Solving for F2 in the Turning Effect of Force Problem

In summary: What is the final answer to the problem?In summary, the homework equations ask for the distance from the wall, not from the side of the slab. The distance from the wall will be 1.5 - 0.5 = 1 m. To solve for the downward force, we will use the formula F1D1 = F2D2. The final answer is 18000 N + 1 m = 19000 N.
  • #1
Muhammad Danish
70
7

Homework Statement


upload_2018-4-15_22-54-34.png


Homework Equations



How will we find the perpendicular distance between the wall and the centre of mass of the slab? How will we calculate the total downward force exerted by the cables on the slab? I thought much over both the cases but was not able to find any solution. Your explanations will be seriously appreciable.

The Attempt at a Solution


Not able to make any attempt.[/B]
 

Attachments

  • upload_2018-4-15_22-54-34.png
    upload_2018-4-15_22-54-34.png
    35.1 KB · Views: 1,219
  • upload_2018-4-15_22-55-8.png
    upload_2018-4-15_22-55-8.png
    2.5 KB · Views: 488
  • upload_2018-4-15_22-55-18.png
    upload_2018-4-15_22-55-18.png
    2.5 KB · Views: 487
Physics news on Phys.org
  • #2
Where do you think the centre of mass of the slab is located?
 
  • Like
Likes Muhammad Danish
  • #3
kuruman said:
Where do you think the centre of mass of the slab is located?
Between 2.5 m and 0.5 m
 
  • #4
Muhammad Danish said:
Between 2.5 m and 0.5 m
At the top of the wall I mean..
 
  • #5
Muhammad Danish said:
Between 2.5 m and 0.5 m
Can you be more specific? If I were to draw a line parallel to the long dimension of the concrete slab and perpendicular to the 3.0 m side, at what distance from where the cables are attached would I have to draw it so that it passes through the CM? The answer to this question is the answer to part (a). Have you done this much?
 
  • #6
Muhammad Danish said:
At the top of the wall I mean..
kuruman said:
Can you be more specific? If I were to draw a line parallel to the long dimension of the concrete slab and perpendicular to the 3.0 m side, at what distance from where the cables are attached would I have to draw it so that it passes through the CM? The answer to this question is the answer to part (a). Have you done this much?
Since the slab is 3.0 m wide, the center of mass should lie at 1.5 m from the side. Am I right?
 
  • #7
You are right. That's the answer to part (a). You can assume that the external force of gravity acts at the CM. This is a static equilibrium problem. Do you know how to proceed with problems of this kind?
 
  • #8
Muhammad Danish said:
the center of mass should lie at 1.5 m from the side.

kuruman said:
That's the answer to part (a)
Not quite... the question asks for the distance from the wall, not the side of the slab.
 
  • #9
haruspex said:
Not quite... the question asks for the distance from the wall, not the side of the slab.
Yes, of course. Thanks for pointing this out.
 
  • #10
haruspex said:
Not quite... the question asks for the distance from the wall, not the side of the slab.
Distance from the wall will be 1.5 - 0.5 = 1 ?
 
  • #11
Muhammad Danish said:
Distance from the wall will be 1.5 - 0.5 = 1 ?
Yes, 1 m to be precise. Do you know where to go next in this static equilibrium problem?
 
  • #12
kuruman said:
Yes, 1 m to be precise. Do you know where to go next in this static equilibrium problem?
To calculate downward force will we use the formula F1D1 = F2D2 ?
 
  • #13
Muhammad Danish said:
To calculate downward force will we use the formula F1D1 = F2D2 ?
That looks about right for this case. What do you think F1, D1,F2 and D2 stand for? To use this equation you need numbers for three of them, the fourth one being the unknown. What are the three numbers?
 
  • #14
kuruman said:
That looks about right for this case. What do you think F1, D1,F2 and D2 stand for? To use this equation you need numbers for three of them, the fourth one being the unknown. What are the three numbers?
They will be, F1= 18000 N, D1= 1 m and F2= ?, D2= 0.5 m ?
 
  • #15
Muhammad Danish said:
They will be, F1= 18000 N, D1= 1 m and F2= ?, D2= 0.5 m ?
Yes. Now can you solve for F2?
 
  • #16
jbriggs444 said:
Yes. Now can you solve for F2?
Yes. Thanks.
 

What is the definition of "turning effect of force"?

The turning effect of force, also known as torque, is the measure of the ability of a force to cause an object to rotate around an axis.

What is the formula for calculating turning effect of force?

The formula for calculating turning effect of force is torque = force x perpendicular distance from the axis of rotation.

How does the direction of force affect the turning effect?

The direction of the force applied determines the direction of the turning effect. If the force is applied perpendicular to the axis of rotation, it will produce the maximum turning effect. If the force is applied at an angle, the turning effect will be less.

What factors affect the turning effect of force?

The factors that affect the turning effect of force are the magnitude of the force, the distance from the axis of rotation, and the direction of the force.

What is the difference between clockwise and counterclockwise turning effect of force?

Clockwise turning effect of force is when the rotation of the object is in the same direction as the hands of a clock, while counterclockwise turning effect of force is when the rotation is in the opposite direction.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
93
Replies
6
Views
797
  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
182
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
555
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
1K
Back
Top