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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