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lesya692
21 day ago
8

15.27 An 18-tooth pinion with a diametral pitch of 6 rotates 1800 rpm and drives a 36-tooth gear at 900 rpm in a gear speed redu

cer. The pinion and gear with 20 ° full-depth involute teeth are keyed to shafts that are simply supported by bearings. The bearings on each shaft are 2.0 in. from the gear center. If the gears transmit 0.5 hp, what are the forces on the pinion, gear, and shafts?

Engineering
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A 10-ft-long simply supported laminated wood beam consists of eight 1.5-in. by 6-in. planks glued together to form a section 6 i
pantera1 [306]
Point B is where Q_B = 101.25 \ in^3 shows the highest Q value at section a–a. The missing diagram that was supposed to accompany this question can be found in the attached file. Based on the given information, we must identify the point with the greatest Q value at section a–a. To achieve this, we're working with the attached image. The image reveals that there are 8 blocks stacked vertically on the right-hand side, totaling 12 inches; thus, each block measures 1.5 inches in height. Additionally, the blocks are divided into sections marked as A, B, C, and D. For point A, we use the formula Q representing the moment of Area A. For point B, we set Q as the moment of Area B, and similarly for points C and D. However, point D has no area, resulting in a moment of 0.
3 0
2 months ago
The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr
iogann1982 [368]

Answer:

153.2 J

Explanation:

Let’s first identify our known variables:

mass (m) of the block = 10 kg

distance slid down (i.e., displacement) = 2 m

coefficient of kinetic friction (μk) = 0.2

In the following diagram, if we analyze the force component directed along the displacement, we find

F_x= Fcos 40°

F_x= 100 (cos 40°)

F_x= 76.60 N

The work done by the frictional force is then calculated as:

W = F_x × displacement

W = 76.60 × 2

W = 153.2 J

Therefore, the work done by the frictional force equals 153.2 J

7 0
3 months ago
A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7
mote1985 [299]

Answer:

r=0.31

Ф=18.03°

Explanation:

Provided:

Original diameter of bar = 75 mm

Diameter post-cutting = 73 mm

Average diameter of the bar d= (75+73)/2=74 mm

Average length of uncut chip = πd

Average length of uncut chip = π x 74 =232.45 mm

Thus, cutting ratio r

Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}

r=\dfrac{73.5}{232.45}   r=0.31

Therefore, the cutting ratio equals 0.31.

Now, the shearing angle is given as

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

Next by substituting the values

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }\

Ф=18.03°

Concluding, the shearing angle is 18.03°.

4 0
4 months ago
Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or
Kisachek [356]

Answer:

Here is the JAVA program:

import java.util.Scanner; // to take input from user

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4; // size is fixed to 4 assigned to NUM_VALS

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

//two arrays origList[] and offsetAmount[] get assigned their values

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

String product=""; // variable for storing the product results

for(i = 0; i <= origList.length - 1; i++){

/* iterates from 0 to the end of origList */

/* multiplies each origList entry with the corresponding offsetAmount entry, stores results in product */

product+= Integer.toString(origList[i] *= offsetAmount[i]) + " "; }

System.out.println(product); }}

Output:

80 180 80 400

Explanation:

If you wish to print the product of origList alongside offsetAmount values vertically, this can be done in this manner:

import java.util.Scanner;

public class VectorElementOperations {

public static void main(String[] args) {

final int NUM_VALS = 4;

int[] origList = new int[NUM_VALS];

int[] offsetAmount = new int[NUM_VALS];

int i;

origList[0] = 20;

origList[1] = 30;

origList[2] = 40;

origList[3] = 50;

offsetAmount[0] = 4;

offsetAmount[1] = 6;

offsetAmount[2] = 2;

offsetAmount[3] = 8;

for(i = 0; i <= origList.length - 1; i++){

origList[i] *= offsetAmount[i];

System.out.println(origList[i]);}

}}

Output:

80

180

80

400

The program is shown with the output as a screenshot along with the example's input.

8 0
3 months ago
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