All categories

3 years ago

How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.

Engineering

1 answer:

Snowcat [4.5K]3 years ago

3 0

Answer:

$N_3=\dfrac{T_1}{T_3}N_1$

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

Lets take

$Speed\ of\ gear 1=N_1$

$Number\ of\ teeth\ of\ gear 1=T_1$

$Speed\ of\ gear 3=N_3$

$Number\ of\ teeth\ of\ gear 3=T_3$

All external matting gears will rotates in opposite direction with respect to each other.

So the speed of gear third can be given as follows

$\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}$

$N_3=\dfrac{T_1}{T_3}N_1$

You might be interested in

Why is it better for a CPU to have more than one cache?

Tomtit [17]

Answer:

In general a cache memory is useful because the speed of the processor is higher than the speed of the ram . so reducing the number of memory is desirable to increase performance .

Explanation:

#hope it helps you ..

(◕ᴗ◕)

3 0

2 years ago

"Using your favorite search engine and the resources of your library, develop a set of recommendations regarding the possible us

alina1380 [7]

Answer:

Cloud computing services are going to be very important to supporting Ashville's mobile app since it has different uses. These applications uses include; data storage, as well as helping to provide network to any business data related work. This will improve the city of Ashville's mobile app operations considerably.

The benefits of using cloud computing are; it allows a business managers to be more concerned with running the business themselves rather than maintaining the data center. The use of cloud computing allows a company's IT administrators to focus on managing the company's operations, thereby allowing performance to be enhanced of the business enterprise.

A commercial company is able to create new technologies more rapidly with the use of cloud computing. Furthermore, the enterprise will be able to automate its activities using cloud computing. Cloud computing is also very important as it is more affordable and thus promotes the company's growth in the market. Cloud computing is also enhancing the global presence of the Ashville mobile app. The drawbacks of using cloud computing include the following; due to using the cloud infrastructure, the company's performance may be unreliable. People believe that cloud computing is not reliable and that cloud computing may not be secure at last and can't always be right for all workloads.

Explanation:

5 0

3 years ago

(40 points) Program the following sorting algorithms: InsertionSort, MergeSort, and QuickSort. There are 9 test les uploaded for

babunello [35]

Answer:

Explanation:

MERGE SORT

#include<stdlib.h>

#include<stdio.h>

#include<string.h>

void merge(int arr[], int l, int m, int r)

{

int i, j, k;

int n1 = m - l + 1;

int n2 = r - m;

int L[n1], R[n2];

for (i = 0; i < n1; i++)

L[i] = arr[l + i];

for (j = 0; j < n2; j++)

R[j] = arr[m + 1+ j];

i = 0;

j = 0;

k = l;

while (i < n1 && j < n2)

{

if (L[i] <= R[j])

{

arr[k] = L[i];

i++;

}

else

{

arr[k] = R[j];

j++;

}

k++;

}

while (i < n1)

{

arr[k] = L[i];

i++;

k++;

}

while (j < n2)

{

arr[k] = R[j];

j++;

k++;

}

void mergeSort(int arr[], int l, int r)

{

if (l < r)

{

int m = l+(r-l)/2;

mergeSort(arr, l, m);

mergeSort(arr, m+1, r);

merge(arr, l, m, r);

}

void printArray(int A[], int size)

{

int i;

for (i=0; i < size; i++)

printf("%d ", A[i]);

printf("\n");

}

int main()

{

int arr[1000] = {0};

int arr_size =0;

int data;

char file1[20];

strcpy(file1,"data.txt");

FILE *fp;

fp = fopen(file1,"r+");

if (fp == NULL) // if file not opened return error

{

perror("Unable to open file");

return -1;

}

else

{

fscanf (fp, "%d", &data);

arr[arr_size]=data;

arr_size++;

while (!feof (fp))

{

fscanf (fp, "%d", &data);

arr[arr_size]=data;

arr_size++;

}

printf("Given array is \n");

printArray(arr, arr_size);

mergeSort(arr, 0, arr_size - 1);

printf("\nSorted array Using MERGE SORT is \n");

printArray(arr, arr_size);

return 0;

}

3 0

3 years ago

Engineering is the use of scientific principles to design and build machines, structures, and other items, including bridges, tu

Thepotemich [5.8K]

Is this a question or a statement?

4 0

3 years ago

A 5-in.-diameter pipe is supported every 9 ft by a small frame consisting of two members asshown. Knowing that the combined weig

jarptica [38.1K]

Answer:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

Explanation:

Given:

- The diameter of the pipe d = 5-in

- The pipe is supported every L = 9 ft of pipe in length

- The weight if the pipe + contents W = 10 lb/ft

Find:

determine the magnitude and location of the maximum bending moment in members AC and BC.

Solution:

- The figure (missing) is given in the attachment.

- We will first determine the external forces acting on each member:

Section: 9-ft section of pipe.

Sum of forces perpendicular to member AC = 0

F_d - 0.8*W*L = 0

F_d = 0.8*10*9 = 72 lb

Sum of forces perpendicular to member BC = 0

F_e - 0.6*W*L = 0

F_e = 0.6*10*9 = 54 lb

F_d = 72 lb , F_e = 54 lb

- Then we will determine the support reactions for each member AC point A and BC point B.

Section: Entire Frame.

Sum of moments about point B = 0

-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0

-A_y*(1.5625) + 15 + 39.375 = 0

A_y = 34.8 lb

Sum of forces in vertical direction = 0

A_y + B_y - 0.8*F_d - 0.6*F_e = 0

B_y = 0.8*(72) + 0.6*(54) - 34.8

B_y = 55.2 lb

Sum of forces in horizontal direction = 0

A_x + B_x - 0.6*F_d + 0.8*F_e = 0

A_x + B_x = 0

Section: Member AC

Sum of moments about point C = 0

F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0

72*2.5 - 34.8*12 - 9*A_x = 0

A_x = -237.6 / 9 = - 26.4 lb

B_x = - A_x = 26.4 lb

A_x = -26.4 lb , B_x = 26.4 lb

- Now we will calculate bending moment for each member at different sections.

Member AC:

From point A till just before point D

-0.6*A_x*x - A_y*0.8*x + M = 0

15.84*x - 27.84*x + M = 0

M = 12*x ..... max value at D, x = 12.25 in

M_max = 12*12.25/12 = 12.25 lb-ft

Member BC:

From point B till just before point E

-0.8*B_x*x + B_y*0.6*x + M = 0

-21.12*x + 33.12*x + M = 0

M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in

M_max = -12*8.75/12 = -8.75 lb-ft

- The maximum bending moments and their locations are:

AC: at D , M_max = 12.25 lb-ft

BC: at E , M_max = 8.75 lb-ft

5 0

3 years ago