Conditions of equilibrium problem

AI Thread Summary
The discussion revolves around solving a conditions of equilibrium problem involving a cantilevered beam with a mass of 25 kg and a suspended block of 280 kg. Participants calculate the downward forces and establish the need for three equations to determine the unknowns: tension in the supporting cable and the vertical and horizontal forces exerted by the hinge. The importance of correctly applying equilibrium equations for both vertical and horizontal forces is emphasized, along with the need to account for torque generated by the tension. Clarifications are made regarding the correct formulation of equations to ensure that all forces and torques balance. The conversation highlights the collaborative effort to arrive at a comprehensive solution for the problem.
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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