Equilibrium : normal reaction of a rotational center?

In summary, the problem involves a force of 90 N applied to a member supported by a frictionless pin and a cable passing over a pulley. The tension in the cable is 117 N and the reaction at D is 129.8 N. To solve for the reaction, assumptions must be made about the weight of the system and the L-shaped piece.
  • #1
Doomski
2
0

Homework Statement


IMG-20190104-WA0005.jpg

D2NwDUI.jpg

I'm trying to figure out the reaction at D,that's where the system can rotate.

Problem:
A force P of magnitude 90 N is applied to member ACDE, which is supported by a frictionless pin at D and by the cable ABE. Since the cable passes over a small pulley at B, the tension may be assumed to be the same in portions AB and BE of the cable. For the case when a=3m, determine:
a)the tension in the cable
b) the reaction at D

2. The attempt at a solution

I've calculated the tension in the cable through the sum of momentum=0, T= 117N. For the reaction, I used the sum of forces=0; We don't have the mass of the system, so I assumed there's three forces acting on D, a normal force and the weight of the system that cancel each other (probably wrong?) and the reaction. From that,I got R = 129.8N. Is that right?
Are there a difference between a normal reaction of a table, for example, and Rd ??

Thanks.

Edit: I had to paint the figure because the photo wasn't clear.
 

Attachments

  • IMG-20190104-WA0005.jpg
    IMG-20190104-WA0005.jpg
    16.2 KB · Views: 500
  • D2NwDUI.jpg
    D2NwDUI.jpg
    7.4 KB · Views: 509
Last edited:
Physics news on Phys.org
  • #2
Doomski said:
I've calculated the tension in the cable through the sum of momentum=0, T= 117N
To get that, I have to assume the horizontal part of the L piece is weightless. Doesn't your method assume that?
To find the reaction at D, you will need to assume the same for the vertical part. You have three unknown scalars, tension and the components of the reaction force, and only three available equations. Any more unknowns make it unsolvable.
I agree with your answer.
 
  • Like
Likes Doomski
  • #3
haruspex said:
To get that, I have to assume the horizontal part of the L piece is weightless. Doesn't your method assume that?
To find the reaction at D, you will need to assume the same for the vertical part. You have three unknown scalars, tension and the components of the reaction force, and only three available equations. Any more unknowns make it unsolvable.
I agree with your answer.

Now that makes sense, thank you!
 

Related to Equilibrium : normal reaction of a rotational center?

1. What is equilibrium in terms of rotational motion?

Equilibrium in rotational motion refers to the state in which an object's rotational center, also known as its center of mass, remains at a fixed position and does not experience any rotational acceleration or change in angular velocity.

2. How is the normal reaction of a rotational center defined?

The normal reaction of a rotational center is the force exerted on an object's rotational center by a supporting surface or pivot point, perpendicular to the surface or pivot. It is also known as the support force or pivot force.

3. What factors affect the normal reaction of a rotational center?

The normal reaction of a rotational center depends on the object's mass, the distance of its center of mass from the pivot point, and the acceleration due to gravity. It also depends on the type of support or pivot, such as a fixed point, hinge, or rolling contact.

4. How does the normal reaction of a rotational center relate to the object's stability?

The normal reaction of a rotational center is directly related to an object's stability. A larger normal reaction force means a greater resistance to tipping or toppling over, while a smaller normal reaction force indicates a lower stability and a higher risk of falling or rotating out of equilibrium.

5. Can the normal reaction of a rotational center be zero?

Yes, the normal reaction of a rotational center can be zero if the object's center of mass is directly above or below the pivot point, resulting in a balanced torque and no net force acting on the object's rotational center. This is known as a neutral equilibrium.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
995
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
7K
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
455
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
3K
Back
Top