EGR Homework 1

7/23/2019 EGR Homework 1

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Introduction to Engineering

EGR 1301 Section 6

Homework Assignment 1

Robin Vo

Baylor ID 892008866

16 September 2015



Robin VoEGR 1301-

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Problem 4.1

The Cartesian components of a vectorB are shown in Fig. 4.5. If Bx, = 8.6 m

andΔ= 35°, findα, By andB.

Assume∠α+Δis a right angle.

α = 90°− ΔwhereΔ is 35°

α = 90°− 35°

α = 55°

TanΔ =By

B x

By = Bx(tanΔ)

By= 8.6 x Tan35°

By= 6.02

CosΔ=By

B

B = Bycos Δ

= 8.6mcos35

B = 10.49 m

Conclusion: B = 10.49m, By = 6.02 m, andα = 55°

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Robin VoEGR 1301-

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Problem 4.8

A pilot of an aircraft knows that the vehicle in landing configuration will glide

2.0 x 101 km from a height of 2.00 x 10

3 m. A TV transmitting tower is located

in a direct line with the local runway. lf the pilot glides over the tower with

3.0 x 101 m to spare and touches down on the runway at a point 6.5 km from

the base of the tower, how high is the tower?

Convert all measurements to one base unit (meters)

*Note: Not Drawn toScale

20 x 103m

30 m

6500 m

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Robin VoEGR 1301-

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Plane glide distance = 2.0 x 101km (

1000m

1km ¿ = 20 x 10

3m

Plane height = 2.00 x 103m

Extra air space = 3.0 x 10

1

mDistance from base of tower = 6.5km (

1000m

1km) = 6500 m or 6.5 x 10

3m

Ratio of glide distance : Height of plane = Distance from where plane lands :

Height of tower

2.0 x 103m : 20 x 10

3m = 6.5 x 10

3m : h+30m, where h is the height of the

tower

Ratio of glide distance : Height of plane = 1:10

Use the known ratio (1

10¿ to find height of tower.

6500m (1

10¿ = 650m − 30m (spare space) = 620m

Conclusion: The height of the tower is 0.62 km.

Problem 4.16

A block of metal has a 90° notch cut from its lower surface. The notched part

rests on a circular cylinder of diameter 2.0 cm, as shown in Fig. 4.15. If the

lower surface of the part is 1.3 cm above the base plane, how deep is the notch



Robin VoEGR 1301-

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∠ BAC = 90°

O is the center,∴OB and OC = 1cm since diameter is 2cm , radius has to be

1cm

∠ ABO = ∠ ACO=¿ 90°

Δ ABO andΔ ACO are isosceles, each with 2 legs that are 1cm each,∴ the

remaining leg must be 1 x √ 2 .

AO = √ 2 cm

AX= √ 2 + 1(radius)

Depth of notch= √ 2 + 1 – 1.3 cm = 1.1cm

Conclusion:The depth of the notch is approximately 1.1 cm.

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Robin VoEGR 1301-

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Problem 4.18

An aircraft moves through the air with a relative velocity of 3.00 x 102

km/ hat a heading of N30°E. In a 35 km/h wind from the west,

(a) Calculate the true ground speed and heading of the aircraft.

(b) What heading should the pilot fly so that the true heading is N30°E?N

&

!

'

60°

300(m)r

30°

%++l, law o- co.ine. to /n tre ron .+ee !4

!2= 300x1024235272300x1024354co.60°4

!2= 81225710500!2=9025!= 291 (m)r

;re eain: sin 60°284.1 km/hr

= sinθ300km /hr

Sin60 ° 300(m)r4 = 291 Sin θ¿

25891 = 291 Sin θ¿

0815= Sin θ Sin 7108154= 66 °

Conclusion: ;e ron .+ee o- te +lane i. 291 (m)r an te tre

eain i. N66°E

35 (m)r

S

<

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Robin VoEGR 1301-

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Problem 4.30

The light striking a pane of glass is refracted as shown in Fig. 4.20. The law

of refraction states that nasin&a= nb sin&b, where na and nb are the refractive

indexes of the materials through which the light is passing and the angles are

from a Line that is normal to the surface. The refractive index of air is 1.00.

What is the refractive index of the glass?

Apply Snell’s law to find the refractive index of the glass.

Snell’s law: nasin&a= nb sin&b

1.00sin(38 ° ) = nb sin(28 °¿

nb= sin38 °

sin28 °

nb= 1.57

Conclusion:The refraction index of the glass is 1.57.



Robin VoEGR 1301-

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Problem 5.8

Electrical resistance for a given material can be a function of both area/unit

thickness and material temperature. Holding temperature constant, a range

of areas are tested to determine resistance. The measured resistance recorded

in the table is expressed in milliohms per meter of conductor length.

(e) Plot and find equations for parts (a), (b), and (c) using computer-assisted

methods.

(f) What would be the best curve fit for this application?

Conclusion:E) For part A,B, and C the equation for the line would be

y = 122.91e-145x

F) The best curve fit for this application would be an exponential curve as the

curve with the plots shows that there is a trend of the line approaching 0.

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Robin VoEGR 1301-

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Problem 5.18

The rate of absorption of radiation by metal plates varies with the plate

thickness and the nature of the source of radiation. A test was conducted at

Ames Labs on October 11, 2005, using a Geiger counter and a constant source

of radiation; the results are shown in the following table.

C) Repeat parts (a) and (b) using a computer-assisted method.

D) What would you expect the counts per second to be for a 2-in.-thick plate of

the metal used in the test?

C)

Rate of Absorption for Dierent Plate Tic!ness

Equation: y = -144.88x + 4677

D) Convert inches to millimeters

2.00 in (2.54 cm

1 inch)(

10mm

1cm) = 50.8mm

y = -144.88x+4677

y= -144.88(50.8)+4677

y = -2682.904

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Robin VoEGR 1301-

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Conclusion:The expected count would be 0, since you cannot count the

negative count.

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EGR Homework 1