Chapter 6 Student Solutions Manual 1. We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion problem (with the person's push F in the +x direction). Applying Newton's second law to the x and y axes
13168-12-26P
AID: 5204 | 19/09/2014
Draw the free body diagram with elastic curve.
Figure-1
Calculate the reactions at A and B by using equilibrium equations.
Take moment about point A.
MA 0:
RB 12 600 8 0
RB 400 lb
Consider equilibrium equation.
Fy 0 :
13168-14-116P
AID: 5204 | 23/09/2014
Determine the slope at
A
of the beam.
Draw the cross-section of the beam.
Figure a
The slope at point A is obtained by placing a virtual unit couple moment at A. By
inspection there are discontinuities of moment on the
13168-14-92P
AID: 5204 | 22/09/2014
Draw the cross-section of the beam.
Real moment functions on the beam
M ( x1 )
.
The real moment functions on the beam by considering the
x1
co-ordinate as shown in
the figure below.
Take the moments about point A.
M 8
13168-14-106P
AID: 5204 | 22/09/2014
Draw the cross-section of the beam.
Real moment functions on the beam
M
.
The real moment functions on the beam by considering the
x
co-ordinate as shown in the
figure below.
Take the moments about point A.
M 8 kip 60
13168-14-98P
AID: 5204 | 22/09/2014
Draw the cross-section of the beam.
Figure a
Real moment functions on the beam
M ( x)
.
The real moment functions on the beam by considering the
x
co-ordinate as shown in the
figure below.
Take the moments about point B
13168-14-93P
AID: 5204 | 22/09/2014
Draw the cross-section of the beam.
Figure a
Real moment functions on the beam
M ( x1 )
.
The real moment functions on the beam by considering the
x1
co-ordinate as shown in
the figure below.
Take the moments about poin
13168-14-115P
AID: 5204 | 22/09/2014
Determine the vertical displacement at
B
of the beam.
Draw the cross-section of the beam.
Figure a
Virtual moment functions on the beam
m( x )
.
The displacement of point B is obtained by placing a virtual unit load at
13168-14-97P
AID: 5204 | 22/09/2014
Draw the cross-section of the beam.
Figure a
Real moment functions on the beam
M ( x1 )
.
The real moment functions on the beam by considering the
x1
co-ordinate as shown in
the figure below.
Take the moments about poin
13168-14-110P
AID: 5204 | 22/09/2014
Determine the displacement at
D
of the beam.
Draw the cross-section of the beam.
Figure a
The displacement of point D is obtained by placing a virtual unit load at D. By inspection
there are discontinuities of moment o
13168-12-60P
AID: 5204 | 18/09/2014
Draw the free body diagram of the shaft.
Figure-1
Draw the elastic curve diagram.
The point load acting on both sides of point A. Therefore, the curve occurs as shown in
Figure-2.
Figure-2
Calculate the deflection at C
Chapter 11 Student Solutions Manual 5. By Eq. 10-52, the work required to stop the hoop is the negative of the initial kinetic 1 1 energy of the hoop. The initial kinetic energy is K = 2 I 2 + 2 mv 2 (Eq. 11-5), where I = mR2 is its rotational inert
Chapter 3 Student Solutions Manual 1. A vector a can be represented in the magnitude-angle notation (a, ), where
2 2 a = ax + a y
is the magnitude and
= tan - 1
is the angle a makes with the positive x axis.
ay ax
(a) Given Ax = -25.0 m a
Chapter 1 Student Solutions Manual 3. Using the given conversion factors, we find (a) the distance d in rods to be d = 4.0 furlongs =
( 4.0 furlongs )( 201.168 m furlong )
5.0292 m rod
= 160 rods,
(b) and that distance in chains to be d =
( 4.0
Chapter 10 Student Solutions Manual 13. We take t = 0 at the start of the interval and take the sense of rotation as positive. 1 Then at the end of the t = 4.0 s interval, the angular displacement is = 0t + 2 t 2 . We solve for the angular velocit
Chapter 15 Student Solutions Manual 3. (a) The amplitude is half the range of the displacement, or xm = 1.0 mm. (b) The maximum speed vm is related to the amplitude xm by vm = xm, where is the angular frequency. Since = 2f, where f is the frequenc
Chapter 5 Student Solutions Manual 5. We denote the two forces F1 and F2 . According to Newton's second law, F1 + F2 = ma , so F2 = ma - F1 . (a) In unit vector notation F1 = 20.0 N i and
a = - (12.0 sin 30.0 m/s 2 ) ^ - (12.0 cos 30.0 m/s 2 ) ^ = -
Chapter 9 Student Solutions Manual 15. We need to find the coordinates of the point where the shell explodes and the velocity of the fragment that does not fall straight down. The coordinate origin is at the firing point, the +x axis is rightward, a
Chapter 14 Student Solutions Manual 1. The pressure increase is the applied force divided by the area: p = F/A = F/r2, where r is the radius of the piston. Thus p = (42 N)/(0.011 m)2 = 1.1 105 Pa. This is equivalent to 1.1 atm. 3. The air inside pu
Chapter 16 Student Solutions Manual 15. The wave speed v is given by v = , where is the tension in the rope and is the linear mass density of the rope. The linear mass density is the mass per unit length of rope:
= m/L = (0.0600 kg)/(2.00 m) =
Chapter 17 Student Solutions Manual 5. Let tf be the time for the stone to fall to the water and ts be the time for the sound of the splash to travel from the water to the top of the well. Then, the total time elapsed from dropping the stone to hear
Chapter 19 Student Solutions Manual 7. (a) In solving pV = nRT for n, we first convert the temperature to the Kelvin scale: T = (40.0 + 273.15) K = 313.15 K . And we convert the volume to SI units: 1000 cm3 = 1000 106 m3. Now, according to the idea
Chapter 18 Student Solutions Manual 8. The change in length for the aluminum pole is
=
0
A1T = (33m)(23 10 -6 / C)(15 C) = 0.011m.
15. If Vc is the original volume of the cup, a is the coefficient of linear expansion of aluminum, and T is the t
Chapter 4 Student Solutions Manual 11. We apply Eq. 4-10 and Eq. 4-16. (a) Taking the derivative of the position vector with respect to time, we have, in SI units (m/s),
v= d ^ ^ (i + 4t 2 ^ + t k) = 8t ^ + k . j j ^ dt
(b) Taking another derivativ
13168-12-115P
AID: 5204 | 29/09/2014
Draw the free body diagram of the beam.
Figure-1
Draw the elastic curve diagram.
The point load is acting between the points B and C of the beam. Thus the elastic curve
changes at B. Therefore, the elastic curve occurs