 2.0 m
Fx = 0 , n - T cos 37 = 0 , or n = 0.799T .
Thus, f = 0.50 0.799T = 0.399T From Using
a
f
FIG. P12.23
Fy = 0 , = 0
f + T sin 37-2 Fg = 0 , or 0.399T - 0.602T - 2 Fg = 0 , giving T = 2.00 Fg .
for an axis perpendicular to the page and through the left end of the beam gives
- Fg x min - Fg 2.0 m + 2 Fg sin 37 4.0 m = 0 , which reduces to x min = 2.82 m .
a
f e j
a
f
360 P12.24
Static Equilibrium and Elasticity
x=
3L 4
L
If the CM of the two bricks does not lie over the edge, then the bricks balance. If the lower brick is placed L over the edge, then the 4
x
second brick may be placed so that its end protrudes over the edge. P12.25
3L 4
FIG. P12.24
To find U, measure distances and forces from point A. Then, balancing torques,
a0.750fU = 29.4a2.25f a0.750fD = a1.50fa29.4f
Also, notice that U = D + Fg , so *P12.26
U = 88.2 N
To find D, measure distances and forces from point B. Then, balancing torques,
D = 58.8 N
Fy = 0 .
Consider forces and torques on the beam.
Fx = 0 : Fy = 0 : = 0 :
(a)
R cos - T cos 53 = 0
aT sin 53f8 m - a600 Nfx - a200 Nf4 m = 0
R sin + T sin 53-800 N = 0
Then T =
600 Nx + 800 N m = 93.9 N m x + 125 N . As x increases from 2 m, this expression 8 m sin 53 grows larger.
b
g
(b)
From substituting back, R cos = 93.9 x + 125 cos 53 R sin = 800 N - 93.9 x + 125 sin 53 Dividing, tan = 800 N R sin = - tan 53+ 93.9 x +125 cos 53 R cos
f F 32 - 1IJ tan = tan 53 G H 3x + 4 K
a
As x increases the fraction decreases and decreases . continued on next page
Chapter 12
361
(c)
To find R we can work out R 2 cos 2 + R 2 sin 2 = R 2 . From the expressions above for R cos and R sin , R 2 = T 2 cos 2 53+T 2 sin 2 53-1 600 NT sin 53+ 800 N R 2 = T 2 - 1 600T sin 53+640 000 R 2 = 93.9 x + 125
a
f
2
a
f
2
- 1 278 93.9 x + 125 + 640 000
a
f
R = 8 819 x 2 - 96 482 x + 495 678
e
j
12
At x = 0 this gives R = 704 N . At x = 2 m , R = 581 N . At x = 8 m , R = 537 N . Over the range of possible values for x, the negative term -96 482x dominates the positive term
8 819 x 2 , and R decreases as x increases.
Section 12.4 P12.27
Elastic Properties of Solids
L F =Y A Li L =
200 9.80 4.00 FLi = = 4.90 mm AY 0.200 10 -4 8.00 10 10 stress = F F = 2 A r
a fa fa f e je j a e f FGH d IJK 2
8 2
P12.28
(a)
F = stress
F = 1.50 10 F = 73.6 kN (b)
F 2.50 10 m I N m j G H 2 JK
2 -2
2
stress = Y strain =
i
a
f
YL Li
8
L =
astressfL = e1.50 10
Y
N m 2 0.250 m
10
ja
1.50 10
N m
2
f=
2.50 mm
*P12.29
The definition of Y =
stress means that Y is the slope of the graph: strain Y= 300 10 6 N m 2 = 1.0 10 11 N m 2 . 0.003
362 P12.30
Static Equilibrium and Elasticity
Count the wires. If they are wrapped together so that all support nearly equal stress, the number should be 20.0 kN = 100 . 0.200 kN Since cross-sectional area is proportional to diameter squared, the diameter of the cable will be
a1 mmf
P12.31
100 ~ 1 cm .
From the defining equation for the shear modulus, we find x as 5.00 10 -3 m 20.0 N hf x = = = 2.38 10 -5 m -4 6 2 2 SA 3.0 10 N m 14.0 10 m
e
e
je
ja
f
j
or x = 2.38 10 -2 mm . P12.32 The force acting on the hammer changes its momentum according to
mvi + F t = mv f so F =
Hence, F = 30.0 kg -10.0 m s - 20.0 m s 0.110 s = 8.18 10 3 N .
a f
m v f - vi t
.
By Newton's third law, this is also the magnitude of the average force exerted on the spike by the hammer during the blow. Thus, the stress in the spike is: stress = F 8.18 10 3 N = = 1.97 10 7 N m 2 b0.023 0 mg2 A
4
and the strain is: strain = P12.33 (a) F = A stress
stress 1.97 10 7 N m 2 = = 9.85 10 -5 . Y 20.0 10 10 N m 2
F
a fa f = e5.00 10
-3
m
j e4.00 10
2
8
N m2
j
3.0 ft
= 3.14 10 4 N (b) The area over which the shear occurs is equal to the circumference of the hole times its thickness. Thus, A = 2r t = 2 5.00 10 -3 m 5.00 10 -3 m = 1.57 10
-4
t A
a f
e
je
j
j
FIG. P12.33
m
2
So, F = A Stress = 1.57 10 -4 m 2 4.00 10 8 N m 2 = 6.28 10 4 N .
af
e
je
Chapter 12
363
P12.34
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: m1 a = T - m1 g where T is the tension in the wire. Solving equation (1) for the acceleration gives: a = and substituting this into equation (2) yields: Solving for the tension T gives T=
2 2m1 m 2 g 2 3.00 kg 5.00 kg 9.80 m s = = 36.8 N . 8.00 kg m 2 + m1
(1)
and
m2 a = m2 g - T
(2)
T - g, m1
m2 T - m2 g = m2 g - T . m1
b
gb
ge
j
From the definition of Young's modulus, Y =
i
a f a36.8 Nfa2.00 mf TL L = = YA e2.00 10 N m j e2.00 10
11 2 9 2
FLi , the elongation of the wire is: A L
-3
m
j
2
= 0.029 3 mm .
P12.35
Consider recompressing the ice, which has a volume 1.09V0 .
F V IJ = -e2.00 10 N m ja-0.090f = P = - BG 1.09 HV K
i
1.65 10 8 N m 2
*P12.36
B=-
P
V Vi
=-
PVi V 1.13 10 8 N m 2 1 m3 PVi =- = -0.053 8 m 3 B 0.21 10 10 N m 2 1.03 10 3 kg 0.946 m
3
(a) (b)
V = -
e
j
The quantity of water with mass 1.03 10 3 kg occupies volume at the bottom 1 m 3 - 0.053 8 m 3 = 0.946 m 3 . So its density is = 1.09 10 3 kg m3 .
(c) *P12.37
With only a 5% volume change in this extreme case, liquid water is indeed nearly incompressible.
Part of the load force extends the cable and part compresses the column by the same distance : F= = YA A A
A
+
Ys As
s
F
YA A A
A
+
Ys As
s
=
7 10 10 0.162 4 2 - 0 .161 4 2 4 3. 25
e
8 500 N
a f
j + 20 10 b0.012 7 g 4a 5.75 f
10
2
= 8.60 10
-4
m
364
Static Equilibrium and Elasticity
Additional Problems *P12.38 (a) (b) The beam is perpendicular to the wall, since 3 2 + 4 2 = 5 2 . Then sin = 4m ; = 53.1 . 5m
hinge = 0 :
+T sin 3 m - 250 N 10 m = 0
T= 2 500 Nm = 1.04 10 3 N 3 m sin 53.1
a f
a
f
(c)
1.04 10 3 N T = = 0.126 m k 8.25 10 3 N m The cable is 5.126 m long. From the law of cosines, x= 4 2 = 5.126 2 + 3 2 - 2 3 5.126 cos
4m
5.126 m
a fa
2
f
3m
= cos -1
(d)
3 + 5.126 - 4 = 51.2 2 3 5.126
2
2
a fa
f
FIG. P12.38 sin sin 51.2 = 5.126 m 4m
From the law of sines, the angle the hinge makes with the wall satisfies sin = 0.998 58
hinge = 0
+T 3 m sin 51.2-250 N 10 m 0.998 58 = 0 T = 1.07 10 3 N (e) x= 1.07 10 3 N = 0.129 m 8.25 10 3 N m 3 2 + 5.129 2 - 4 2 = 51.1 2 3 5.129
a f
a
fa
f
= cos -1
(f)
a fa
f
Now the answers are self-consistent: sin 51.1 = 0.998 51 4m T 3 m sin 51.1-250 N 10 m 0.998 51 = 0 sin = 5.129 m
a f
a
fa
f
T = 1.07 10 N x = 0.129 5 m
3
= 51.1
P12.39 Let n A and n B be the normal forces at the points of support. Choosing the origin at point A with we find: n A + n B - 8.00 10 4 g - 3.00 10 4 g = 0 and
4 4 B
Fy = 0 and = 0,
A
B
j e j -e3.00 10 jb g g15.0 - e8.00 10 jb g g25.0 + n a50.0 f = 0
e
15.0 m 50.0 m
FIG. P12.39
The equations combine to give n A = 5.98 10 5 N and B b = 4.80 10 5 N .
Chapter 12
365
P12.40
When the concrete has cured and the pre-stressing tension has been released, the rod presses in on the concrete and with equal force, T2 , the concrete produces tension in the rod. (a) In the concrete: stress = 8.00 10 6 N m 2 = Y strain = Y Thus, L = or (b)
a
astressfL = e
i
8.00 10 6 N m 2 1.50 m 30.0 10 N m
9 2
ja
f
f FGH L IJK L
i
Y
L = 4.00 10 -4 m = 0.400 mm . T2 = 8.00 10 6 N m 2 , so Ac
In the concrete: stress =
T2 = 8.00 10 6 N m 2 50.0 10 -4 m 2 = 40.0 kN (c) For the rod: T2 Li T2 L = Ysteel so L = AR Li A R Ysteel
e
je
j
FG IJ H K
e4.00 10 Nja1.50 mf L = = 2.00 10 e1.50 10 m je20.0 10 N m j
4 -4 2 10 2
-3
m = 2.00 mm
(d)
The rod in the finished concrete is 2.00 mm longer than its unstretched length. To remove stress from the concrete, one must stretch the rod 0.400 mm farther, by a total of 2. 40 mm . For the stretched rod around which the concrete is poured:
(e)
FG L IJ A Y H L K F 2.40 10 m I e1.50 10 m je20.0 10 T =G H 1.50 m JK
Ltotal T1 = Ysteel AR Li
-3 1
FG H
IJ K
or T1 =
total i
R steel
-4
2
10
N m 2 = 48.0 kN
j
*P12.41
With as large as possible, n1 and n 2 will both be large. The equality sign in f 2 sn 2 will be true, but the less-than sign in f1 < sn1 . Take torques about the lower end of the pole. n 2 cos + Fg
n2
f2 d
FG 1 IJ cos - f H2 K f
2
sin = 0
Fg
n1
f1 FIG. P12.41
Setting f 2 = 0.576n 2 , the torque equation becomes n 2 1 - 0.576 tan + Since n 2 > 0 , it is necessary that 1 - 0.576 tan < 0 1 = 1.736 0.576 > 60.1 tan > = 7.80 ft d < = 9.00 ft sin sin 60.1
a
1 Fg = 0 2
366 P12.42
Static Equilibrium and Elasticity
Call the normal forces A and B. They make angles and with the vertical.
Mg
Fx = 0: Fy = 0:
Substitute B = A sin sin
A sin - B sin = 0 A cos - Mg + B cos = 0
A
B
A cos + A cos
A cos sin + sin cos = Mg sin sin A = Mg sin +
A sin A cos
b
sin = Mg sin
g
Mg B sin B cos
b
g
sin B = Mg sin +
b
g
Ry Rx O x 700 N 3.00 m
FIG. P12.42
T 60.0
P12.43
(a) (b)
See the diagram. If x = 1.00 m , then
O = a-700 N fa1.00 mf - a200 N fa3.00 mf
a fa f +aT sin 60.0fa6.00 mf = 0
- 80.0 N 6.00 m
200 N 3.00 m
80.0 N
FIG. P12.43
Solving for the tension gives: T = 343 N . From From (c)
Fx = 0 , R x = T cos 60.0 =
171 N . 683 N .
Fy = 0 , R y = 980 N - T sin 60.0 =
If T = 900 N :
O = a-700 N fx - a 200 N fa3.00 mf - a80.0 N fa6.00 mf +
Solving for x gives: x = 5.13 m .
a900 Nf sin 60.0 a6.00 mf = 0 .
Chapter 12
367
P12.44
(a)
Sum the torques about top hinge:
D C
T sin 30.0 T cos 30.0
= 0:
C 0 + D 0 + 200 N cos 30.0 0
af af
+200 N sin 30.0 3.00 m
Giving A = (b)
af 160 N bright g
+B 0 = 0
-392 N 1.50 m + A 1.80 m
a
f a
a
af
f
1.80 m
392 N
f
A 1.50 m 1.50 m
.
B
Fx = 0 :
-C - 200 N cos 30.0+ A = 0 C = 160 N - 173 N = -13.2 N In our diagram, this means 13.2 N to the right .
FIG. P12.44
(c)
Fy = 0 : +B + D - 392 N + 200 N sin 30.0 = 0
B + D = 392 N - 100 N = 292 N up
b g
(d)
Given C = 0: Take torques about bottom hinge to obtain
A 0 + B 0 + 0 1.80 m + D 0 - 392 N 1.50 m + T sin 30.0 3.00 m + T cos 30.0 1.80 m = 0
so T = P12.45 Using
af af a a
f af
a
f
a
f
a
f
588 N m = 192 N . 1.50 m + 1.56 m
f
Fx = Fy = = 0, choosing the origin at the left end
of the beam, we have (neglecting the weight of the beam)
Fx = Rx - T cos = 0 , Fy = Ry + T sin - Fg = 0 ,
and
= - Fg aL + d f + T sin a2L + d f = 0.
Solving these equations, we find: (a) T=
(b)
Rx
a f sin a 2L + d f F aL + d f cot =
Fg L + d
g
2L + d
Ry =
Fg L 2L + d
FIG. P12.45
368 P12.46
Static Equilibrium and Elasticity
point 0 = 0 gives
aT cos 25.0fFGH 34 sin 65.0IJK + aT sin 25.0fFGH 34 cos 65.0IJK F I = b 2 000 N ga cos 65.0f + b1 200 N gG cos 65.0J H2 K
From which, T = 1 465 N = 1.46 kN From
T sin 25.0 l T cos 25.0 3l 4 1 200 N H 65.0 V FIG. P12.46 2 000 N
Fx = 0 ,
H = T cos 25.0 = 1 328 N toward right = 1.33 kN From
b
g
Fy = 0 ,
V = 3 200 N - T sin 25.0 = 2 581 N upward = 2.58 kN P12.47 We interpret the problem to mean that the support at point B is frictionless. Then the support exerts a force in the x direction and FBy = 0
b
g
g and = -b3 000 g ga 2.00f - b10 000 g ga6.00f + F a1.00 f = 0 .
FAy - 3 000 + 10 000 g = 0
Bx
Fx = FBx - FAx = 0
b
These equations combine to give FAx = FBx = 6.47 10 N FAy = 1.27 10 5 N P12.48
5
FIG. P12.47
n= M+m g H = f
a
f
H
H max = f max = s m + M g
a
f
A = 0 =
mgL cos 60.0+ Mgx cos 60.0- HL sin 60.0 2 m + M tan 60.0 m x H tan 60.0 m = - = s - 2M 2M L Mg M
x Mg mg 60.0 A f
a
f
3 1 = s tan 60.0- = 0.789 2 4
n
FIG. P12.48
Chapter 12
369
P12.49
From the free-body diagram, the angle T makes with the rod is
T 20
= 60.0+20.0 = 80.0
and the perpendicular component of T is T sin 80.0. Summing torques around the base of the rod,
= 0: Fx = 0 :
Fy = 0 :
- 4.00 m 10 000 N cos 60+T 4.00 m sin 80 = 0 T=
a
b10 000 Ng cos 60.0 =
sin 80.0
fb
g
a
f
10 000 N FV 60 FH
5.08 10 3 N
FH - T cos 20.0 = 0
FH = T cos 20.0 = 4.77 10 3 N
FV + T sin 20.0-10 000 N = 0
FIG. P12.49
and FV = 10 000 N - T sin 20.0 = 8.26 10 3 N P12.50 Choosing the origin at R, (1) (2) (3) T 90
b
g
Fx = + R sin 15.0-T sin = 0 Fy = 700 - R cos 15.0+T cos = 0 = -700 cos a0.180f + T b0.070 0g = 0
1 800 sin cos sin 15.0
R 15.0
Solve the equations for from (3), T = 1 800 cos from (1), R =
n
18.0 cm 25.0 cm
1 800 sin cos cos 15.0 Then (2) gives 700 - + 1 800 cos 2 = 0 sin 15.0 or Let cos 2 + 0.388 9 - 3.732 sin cos = 0 u = cos 2 then using the quadratic equation, u = 0.011 65 or 0.869 3 Squaring, cos 4 - 0.880 9 cos 2 + 0.010 13 = 0
FIG. P12.50
Only the second root is physically possible, = cos -1 0.869 3 = 21.2 T = 1.68 10 3 N P12.51 and R = 2.34 10 3 N
Choosing torques about R, with -
= 0
2L L 350 N + T sin 12.0 - 200 N L = 0 . 2 3
a
f a
fFGH IJK a
f
From which, T = 2.71 kN . Let R x = compression force along spine, and from R x = Tx = T cos 12.0 = 2.65 kN .
Fx = 0
FIG. P12.51
370 P12.52
Static Equilibrium and Elasticity
(a)
Just three forces act on the rod: forces perpendicular to the sides of the trough at A and B, and its weight. The lines of action of A and B will intersect at a point above the rod. They will have no torque about this point. The rod's weight will cause a torque about the point of intersection as in Figure 12.52(a), and the rod will not be in equilibrium unless the center of the rod lies vertically below the intersection point, as in Figure 12.52(b). All three forces must be concurrent. Then the line of action of the weight is a diagonal of the rectangle formed by the trough and the normal forces, and the rod's center of gravity is vertically above the bottom of the trough. In Figure (b), AO cos 30.0 = BO cos 60.0 and L2 = AO + BO = AO + AO AO = L 1+
cos 2 30.0 cos 2 60.0 2 2 2 2
B A Fg
O
FIG. P12.52(a)
(b)
F cos GH cos
2
30.0 2 60.0
I JK
A 30.0
O
B Fg 60.0
L = 2
So cos = P12.53 (a)
AO 1 = and = 60.0 . 2 L
FIG. P12.52(b)
Locate the origin at the bottom left corner of the cabinet and let x = distance between the resultant normal force and the front of the cabinet. Then we have
Fx = 200 cos 37.0- n = 0 Fy = 200 sin 37.0+n - 400 = 0 = na0.600 - xf - 400a0.300f + 200 sin 37.0 a0.600 f
-200 cos 37.0 0.400 = 0
From (2), From (3), n = 400 - 200 sin 37.0 = 280 N 72.2 - 120 + 280 0.600 - 64.0 280 x = 20.1 cm to the left of the front edge x=
(1) (2)
a
f
(3)
a
f
From (1), (b)
k =
200 cos 37.0 = 0.571 280 FIG. P12.53
In this case, locate the origin x = 0 at the bottom right corner of the cabinet. Since the cabinet is about to tip, we can use = 0 to find h:
= 400a0.300f - a300 cos 37.0fh = 0
h=
120 = 0.501 m 300 cos 37.0
Chapter 12
371
P12.54
(a), (b) Use the first diagram and sum the torques about the lower front corner of the cabinet. = 0 - F 1.00 m + 400 N 0.300 m = 0
0.300 m F 400 N 1.00 m
f a fa a400 Nfa0.300 mf = 120 N yielding F =
1.00 m Fx = 0 - f + 120 N = 0 , or so
a
f
Fy = 0 -400 N + n = 0 ,
f 120 N Thus, s = = = 0.300 . n 400 N (c)
f = 120 N n = 400 N
n
f
Apply F at the upper rear corner and directed so + = 90.0 to obtain the largest possible lever arm.
F'
= tan -1
FG 1.00 m IJ = 59.0 H 0.600 m K
2
Thus, = 90.0-59.0 = 31.0 . Sum the torques about the lower front corner of the cabinet: -F
2
1.00 m 400 N
f n 0.600 m
a1.00 mf + a0.600 mf + a400 Nfa0.300 mf = 0
120 N m = 103 N . so F = 1.17 m Therefore, the minimum force required to tip the cabinet is 103 N applied at 31.0 above the horizontal at the upper left corner . P12.55 (a) We can use Then
FIG. P12.54
Fx = Fy = 0 and = 0 with pivot point at
P T L/2
the contact on the floor.
Fx = T - sn = 0 ,
Fy = n - Mg - mg = 0, and
= MgaL cos f + mg G 2 cos J - T aL sin f = 0 H K
Solving the above equations gives M= m 2 s sin - cos 2 cos - s sin
FL
I
Mg L/2 mg n
FG H
IJ K
f
FIG. P12.55
This answer is the maximum vaue for M if s < cot . If s cot , the mass M can increase without limit. It has no maximum value, and part (b) cannot be answered as stated either. In the case s < cot , we proceed. (b) At the floor, we have the normal force in the y-direction and frictional force in the xdirection. The reaction force then is R = n 2 + sn
b g
2
2
=
a M + mf g
a
2 1 + s .
At point P, the force of the beam on the rope is F = T 2 + Mg
b g
2 = g M2 + s M + m
f
2
.
372 P12.56
Static Equilibrium and Elasticity
(a)
The height of pin B is
1000 N B 10.0 m nA 30.0 45.0 C A nC
a10.0 mf sin 30.0 = 5.00 m .
The length of bar BC is then BC = 5.00 m = 7.07 m. sin 45.0
Consider the entire truss:
FIG. P12.56(a)
Fy = n A - 1 000 N + nC = 0 A = -b1 000 N g10.0 cos 30.0+nC 10.0 cos 30.0+7.07 cos 45.0
Which gives nC = 634 N . Then, n A = 1 000 N - nC = 366 N . (b) Suppose that a bar exerts on a pin a force not along the length of the bar. Then, the pin exerts on the bar a force with a component perpendicular to the bar. The only other force on the bar is the pin force on the other end. For F = 0 , this force must also have a component perpendicular to the bar. Then, the total torque on the bar is not zero. The contradiction proves that the bar can only exert forces along its length. Joint A:
=0
FIG. P12.56(b)
(c)
CAB A TAC nA = 366 N
Fy = 0 : -C AB sin 30.0+366 N = 0 ,
so C AB = 732 N
Fx = 0 : -C AB cos 30.0+TAC = 0
TAC = 732 N cos 30.0 = 634 N Joint B:
30.0
a
f
1000 N B 45.0 CBC
Fx = 0 : 732 N cos 30.0-CBC cos 45.0 = 0
C BC =
a732 Nf cos 30.0 =
cos 45.0
a
f
CAB = 732 N
897 N
FIG. P12.56(c)
Chapter 12
373
P12.57
From geometry, observe that cos = 1 4 and
= 75.5
For the left half of the ladder, we have
Fx = T - R x = 0 Fy = R y + n A - 686 N = 0 top = 686 Na1.00 cos 75.5f + T a2.00 sin 75.5f
-n A 4.00 cos 75.5 = 0
For the right half of the ladder we have
(1) (2) (3) FIG. P12.57
a
f
Fx = Rx - T = 0 Fy = nB - Ry = 0 top = nB a 4.00 cos 75.5f - T a 2.00 sin 75.5f = 0
Solving equations 1 through 5 simultaneously yields: (a) (b) (c) T = 133 N n A = 429 N R x = 133 N and and n B = 257 N R y = 257 N
(4) (5)
The force exerted by the left half of the ladder on the right half is to the right and downward. P12.58 (a) x CG = = yCG =
b1 000 kg g10.0 m + b125 kgg0 + b125 kg g0 + b125 kg g20.0 m =
1 375 kg 1 375 kg
m i xi mi
9.09 m
b1 000 kg g10.0 m + b125 kgg20.0 m + b125 kgg20.0 m + b125 kg g0
= 10.9 m (b) By symmetry, x CG = 10.0 m There is no change in yCG = 10.9 m (c) P12.59 vCG =
FG 10.0 m - 9.09 m IJ = H 8.00 s K
0.114 m s
Considering the torques about the point at the bottom of the bracket yields:
b0.050 0 mga80.0 Nf - Fb0.060 0 mg = 0 so
F = 66.7 N .
374 P12.60
Static Equilibrium and Elasticity
When it is on the verge of slipping, the cylinder is in equilibrium. and f1 = n 2 = sn1 f 2 = sn 2 Fx = 0 :
Fy = 0 : = 0:
P + n1 + f 2 = Fg P = f1 + f 2 and and becomes n 2 n1 = 2 4 n n 3 P = 1 + 1 = n1 2 4 4 5 4 P+ P = Fg 4 3 f2 =
As P grows so do f1 and f 2 n 1 Therefore, since s = , f1 = 1 2 2 n then (1) P + n1 + 1 = Fg 4 5 So P + n1 = Fg 4 3 Therefore, P = Fg 8 P12.61 (a)
FIG. P12.60 (2) or 8 P = Fg 3
FG IJ H K
F = k L , Young's modulus is Y =
Thus, Y =
L 0
a f
F A L Li
=
FLi A L
a f
a L f
2 Li
2
kLi YA and k = A Li
L 0
(b) P12.62 (a)
W = - Fdx = -
z
za
- kx dx =
f
YA Li
L 0
z
xdx = YA
Take both balls together. Their weight is 3.33 N and their CG is at their contact point.
Fx = 0 : + P3 - P1 = 0 P2 = 3.33 N Fy = 0 : + P2 - 3.33 N = 0 A = 0: - P3 R + P2 R - 3.33 NaR + R cos 45.0f
+ P1 R + 2 R cos 45.0 = 0
P1
a
f
3.33 N P3 Fg
Substituting,
- P1 R + 3.33 N R - 3.33 N R 1 + cos 45.0
a
a3.33 Nf cos 45.0 = 2 P cos 45.0
1
fa + P Ra1 + 2 cos 45.0f = 0
1
f a
f
P2
P1 = 1.67 N so P3 = 1.67 N (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.
FIG. P12.62(a)
1.67 N
P1
Fx = 0 : n cos 45.0- P1 = 0
n=
n cos 45.0
1.67 N = 2.36 N cos 45.0 Fy = 0 : n sin 45.0-1.67 N = 0 gives the same result
n sin 45.0
FIG. P12.62(b)
Chapter 12
375
P12.63
Fy = 0 :
+380 N - Fg + 320 N = 0 Fg = 700 N
Take torques about her feet:
= 0:
-380 N 2.00 m + 700 N x + 320 N 0 = 0
x = 1.09 m FIG. P12.63
a
f a
f a
f
P12.64
The tension in this cable is not uniform, so this becomes a fairly difficult problem. dL F = L YA At any point in the cable, F is the weight of cable below that point. Thus, F = gy where is the mass per unit length of the cable. Then, y =
Li 0
z FGH
2 g i 1 gLi dL dy = ydy = 2 YA L YA 0 2
IJ K
L
z
y =
P12.65
(a)
a fa fa f e je j a10.0 - 1.00f m s = F v I F = mG J = b1.00 kg g H t K 0.002 s
a fa f
2. 40 9.80 500 1 = 0.049 0 m = 4.90 cm 2 2.00 10 11 3.00 10 -4 4 500 N
(b) (c)
stress =
4 500 N F = = 4.50 10 6 N m 2 0.010 m 0.100 m A
Yes . This is more than sufficient to break the board.
376 P12.66
Static Equilibrium and Elasticity
The CG lies above the center of the bottom. Consider a disk of water at height y above the bottom. Its radius is y fFGH 30.0y cm IJK = 25.0 cm + 3 yI yI yI F F F Its area is G 25.0 cm + J . Its volume is G 25.0 cm + J dy and its mass is G 25.0 cm + J H K H K H 3 3 3K 25.0 cm + 35.0 - 25.0 cm
a
2
2
2
dy . The
whole mass of the water is
M=
30 .0 cm y =0
z
dm =
30 .0 cm 0
z
625 +
2 3
L 50.0 y y O + P M = M625 y + 6 27 P MN Q L 50.0a30.0f a30.0f OP + M = M625a30.0f + 6 27 P MN Q M = e10 kg cm je 27 250 cm j = 85.6 kg
30.0 0 2 3 -3 3 3
F GH
50.0 y y 2 + dy 3 9
I JK
The height of the center of gravity is yCG =
30 .0 cm y =0
z
ydm M
=
30 .0 cm 0
z
LM OP MN PQ .0 L 625a30.0f = MM 2 + 50.0a30.0f + a3036 f 9 MN e10 kg cm j 453 750 cm =
625 y 2 50.0 y 3 y 4 = + + 2 9 36 M
2 30 .0 cm 0 3 -3 3 4
F 625y + 50.0 y GH 3
2
+
y 3 dy 9 M
I JK
4
OP PQ
yCG
M 1.43 10 3 kg cm = = 16.7 cm 85.6 kg
Chapter 12
377
P12.67
Let represent the angle of the wire with the vertical. The radius of the circle of motion is r = 0.850 m sin . For the mass:
a
f
r
T
mg
v2 = mr 2 r T sin = m 0.850 m sin 2
Fr = mar = m
a
f
T Further, = Y strain or T = AY strain A Thus, AY strain = m 0.850 m 2 , giving
-4
a f f a f a f e3.90 10 mj e7.00 10 N m je1.00 10 j AY astrainf = = ma0.850 mf b1.20 kg ga0.850 mf a
2 10 2 -3
FIG. P12.67
or P12.68
= 5.73 rad s .
B D
For the bridge as a whole:
A = n A a0f - a13.3 kNfa100 mf + nE a 200 mf = 0
so
nE =
a13.3 kNfa100 mf =
200 m
6.66 kN
A C
nA
E
100 m 100 m 13.3 kN
nE
Fy = n A - 13.3 kN + n E = 0 gives
n A = 13.3 kN - n E = 6.66 kN At Pin A:
FAB =
6.66 kN = 10.4 kN compression sin 40.0 Fx = FAC - 10.4 kN cos 40.0 = 0 so
Fy = - FAB sin 40.0+6.66 kN = 0 or
FAB
FAC = 10.4 kN cos 40.0 = 7.94 kN tension At Pin B:
a
f
a
f
b
g f
40.0
40.0 FAC nA = 6.66 kN
a
Fy = a10.4 kN f sin 40.0- FBC sin 40.0 = 0
FBD 40.0 FBC FAB = 10.4 kN
Thus, FBC = 10. 4 kN tension
a
f b g
Fx = FAB cos 40.0+ FBC cos 40.0- FBD = 0
FBD = 2 10.4 kN cos 40.0 = 15.9 kN compression By symmetry: FDE = FAB = 10.4 kN compression FDC = FBC and FEC = FAC
a
f
b g = 10.4 kN atensionf = 7.94 kN atensionf
10.4 kN 40.0 7.94 kN
10.4 kN 40.0 7.94 kN
We can check by analyzing Pin C:
13.3 kN
Fx = +7.94 kN - 7.94 kN = 0 or 0 = 0 Fy = 2a10.4 kNf sin 40.0-13.3 kN = 0
which yields 0 = 0 .
FIG. P12.68
378 P12.69
Static Equilibrium and Elasticity
Member AC is not in pure compression or tension. It also has shear forces present. It exerts a downward force S AC and a tension force FAC on Pin A and on Pin C. Still, this member is in equilibrium. Fx = FAC - FAC = 0 FAC = FAC A = 0: -a14.7 kNfa25.0 mf + S AC a50.0 mf = 0 or
SAC FAC A
25.0 m
25.0 m
SAC
FAC C 14.7 kN
S = 7.35 kN AC Fy = S AC - 14.7 kN + 7.35 kN = 0 S AC = 7.35 kN
B
D
Then S AC = S and we have proved that the loading by the car A AC is equivalent to one-half the weight of the car pulling down on n A each of pins A and C, so far as the rest of the truss is concerned. For the Bridge as a whole:
E C
75.0 m 25.0 m 14.7 kN nE
A = 0:
- 14.7 kN 25.0 m + n E 100 m = 0 n E = 3.67 kN
a
fa
f a
f
7.35 kN
FAB 30.0 FAC nA = 11.0 kN
Fy = n A - 14.7 kN + 3.67 kN = 0
n A = 11.0 kN At Pin A:
Fy = -7.35 kN + 11.0 kN - FAB sin 30.0 = 0 Fx = FAC - a7.35 kN f cos 30.0 = 0
FAC = 6.37 kN tension FAB = 7.35 kN compression
FBD 30.0 7.35 kN 60.0 FBC
b
g
a
f
At Pin B:
4.24 kN
Fy = -a7.35 kN f sin 30.0- FBC sin 60.0 = 0
FBC = 4.24 kN tension
x
FCD 60.0 FCE
60.0 6.37 kN
a f F = a7.35 kNf cos 30.0+a 4.24 kN f cos 60.0- F F = 8.49 kN bcompressiong
BD
BD
=0
7.35 kN
At Pin C:
Fy = a 4.24 kNf sin 60.0+ FCD sin 60.0-7.35 kN = 0 Fx = -6.37 kN - a4.24 kNf cos 60.0+a 4.24 kNf cos 60.0+ FCE = 0
FCE = 6.37 kN tension FCD = 4.24 kN tension
FDE 30.0 6.37 kN 3.67 kN
a
f
a
f
At Pin E:
FIG. P12.69
Fy = - FDE sin 30.0+3.67 kN = 0
FDE = 7.35 kN compression
b
g
or Fx = -6.37 kN - FDE cos 30.0 = 0 which gives FDE = 7.35 kN as before.
Chapter 12
379
P12.70
(1) (2)
ph = I p = MvCM 2 I I I = = = R 5 p MvCM MR If the acceleration is a, we have Px = ma and Py + n - Fg = 0 . Taking the origin at the center of gravity, the torque equation gives Py L - d + Px h - nd = 0 . Solving these equations, we find
d
p h vCM
If the ball rolls without slipping, R = vCM So, h = P12.71 (a) FIG. P12.70
H L
a
f
CG P h
(b) (c)
F d - ah I . L G H g JK ah e 2.00 m s ja1.50 mf = = If P = 0 , then d =
Py = Fg
2 y
n
Fy g
FIG. P12.71 0.306 m .
g
9.80 m s 2
Using the given data, Px = -306 N and Py = 553 N . Thus, P = -306 i + 553 j N .
e
j
*P12.72
When the cyclist is on the point of tipping over forward, the normal force on the rear wheel is zero. Parallel to the plane we have f1 - mg sin = ma . Perpendicular to the plane, n1 - mg cos = 0 . Torque about the center of mass:
mg
mg 0 - f1 1.05 m + n1 0.65 m = 0 .
Combining by substitution, ma = f1 - mg sin = a = g cos 20
af a
f a
f
f1
n1
FIG. P12.72 n1 0.65 m 0.65 m - mg sin = mg cos - mg sin 1.05 m 1.05 m
*P12.73
0.65 - sin 20 = 2.35 m s 2 1.05 When the car is on the point of rolling over, the normal force on its inside wheels is zero.
FG H
IJ K
Fy = ma y :
n - mg = 0 f= mv 2 R d = 0. 2 v max = gdR 2h
mg h f d FIG. P12.73 mg
Fx = ma x :
Take torque about the center of mass: fh - n Then by substitution
2 mv max
R
h-
mgd =0 2
A wider wheelbase (larger d) and a lower center of mass (smaller h) will reduce the risk of rollover.
380
Static Equilibrium and Elasticity
ANSWERS TO EVEN PROBLEMS
P12.2 Fy + R y - Fg = 0 ; Fx - R x = 0 ; Fy cos - Fg P12.4 P12.6 P12.8 P12.10 P12.12 P12.14
FG IJ cos - F H 2K
P12.40
x
sin = 0 P12.42 P12.44
(a) 0.400 mm; (b) 40.0 kN; (c) 2.00 mm; (d) 2.40 mm; (e) 48.0 kN at A: Mg sin sin ; at B: Mg sin + sin +
see the solution 0.750 m
b
g
b
g
a2.54 m, 4.75 mf
(a) 9.00 g; (b) 52.5 g; (c) 49.0 g (a) 392 N; (b) 339 i + 0 j N (a) f = ng P12.46 P12.48 P12.50 P12.52 P12.54 P12.56
(a) 160 N to the right; (b) 13.2 N to the right; (c) 292 N up; (d) 192 N 1.46 kN ; 1.33 i + 2.58 j kN 0.789 T = 1.68 kN ; R = 2.34 kN; = 21.2 (a) see the solution; (b) 60.0 (a) 120 N; (b) 0.300; (c) 103 N at 31.0 above the horizontal to the right (a), (b) see the solution; (c) C AB = 732 N ; TAC = 634 N ; C BC = 897 N (a) 9.09 m, 10.9 m ; (b) 10.0 m, 10.9 m ; (c) 0.114 m s to the right 3 Fg 8 (a) P1 = 1.67 N ; P2 = 3.33 N ; P3 = 1.67 N ; (b) 2.36 N 4.90 cm 16.7 cm above the center of the bottom C AB = 10.4 kN ; TAC = 7.94 kN ; TBC = 10.4 kN ; C BD = 15.9 kN ; C DE = 10.4 kN ; TDC = 10. 4 kN ; TEC = 7.94 kN 2 R 5 2.35 m s 2
e
j
e
j
LM m g + m gx OP cot ; N2 L Q e + j cot = bm + m g g ; (b) = m +m
1 2 m1 2 m2d L 1 2 1 2
P12.16 P12.18 P12.20 P12.22
see the solution; 0.643 m 36.7 N to the left ; 31.2 N to the right (a) 35.5 kN; (b) 11.5 kN to the right; (c) 4.19 kN down (a) 859 N; (b) 104 kN at 36.9 above the horizontal to the left 3L 4 (a) see the solution; (b) decreases ; (c) R decreases (a) 73.6 kN; (b) 2.50 mm ~ 1 cm 9.85 10 -5 0.029 3 mm (a) -0.053 8 m3 ; (b) 1.09 10 3 kg m3 ; (c) Yes, in most practical circumstances (a) 53.1; (b) 1.04 kN; (c) 0.126 m, 51.2; (d) 1.07 kN; (e) 0.129 m, 51.1; (f) 51.1 P12.70 P12.72
P12.58
a
f
a
f
P12.60 P12.62
P12.24 P12.26 P12.28 P12.30 P12.32 P12.34 P12.36
P12.64 P12.66 P12.68
P12.38