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Stels [109]
3 years ago
12

Draw the free-body diagram of the beam which supports the 80-kg load and is supported by the

Engineering
1 answer:
Sauron [17]3 years ago
5 0

The free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.

The first image shows the diagram of the beam and the second image shows the free-body diagram of the beam.

The resolution of forces in the system is well understood by the principle of equilibrium where a stationary body will remain balanced when subject to parallel forces provided that the total sum of the overall external forces is zero.

The free-body diagram is a graphical representation used to visualize the forces applied to an object.

The equilibrium of forces on the x-axis is:

\mathbf{\sum F_x  = 0}

The equilibrium of forces on the y-axis is:

\mathbf{\sum F_y = 0}

The equilibrium condition at any point is:

\mathbf{\sum M = 0}

From the free body diagram attached in the second image below,

  • the horizontal reaction is located at point A as \mathbf{ A_x}
  • the vertical reaction is  located at point A  as \mathbf{A_y}
  • the tension =  T
  • the weight = W

Therefore, we can conclude that the free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.

Learn more about the free-body diagram here:

brainly.com/question/19345060?referrer=searchResults

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In part A you are asked to write the pseudocode for the program. In part B you are asked to write the syntax of the code for the
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Answer:

C++.

Explanation:

#include <iostream>

#include <string>

using namespace std;

///////////////////////////////////////////////////////////////

int main() {

   string quote, book;

   int page;

   

   cout<<"What is your favorite quote from a book?"<<endl;

   getline(cin, quote);

   cout<<endl;

   /////////////////////////////////////////////

   cout<<"What book was that quote from?"<<endl;

   getline(cin, book);

   cout<<endl;

   /////////////////////////////////////////////

   cout<<"What page was that quote from?"<<endl;

   cin>>page;

   cout<<endl;

   /////////////////////////////////////////////

   int no_of_upper_characters = 0;

   for (int i=0; i<quote.length(); i++) {

       if (isupper(quote[i]))

          no_of_upper_characters++;

   }

   

   cout<<"No. of upper case characters: "<<no_of_upper_characters<<endl;

   /////////////////////////////////////////////

   int no_of_characters = quote.length();

   cout<<"No. of characters: "<<no_of_characters<<endl;

   /////////////////////////////////////////////

   bool isDog = false;

   for (int i=0; i<quote.length(); i++) {

       if (isDog == true)

           break;

       else if (quote[i] == 'd') {

           for (int j=i+1; j<quote.length(); j++) {

               if (isDog == true)

                   break;

               else if (quote[j] == 'o') {

                   for (int z=j+1; z<quote.length(); z++) {

                       if (quote[z] == 'g') {

                           isDog = true;

                           break;

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                   }

               }

           }

       }

   }

   

   if (isDog == true)

       cout<<"This includes 'd' 'o' 'g' in the quote";

   //////////////////////////////////////////////

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}

3 0
3 years ago
A construction crew lifts approximately 400 lb. of material several times during a day from a flatbed truck to a 25 ft. rooftop.
Irina18 [472]

Answer:

2ib

Explanation:

if you divide 10 divided by 2 it gives you 5 and then subtract it by 2.2 = 2.8

there goes your answer.

5 0
2 years ago
Should the ship breaking business continue why or why not?
Dmitry [639]

Answer:

Ship-breaking or ship demolition is a type of ship disposal involving the breaking up of ships for either a source of parts, which can be sold for re-use, or for the extraction of raw materials, chiefly scrap. It may also be known as ship dismantling, ship cracking, or ship recycling. Modern ships have a lifespan of 25 to 30 years before corrosion, metal fatigue and a lack of parts render them uneconomical to operate.[1] Ship-breaking allows the materials from the ship, especially steel, to be recycled and made into new products. This lowers the demand for mined iron ore and reduces energy use in the steel making process. Fixtures and other equipment on board the vessels can also be reused. While ship-breaking is sustainable, there are concerns about the use of poorer countries without stringent environmental legislation. It is also labor-intensive, and considered one of the world's most dangerous industries.[2]

In 2012, roughly 1,250 ocean ships were broken down, and their average age was 26 years.[3][4] In 2013, the world total of demolished ships amounted to 29,052,000 tonnes, 92% of which were demolished in Asia. As of January 2020, India has the largest global share at 30%;[5] followed by Bangladesh, China and Pakistan.[6] Alang, India currently has the world's largest ship graveyard,[5] followed by Chittagong Ship Breaking Yard in Bangladesh and Gadani in Pakistan.[6]

The largest sources of ships are states of China, Greece and Germany respectively, although there is a greater variation in the source of carriers versus their disposal.[7] The ship-breaking yards of India, Bangladesh, China and Pakistan employ 225,000 workers as well as providing many indirect jobs. In Bangladesh, the recycled steel covers 20% of the country's needs and in India it is almost 10%.[8]

As an alternative to ship-breaking, ships may be sunk to create artificial reefs after legally-mandated removal of hazardous materials, or sunk in deep ocean waters. Storage is a viable temporary option, whether on land or afloat, though all ships will be eventually scrapped, sunk, or preserved for museums.

6 0
3 years ago
It has a piece of 1045 steel with the following dimensions, length of 80 cm, width of 30 cm, and a height of 15 cm. In this piec
Serggg [28]

Answer:

material remove in 3 min is 16790.4 mm³/s

Explanation:

given data

length L = 80 cm = 800 mm

width W = 30 cm

height H = 15 cm

make grove length = 80 cm

width = 8 cm

depth = 10 cm

mill toll diameter = 4 mm

axial cutting depth = 20 mm

to find out

How much material removed in 3 minutes

solution

first we find time taken for length of advance that is

time = \frac{length}{advance}

here advance is given as 0.001166 mts / sec

so  time = \frac{800}}{0.001166*1000}

time = 686.106 seconds

now we find material remove rate that is

remove rate = mill toll rate × axial cutting depth × advance

remove rate = 4 × 20×0.001166 ×1000

remove rate = 93.28 mm³/s

so

material remove in 3 minute = 3 × 60 = 180 sec

so material remove in 3 min = 180 × 93.28

material remove in 3 min is 16790.4 mm³/s

7 0
3 years ago
In your first job with a large U.S based steel company, you have been assigned to a team tasked with developing a new low carbon
nignag [31]

Answer:

Option A

Explanation:

3 0
4 years ago
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