Calculator techniques 2

Time Problem ●

What time after 3:00 o'clock will the minute-hand and the hour-hand of the clock be together for the first time Mode 3 2 Input 0 15 60 -40 0 Shift 1

5

16.36 Answer 3:16.36

4 0X

Interpolation ●

Data

●

–

210

340

–

220

___

–

235

750

Mode 3 2 –

Input ●

●

–

210 235

220Y = 504

340 750

Interpolation ●

Data

●

–

2.10

2.111

–

___

2.123

–

2.20

2.233

Mode 3 2 –

Input ●

●

–

2.10 2.20

2.111 2.233

2.123X = 2.105

Flow ●

Water is poured into a conical tank at the rate of 2.15 cubic meters per minute. The tank is 8 meters in diameter across the top and 10 meters high. How fast the water level rising when the water stands 3.5 meters deep. Courtesy of Mathalino

Flow

Flow ●

Mode 3 –

0

0

–

5

π22

–

10 π42

3

●

2.15 - 3.5Ý

●

.3492

Component forces to resultant ●

Given

●

X = 200

●

Y = 300

●

Mode 2 –

200 + 300i

–

Shift 2 3

–

360.55, 56.31

Multi-Component ●

Given

●

x

y

–

100

200

–

20

52

–

56

-220

●

Mode 2 100 + 200i + 20 + 52i + 56 -220i

●

176 + 32i

●

Shift 2 3

●

178.88, 10.30

Centroid ●

Mode Mode 3 2 –

–

–

-1 -2-1 1

2x2 3x1 3x3/2

–

–

Σxy/Σy = -0.74

–

●

Mode Mode 3 2 –

–

–

1+1 .5 1

2x2 3x1 3x3/2

–

–

Σxy/Σy = 1.21

Components of 3 dimension A force AB of 100 N passes from A(0,0,0) to point B(3,4,5). Find the component of the force AB and the angle

4

3

5

Components of 3 dimension ENTER: Mode 8 1 1 ●

Input [ 3

●

ENTER AC

●

5] 4

ENTER –

●

4

Vct A / Abs(Vct A)

ANSWER –

0.424

0.565

–

Ans x 100

–

42.4

56.5

0.707 70.7

3

5

Sum of Component Forces ●

300

Find the centroid of the following parallel forces

400 100 200 2

2

O

6 4 2

Sum of Component Forces ●

Lets do some more tweak in stat mode:

●

Enable Frequency –

Shift Mode (down) 4

●

Mode 3 2

●

Input

●

–

X Y

Freq

–

2

0

100

–

6

2

200

–

0

6

300

–

4

6

400

400 100 200 2

3000 (Moment on x)

Shift 1 3 4 (sum of Y) –

●

1

O

6 4

Shift 1 3 2 (sum of x) –

●

300

4600 (Moment on Y)

Therefore –

X = 3000/1000 = 3

–

Y = 4600/1000 = 4.6

2

2

Sample Problems ●

Determine the magnitude of the resultant hydrostatic force acting on the surface of a seawall shaped in the form of a parabola as shown The wall is 5 m long: (dw = 1020 kg/m3 .

●

wb = 1020 x 9.81 x 5 = 150.1 kN/m

Sample Problems ●

Fh = ½ (150)(3) = 225.1 KN

●

Fv = Aw(9.81) = 1/3 (3)(1)(5)(9.81) –

= 49.01 kN

●

For Resultant

●

Pol(225.1,49.01) –

(230,12.28)

–

Same as sqrt(225.1^2 + 49.01^2)

Sample Problem ●

Given the following infiltration Data solve for the value of hydraulic conductivity constant K and constant m for the model of the infiltration depth Z.

Z =Kt

m

T

Depth 1

11

3

18

7

26

15

37

Sample Problem ●

Mode 3 –

11

1 3 7 15

3

18

7

26

15

37

●

●

11 18 26 37

Shift 1 5 ●

1

A = 10.99

Shift 1 5 ●

–

Depth 1

●

–

T

Input Data ●

–

7

2

B = 0.44

A = K, m = b

Note Shift 1 if for Stat, Shift 2 for CMPLX, etc

The regular way of solving this problem it to transform the equation to linear equation ln Z = lnK + m ln t Graph solve for slope and intercept I = ln k

Sample Problem ●

What is the future amount of 4500 with interest rate of 12% compounded quarterly in 5 years

●

I forgot the formula

●

Analyze –

1st quarter money 4500 + (4500x.12/4) = 4635 Data ●

●

●

●

●

●

F = P(1 + i/m)^mi 8127

0 4500 1 4635 5x4 ?

Mode 3

6

–

0

4500

–

1

4635

●

20 Y = 8127 (20 shift 1

●

8127

5

5)

Sample Problem ●

What is the amount deposit 5 years agro if the present amount is 15000 if the interest rate is 15% compounded monthly.

●

●

I forgot the formula (na naman)

Isip-isip AHA 5 years is 60 mths In the 59th mth its 15,000 -15,000*.15/12=14,812.5 Data will be 60 15000 59 14,812.5 What is zero Mode stat 3 6 Enter Data 60 15000 59 14,812.5 0Y = 7052.08 –

–

–

●

●

P=[

F

mt

] i 1+ m 1500 12 P=[ ] ∗5 .15 1+ 12 P=7118.81

–

–

●

–

–

●

Sample Problem ●

Replace the system shown by an equivalent force couple force acting on point A, given the F = 100 N and C = 120 N

E

D A 2m

4m

O

C 4m

F B

Sample Problem Two Forces acting on 3 vectors ●

Setting conditions –

Forces ●

●

–

A 2m

4m

O

C=120 F=100

For couple on A (Moment Ax) ●

●

–

For F (-4,0,2) For C (4,4,-2)

E

D

C (none) F (0,4,-2)

4m

B

Sample Problem So we have 3 vectors ●

●

–

Shift 5 2(enter vectors)

–

VctA (-4,0,2) for force F

–

VctB (4,4,-2) for Force C

–

VctC ((0,4,-2) for couple

–

●

A

Summing for couples –

VctC 100(VctA % abs(VctA) + 120(VctB % abs(VctB)

2m

4m

O

C=120 F=100

(258.88 258.88 317.77)

Enter Abs(VctAns) –

E

D

Mode 8

484.79

4m

B

●

P = 1000

●

I = 15%

●

T=3

●

Monthly = m =12

●

●

P(1+i/m)^(mt) Y(1+A%B)^BxX

What you learned are all just Techniques!!! Tools you need to solve some problems... Knowing the problem/ Understanding them Is more critical than using the tools... Always use the right Tools in Solving Problems That's Engineering 101 RMA

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