Conditions of equilibrium problem

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Homework Help Overview

The problem involves a cantilevered beam with a mass of 25 kg and a suspended block of 280 kg, with a supporting cable making a 30-degree angle with the beam. Participants are tasked with determining the force exerted by the wall on the beam at the hinge and the tension in the supporting cable.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conversion of weights to Newtons and the application of equilibrium conditions. There is an exploration of the forces acting on the beam, including vertical and horizontal components exerted by the hinge. Some participants question the equations used to represent these forces and the assumptions made regarding equilibrium.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the relationships between the forces and torques involved. There is recognition of the need for multiple equations to solve for the unknowns, and some guidance is offered regarding the correct forms of the equations. However, there is no explicit consensus on the approach being taken.

Contextual Notes

Participants note the presence of three unknowns: the tension in the cable, the vertical component of the hinge force, and the horizontal component of the hinge force. The discussion includes questioning the assumptions about the equilibrium of forces and torques.

Dragonite
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Homework Statement


Consider the following cantilevered beam:
The beam has a mass of m = 25 kg and is 2.2 meters long. The suspended block has
a mass M = 280 kg and the supporting cable makes an angle of 30 degrees with the beam.
Determine the force that the wall exerts on the beam at the hinge and determine
the tension in the supporting cable.

(There's a horizontal beam making a 90 degree angle with a wall. At the end of the beam, there's a string with the 280 kg block. Also, there's a string connecting the top of the wall to the end of the beam. )


Homework Equations



∑F=0
∑T=0

The Attempt at a Solution



1) I converted the weight of the block and the beam to Newtons.

280 kg x 9.8 m/s^2 = 2744 N
25 kg x 9.8 m/s^2 = 245 N

2) Using the first condition of equilibrium,

245 + 2744 = 2989 N = Downward force = Upward force by string making 30 degree angle with beam

3) Using sine law

Force toward the left of the string = 2989 sin 60/sin 30 = 5177.1 N = Force toward the right by the wall

4)

Squareroot of quantity Fx^2 + Fy^2 = 5978 N = Tension on string
 
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Dragonite said:
∑F=0
∑T=0

The Attempt at a Solution



1) I converted the weight of the block and the beam to Newtons.

280 kg x 9.8 m/s^2 = 2744 N
25 kg x 9.8 m/s^2 = 245 N
So far so good.

2) Using the first condition of equilibrium,

245 + 2744 = 2989 N = Downward force = Upward force by string making 30 degree angle with beam
What about the vertical force exerted by the hinge on the beam? There is also a horizontal force exerted by the hinge on the beam.

Bottom line: You have three unknowns, the tension, the vertical component of the hinge force and the horizontal component of the hinge force. Therefore, you need to write down three equations to solve for the unknowns. What are the three equations?
 
kuruman said:
So far so good.


What about the vertical force exerted by the hinge on the beam? There is also a horizontal force exerted by the hinge on the beam.

Bottom line: You have three unknowns, the tension, the vertical component of the hinge force and the horizontal component of the hinge force. Therefore, you need to write down three equations to solve for the unknowns. What are the three equations?

Let me try...

Horizontal force of hinge = Horizontal component of tension / Tension sin 60
Vertical component of tension + Vertical force of hinge = 2989 N
2744 N x 2.2 + 245 x 1.1 = Vertical component of tension / Tension sin 30

Is that right?
 
Dragonite said:
Let me try...

Horizontal force of hinge = Horizontal component of tension / Tension sin 60
Where did you get this? Where does it say that the sum of the horizontal forces is zero? Your equation must be of the form
F1x+F2x+F3x+... = 0.
Vertical component of tension + Vertical force of hinge = 2989 N
Correct. This is the same as F1y+Ty+(-W1)+(-W2)=0.
Note that it is in the correct form, ΣFy=0.
2744 N x 2.2 + 245 x 1.1 = Vertical component of tension / Tension sin 30
What is the torque generated by the tension? It is not what you have on the right side.
 
kuruman said:
Where did you get this? Where does it say that the sum of the horizontal forces is zero? Your equation must be of the form
F1x+F2x+F3x+... = 0.

Wouldn't the horizontal forces be equal since the tension and the beam are in equilibrium?

kuruman said:
Correct. This is the same as F1y+Ty+(-W1)+(-W2)=0.
Note that it is in the correct form, ΣFy=0.

Ok.

kuruman said:
What is the torque generated by the tension? It is not what you have on the right side.
Oh sorry. I forgot to put the distance.

2744N x 2.2 + 245N x 1.1 - Ty x 2.2 = 0
 
Yes they would. Does your equation

Horizontal force of hinge = Horizontal component of tension / Tension sin 60

say this or does it say something else? Why are you dividing by Tension sin60?
 
kuruman said:
Yes they would. Does your equation

Horizontal force of hinge = Horizontal component of tension / Tension sin 60

say this or does it say something else? Why are you dividing by Tension sin60?

I meant to say horizontal component of tension or tension sin 60. Sorry and thank you very much.
 

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