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kompoz
29 days ago
14

3/63 A 2‐kg sphere S is being moved in a vertical plane by a robotic arm. When the angle θ is 30°, the angular velocity of the a

rm about a horizontal axis through O is 50 deg/s clockwise and its angular acceleration is 200 deg/s2 counterclockwise. In addition, the hydraulic element is being shortened at the constant rate of 500 mm/s. Determine the necessary minimum gripping force P if the coefficient of static friction between the sphere and the gripping surfaces is 0.50. Compare P with the minimum gripping force Ps required to hold the sphere in static equilibrium in the 30° position.

Engineering
1 answer:
Kisachek [217]29 days ago
5 0

Answer:

Ps=19.62N

Explanation:

A thorough explanation of the answer can be found in the attached files.

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3.24 Program: Drawing a half arrow (Java) This program outputs a downwards facing arrow composed of a rectangle and a right tria
alex41 [274]

Answer:

This is the JAVA code:

import java.util.Scanner; // to capture user input

public class DrawHalfArrow{ // class declaration

public static void main(String[] args) { // main function starts here

    Scanner scnr = new Scanner(System.in); //input reading

int arrowBaseHeight = 0; // variable for arrow base height

int arrowBaseWidth  = 0; // variable for arrow base width

int arrowHeadWidth = 0; // variable for arrow head width

// prompts user for height, base width, and head width

System.out.println("Enter arrow base height: ");

arrowBaseHeight = scnr.nextInt(); // reads integer input

System.out.println("Enter arrow base width: ");

arrowBaseWidth = scnr.nextInt();

/* loop for asking user for arrow head width until it exceeds base width */

while (arrowHeadWidth <= arrowBaseWidth) {

    System.out.println("Enter arrow head width: ");

    arrowHeadWidth = scnr.nextInt(); }

//begin nested loop for base

//outer loop for outputting based on arrow base height

 for (int i = 0; i < arrowBaseHeight; i++) {

//inner loop produces the stars

      for (int j = 0; j <arrowBaseWidth; j++) {

          System.out.print("*");        } //prints stars

          System.out.println();          }

//used for the width of the arrowhead

int k = arrowHeadWidth;

//outer loop for the head height

for (int i = 1; i <= arrowHeadWidth; i++)

{     for(int j = k; j > 0; j--)     {//inner loop for stars

       System.out.print("*");    } //print stars

   k = k - 1;

   System.out.println(); } } } //marks end of asterisks for a new line

Explanation:

The code prompts for arrow dimensions, including height, base width, and head width. The program verifies that the head width must be larger than the base width through a condition. This loop instigates until the user provides a valid head width.

The loop manages the output of the arrow base corresponding to the specified height. Thus point (1) is fulfilled.

A nested loop is utilized to present an arrow base matching the defined width. The inner loop creates stars, establishing the width for the base, while the outer loop iterates according to the specified height. This satisfies point (2).

A temporary variable, k, saves the initial arrow head width for necessary adjustments.

The final nested loop outputs an arrow head corresponding to the specified width with stars. Therefore, point (3) is achieved.

The value of k decreases by 1 for each execution of the nested loop, ensuring each subsequent line contains fewer stars.

The output screenshot is attached.

3 0
29 days ago
A test machine that kicks soccer balls has a 5-lb simulated foot attached to the end of a 6-ft long pendulum arm of negligible m
choli [191]

Answer:

a) v₂ = 26.6 ft/s

b) v₂ = 31.9 ft/s

Explanation:

a) Applying the principle of energy conservation for the soccer arm from the point of impact to its lowest point:

mgh=\frac{1}{2} mv^{2} \\v=\sqrt{2gh} =\sqrt{2*32.2*6} =19.66ft/s

The ball's velocity in the tangential direction is given by:

v₂*sinθ = v₁*sin30

With the values:

if v₁ = 0

θ = 0º

The restitution coefficient is:

e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{1}*cos30 )} \\0.8=\frac{v_{2}cos0-v_{2}*cos30 }{19.66*cos30-(0*cos30)} \\v_{2} =\frac{v_{2}-13.6 }{cos30}

where O=θ

The aggregate momentum equation is:

mv-mv_{1} =mv_{2} +mv_{2} cos(O+30)\\, where O = θ

mv-mv_{1} =m(\frac{v_{2}-13.6 }{cos30} )+mv_{2} cos(O+30)

When we incorporate gravitational acceleration:

mgv-mgv_{1} =mg(\frac{v_{2}-13.6 }{cos30} )+mgv_{2} cos(O+30)\\5*19.66-1*0=5*(\frac{v_{2}-13.6 }{cos30} )+1*v_{2} cos(O+30)

Isolating v₂ results in:

v₂ = 26.6 ft/s

b) To find the ball's velocity, we utilize:

v₂ * sinθ = v₁ * sin30

if v₁ = 10 ft/s

then v₂ * sinθ = 10 * sin30

thus sinθ = 5/v₂

The restitution coefficient remains:

e=\frac{v_{2}*cosO-v_{2}cos30 }{v*cos30-(-v_{2}*cos30) } \\0.8=\frac{v_{2}cosO-v_{2}cos30 }{19.66*cos30-(-10*cos30)} \\v_{2} =\frac{v_{2}cos30-20.55}{cos30}

where O = θ

mgv-mgv_{1} =mgv_{2} +mgv_{2} cos(O+30)\\5*19.66-1*0=5v_{2} +v_{2} cos(O+30)\\98.3=5v_{2}+v_{2} cosOcos30-v_{2} sinOsin30

Solving for v₂ yields:

v₂ = 31.9 ft/s

6 0
22 days ago
PDAs with two stacks are strictly more powerful than PDAs with one stack. Prove that 2-stack PDAs are not a valid model for CFLs
iogann1982 [279]

Answer:

?pooooooooooooooooooooooop

Explanation:

?pooooooooooooooooooooooop

6 0
2 days ago
Multiply each element in origList with the corresponding value in offsetAmount. Print each product followed by a space.Ex: If or
Kisachek [217]

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
25 days ago
A technician has been dispatched to assist a sales person who cannot get his laptop to display through a projector. The technici
mote1985 [204]

Answer:Ensure the correct cable is connected between the laptop and the projector. Check for HDMI inputs or 15-pin video output interfaces.

Also, make sure the laptop is set to project to the correct display output.

Explanation:

7 0
21 day ago
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