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Montano1993 [528]
3 years ago
14

Consider the cantilever-beam Wheatstone bridge system that has four strain gages (two in compression and two in tension). Which

of the following statements is not true: (a) the change in resistance in each gage is proportional to the applied force, (b) temperature and torsional effects are automatically compensated for by the bridge, (c) the longitudinal (axial) strain in the beam is proportional to the output voltage of the bridge, (d) a downward force on the beam causes an increase in the resistance of a strain gage placed on its lower (under) side. Final Ans: (d) Compression on a lower side gage causes an increase in its resistance.
Physics
1 answer:
Ivahew [28]3 years ago
3 0

Answer:  (b) temperature and torsional effects are automatically compensated for by the bridge,

Explanation:

As can be seen, strain gages 1 and 4 are on top of the beam and strain gages 2 and 3 are on the bottom of

the beam. Therefore strain gages 1 and 4 experience a tensile strain (are stretched) and strain gages 2

and 3 experience a compressive strain. If the relationship between strain and resistance is linear, then

under some load F the changes in resistance will be

R1= R1 + dR1

R4= R4 + dR4

tensile(4)and

R ¢2= R2 -dR2

R ¢3= R3 -dR3

compressive. (5)

when the four strain gages have an equal nominal

resistance (i.e., R1 = R2 = R3 = R4 = R) then the deflection method Wheatstone bridge equation reduces to the linear equation

Using known weights, a calibration curve can be established that relates the weight W to the output

voltage on a digital meter Eo,

Eo= a0 + a1W , (4)

where a0 and a1 are some constants. Once an unknown weight is known, an unknown mass or density

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Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V=  30 m/s  ( + x direction )

d)  λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

As we know that standard form of wave equation given as

y=A sin(\phi -\omega t)

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

So from above equation we can say that

Amplitude ,A = 2 mm

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ω= 2πf

f= ω /2π

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K= 20 rad/m

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V = f λ

30 = 95.49 x  λ

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U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

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Umax = 1200 mm/s

Umax= 1.2 m/s

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4 years ago
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