GIVEN:
Amplitude, A = 0.1mm
Force, F =1 N
mass of motor, m = 120 kg
operating speed, N = 720 rpm
=
Formula Used:
Solution:
Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K
We know that:
so,
= 75.39 rad/s
Using the given formula:
Damping is negligible, so,
will give the tranfer function
Therefore,
=
=
Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm
Answer:
a. 0.11
b. 110 students
c. 50 students
d. 0.46
e. 460 students
f. 540 students
g. 0.96
Explanation:
(See attachment below)
a. Probability that a student got an A
To get an A, the student needs to get 9 or 10 questions right.
That means we want P(X≥9);
P(X>9) = P(9)+P(10)
= 0.06+0.05=0.11
b. How many students got an A on the quiz
Total students = 1000
Probability of getting A = 0.11 ---- Calculated from (a)
Number of students = 0.11 * 1000
Number of students = 110 students
So,the number of students that got A is 110
c. How many students did not miss a single question
For a student not to miss a single question, then that student scores a total of 10 out of possible 10
P(10) = 0.05
Total Students = 1000
Number of Students = 0.05 * 1000
Number of Students = 50 students
We see that 5
d. Probability that a student pass the quiz
To pass, a student needed to get at least 6 questions right.
So we want P(X>=6);
P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)
=0.08+0.12+0.15+0.06+0.05=0.46
So, the probability of a student passing the quiz is 0.46
e. Number of students that pass the quiz
Total students = 1000
Probability of passing the quiz = 0.46 ----- Calculated from (d)
Number of students = 0.46 * 1000
Number of students = 460 students
So,the number of students that passed the test is 460
f. Number of students that failed the quiz
Total students = 1000
Total students that passed = 460 ----- Calculated from (e)
Number of students that failed = 1000 - 460
Number of students that failed = 540
So,the number of students that failed is 540
g. Probability that a student got at least one question right
This means that we want to solve for P(X>=1)
Using the complement rule,
P(X>=1) = 1 - P(X<1)
P(X>=1) = 1 - P(X=0)
P(X>=1) = 1 - 0.04
P(X>=1) = 0.96
Answer:
F = 8849 N
Explanation:
Given:
Load at a given point = F = 4250 N
Support span = L = 44 mm
Radius = R = 5.6 mm
length thickness of tested material = 12 mm
First compute the flexural strength for circular cross section using the formula below:
σ
σ = FL / π R³
Putting the given values in the above formula:
σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³
= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³
= 187 / 3.141593 (0.0056)³
= 338943767.745358
= 338.943768 x 10⁶
σ = 338 x 10⁶ N/m²
Now we compute the load i.e. F from the following formula:
= 2 σ
d³/3 L
F = 2σd³/3L
= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)
= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)
= 146016 / 125 / 3 ( 44 x 1 / 1000 )
= ( 146016 / 125 ) / (3 ( 11 / 250 ))
= 97344 / 11
F = 8849 N