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8_murik_8 [283]
3 years ago
6

A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can

resist a force of 5000 lb on the other end of the hawser. Determine (a) the coefficient of static friction between the hawser and the bollard, (b) the number of times the hawser should be wrapped around the bollard if a 20,000-lb force is to be resisted by the same 80-lb force
Engineering
1 answer:
Brut [27]3 years ago
4 0

Answer:

μ=0.329, 2.671 turns.

Explanation:

(a)   ln(T2/T1)=μβ         β=angle of contact in radians

take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π  radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π  

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb     T1=80 lb   μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

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Create a program named PaintingDemo that instantiates an array of eight Room objects and demonstrates the Room methods. The Room
Serggg [28]

Answer:

Explanation:

Code used will be like

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

namespace PaintingWall

{

class Room

{

public int length, width, height,Area,Gallons;

public Room(int l,int w,int h)

{

length = l;

width = w;

height = h;  

}

private int getLength()

{

return length;

}

private int getWidth()

{

return width;

}

private int getHeight()

{

return height;

}

public void WallAreaAndNumberGallons()

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Area = getLength() * getHeight() * getWidth();

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}

class PaintingDemo

{

static void Main(string[] args)

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int l, w, h;

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for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room "+(i+1));

Console.Write("Enter Length : ");

l = Convert.ToInt32(Console.ReadLine() );

Console.Write("Enter Width : ");

w = Convert.ToInt32(Console.ReadLine());

Console.Write("Enter Height : ");

h= Convert.ToInt32(Console.ReadLine());

r[i] = new Room(l,w,h);

Console.WriteLine();

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for (int i = 0; i <= 7; i++)

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3 0
3 years ago
A cylindrical metal specimen having an original diameter of 11.34 mm and gauge length of 53.3 mm is pulled in tension until frac
WINSTONCH [101]

Answer:

a) 70.29 %

b) 37%

Explanation:

percent reduction can be found from:

PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2

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     =70.29 %

percent elongation can be found from:

EL =L_f - Lo/Lo*100

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3 years ago
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Answer:

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Hope this helped you!

Explanation:

5 0
3 years ago
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