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0.10 Simultaneous equations  (Page 2/4)

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Diagram of a torsion beam.

Figure 1. Torsion beam.

In this figure, we observe a situation in which forces are applied at each end of the beam. We also note that each force is applied at a right angle to the beam. Under such conditions, torque can be calculated as the product of the force and the length of the beam from its end to its support point atop the pivot point of the beam.

The force ( F 1 ) contributes a counter-clockwise torque ( T CCW ) on the beam. The value of this torque can be calculated

T CCW = F 1 × R 1 size 12{T rSub { size 8{ ital "CCW"} } =F rSub { size 8{1} } times R rSub { size 8{1} } } {}

The force ( F 2 ) contributes a clockwise torque ( T CW ) on the beam. The value of this torque can be calculated as

T CW = F 2 × R 2 size 12{T rSub { size 8{ ital "CW"} } =F rSub { size 8{2} } times R rSub { size 8{2} } } {}

Suppose that the beam is not in motion. Engineers would say that the beam is static or that the beam system is in equillibrium. A necessary condition for equillibrium of the system is that the two torques be equal, that is

T CW = T CCW size 12{T rSub { size 8{ ital "CW"} } =T rSub { size 8{ ital "CCW"} } } {}

We now will apply what we know about simultaneous equations to solve a problem involving a torsion beam.

Example 2: Assume that a a beam of length 56 inches is supported by a fulcrum. Force is applied at each end of the beam. Assume that the force ( F 1 ) on the left hand side of the beam is 24 lbs while the force on the right hand side of the beam ( F 2 ) is 32 lbs. If the beam is static, then what are the values of R 1 and R 2 ?

We know from that the clockwise and counter-clockwise torques must balance as a condition for equilibrium. This leads to the first equation that we will use to solve this problem

24 R 1 = 32 R 2 size 12{"24"`R rSub { size 8{1} } ="32"`R rSub { size 8{2} } } {}

which can be written as

24 R 1 − 32 R 2 = 0 size 12{"24"`R rSub { size 8{1} } - "32"`R rSub { size 8{2} } =0} {}

We also know that the overall length of the beam is 56 inches. Therefore,

R 1 + R 2 = 56 size 12{R rSub { size 8{1} } +R rSub { size 8{2} } ="56"} {}

which can be solved for R 1 as

R 1 = 56 − R 2 size 12{R rSub { size 8{1} } ="56" - R rSub { size 8{2} } } {}

We may substitute the expression for R 1 in equation () to obtain

24 ( 56 − R 2 ) − 32 R 2 = 0 size 12{"24"` \( "56" - R rSub { size 8{2} } \) - "32"`R rSub { size 8{2} } =0} {}

Multiplication and rearrangement yield the equation

1, 344 = 56 R 2 size 12{1,"344"="56"`R rSub { size 8{2} } } {}

which can be solved for R 2 .

R 2 = 24 in size 12{R rSub { size 8{2} } ="24"` ital "in"} {}

The value for R1 can be easily found as

R 1 = 56 − R 2 = 56 − 24 = 32 inches size 12{R rSub { size 8{1} } ="56" - R rSub { size 8{2} } ="56" - "24"="32"~ ital "inches"} {}

Parallel processing

Parallel processing is a term drawn from Compute Science that involves the simultaneous use of more than one central processing unit (CPU) to execute a program. Ideally, parallel processing makes programs run faster because there are more than one processors in use to execute the program. Though most computers have just one CPU, newer computer architectures that feature several CPU’s are becoming the norm more prevalent.

Example 3: Suppose that we are interested in implementing a computer algorithm on two processors. Let us call these processors A and B. The steps involved in the accomplishment of the algorithm are divided among the two processors. In all there are 17,000,000 computations which will need to be performed on the two processors. Working in parallel, the two processors accomplish the algorithm. In doing so, the processor A is in service for 3 seconds while the processor B is in services for 2 seconds.

Suppose that a second algorithm is to be accomplished by processors A and B working in parallel. In all, the second algorithm requires the execution of 15,500,000 computations. In executing the second algorithm, 2 seconds of computing time are required of processor A while 3 seconds of computing time are required of processor B.
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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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