Motherboards have one floppy connector, and many newer motherboards support only one floppy drive. A computer can have no more than two floppy drives. About the only need these days for two floppy drives is if the user has 5.25-inch disks to access. In this case, the 3.5-inch drive should be A and the 5.25-inch should be B. A small power connector (see Figure 6.14) is used for 3.5-inch floppy drives. A standard floppy drive ribbon cable has three connectors: one for A, one for B, and one for the motherboard connector (see Figure 6.15). You’ll see that a floppy cable has a twist. The twist should be nearest the A drive. If there is some compelling reason to switch drive letters, such as having to use the B drive as a boot drive (only A can be a boot drive), this should be possible in the BIOS. The other option is to change jumpers, but they differ from drive to drive and tend to be confusing.

Figure 6.14: The large connector is for hard drives, optical drives, and 5.25-inch floppy drives. The small connector is only for 3.5-inch floppy drives.

Figure 6.15: One floppy cable with a twist.

Just about every case has a space dedicated to a floppy drive. Look for a 3.5-inch bay that corresponds to an opening in the front of the case (shown in Figure 6.16). Just as in hard drives, you’ll have to match up pins 1. Pin 1 on the cable has a stripe, and there will be some type of marking on the drive as shown in Figure 6.17, and on the motherboard, as shown in Figure 6.18. The drive markings can be cryptic, but know that if pins 0 and 34 are marked, pin 1 will be next to pin 0, and on the opposite side from pin 34. Other drives have only a red mark. Once you ascertain which side pin 1 is on, it is a good idea to mark it on the drive with a fine-tipped permanent marker. However, damage won’t occur if you make a mistake with the data cable; if the floppy drive light stays on continuously, it means that one end of the data cable is in backward.

Figure 6.16: A bay usable for floppy drive.

Figure 6.17: Pin 1 markings can be vague.

Figure 6.18: Pin 1 marking on the motherboard.

Finally, make sure that the BIOS is set correctly for the floppy drive(s) in the system.

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Left part:     ?F_x = – F₁ + A_x = 0
(OA)          ?F_y = P₁ + A_y – F₂ = 0
?M₀ = P₁c + A_y(c + d) + M_A – F₂a = 0
Right side: (AB)        ?Fx = – Ax + P3 = 0
?F_y = P₂ – A_y – F₃ = 0
?M_A = -M_A + P₂e + M – F₃? = 0

Six unknowns (P₁, P₂, P₃, A_x, A_y, M_A) are in six equations.
Note: In the ?rst diagram (Figure 1.2.13) the couple M may be moved
anywhere from O to B. M is not shown in the second diagram (O to A)
because it is shown in the third diagram (in which it may be moved
anywhere from A to B).

Example 3

The arm of a factory robot is modeled as three bars (Figure 1.2.15) with
coordinates A: (0.6, 0.3, 0.4) m; B: (1, 0.2, 0) m; and C: (0.9, 0.1, 0.25) m.
The weight of the arm is represented by W_A = 60 Nj at A, and W_B = 40 Nj
at B. A moment M_C = (100i 20j + 50k) N ? m is applied to the arm at C.
Determine the force and moment reactions at O, assuming that all joints
are temporarily ?xed.

FIGURE 1.2.15 Model of a factory robot.

Solution. The free-body diagram is drawn in Figure 1.2.15b, showing the
unknown force and moment reactions at O. From Equation 1.2.13,

?F = 0
F₀ + W_A + W_B = 0
F₀ – 60 Nj – 40 Nj = 0
F_O = 100 Nj
?M₀ = 0
M₀ + M_C + (r_0A x W_A) + (r_0B x W_B) = 0

M₀ + (100i – 20j + 50k)N?m +(0.6i – 0.3j + 0.4k)m x(- 60 Nj) +
(i – 0.2j) m x(- 40 Nj) = 0

M₀ + 100 N?m i – 20 N?m j + 50 N?m k – 36 N?m k + 24 N?m i -
40 N?m k = 0

M₀ = (124i + 20j + 26k) N?m

Example 4

A load of 7 kN may be placed anywhere within A and B in the trailer of
negligible weight. Determine the reactions at the wheels at D, E, and F,
and the force on the hitch H that is mounted on the car, for the extreme
positions A and B of the load. The mass of the car is 1500 kg, and its
weight is acting at C (see Figure 1.2.16).

FIGURE 1.2.16 Analysis of a car with trailer.

Solution. The scalar method is best here.

Put the load at position A ?rst
For the trailer alone, with y as the vertical axis
?M_F = 7(1) H_y(3) = 0, H_y = 2.33 kN
On the car
H_y = 2.33 kN  Ans.
?F_y = 2.33 7 + F_y = 0, F_y = 4.67 kN  Ans.
For the car alone
?M_E = 2.33(1.2) D_y(4) + 14.72(1.8) = 0
D_y = 5.93 kN  Ans.
?F_y = 5.93 + E_y 14.72 2.33 = 0
E_y = 11.12 kN  Ans.
Put the load at position B next
For the trailer alone
?M_F = 0.8(7) H_y(3) = 0, H_y = 1.87 kN
On the car
H_y = 1.87 kN  Ans.
?F_y = 1.87 7 + Ey = 0
E_y = 8.87 kN  Ans.
For the car alone
?M_E = (1.87)(1.2) Dy(4) + 14.72(1.8) = 0
D_y = 7.19 kN  Ans.
?F_y = 7.19 + E_y 14.72 (1.87) = 0
E_y = 5.66 kN  Ans.

Forces and Moments in Beams

Beams are common structural members whose main function is to resist
bending. The geometric changes and safety aspects of beams are analyzed by
?rst assuming that they are rigid.  The preceding sections enable one to
determine (1) the external (supporting) reactions acting on a statically
determinate beam, and (2) the internal forces and moments at any cross
section in a beam.

Classi?cation of Supports

Common supports and  external reactions for two-dimensional loading of
beams are shown in Figure 1.2.17.

FIGURE 1.2.17 Common beam supports.

Internal Forces and Moments

The internal force and moment reactions in a beam caused by external
loading must be determined for evaluating the strength of the beam. If there
is no torsion of the beam, three kinds of internal reactions are possible:
a horizontal normal force H on a cross section,  vertical (transverse) shear
force V, and bending moment M. These reactions are calculated from the
equilibrium equations applied to the left or right part of the beam from the
cross section considered. The process involves free-body diagrams
of the beam and a consistently applied system of signs. The modeling is
illustrated for a cantilever beam in Figure 1.2.18.

FIGURE 1.2.18 Internal forces and moments in a cantilever beam.

Sign Conventions. Consistent sign conventions should be used in any
given problem. These could be arbitrarily set up, but the following is slightly
advantageous. It makes the signs of the answers to the equilibrium equations
correct for the directions of the shear force and bending moment.
A moment that makes a beam concave upward is taken as positive.
Thus, a clockwise moment is positive on the left side of a section, and a
counterclockwise moment is positive on the right side. Ashear force that
acts upward on the left side of a section, or downward on the right side, is
positive (Figure 1.2.19).

FIGURE 1.2.19 Preferred sign conventions.
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