Conditions of equilibrium problem

In summary: I’ll keep that in mind next time.In summary, the cantilevered beam has a mass of 25 kg and is 2.2 meters long. The suspended block has a mass of 280 kg and the supporting cable makes an angle of 30 degrees with the beam. The force that the wall exerts on the beam at the hinge is 5177.1 N and the tension in the supporting cable is 5978 N. Equations were used to determine the unknowns, which were the tension, the vertical component of the hinge force and the horizontal component of the hinge force. Three equations were written to solve for the unknowns, including the horizontal component of tension, the vertical component of tension, and the torque
  • #1
Dragonite
9
0

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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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
 
  • #6
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?
 
  • #7
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.
 

Related to Conditions of equilibrium problem

What is the definition of equilibrium?

Equilibrium is a state of balance or stability in a system where all forces and factors are equal and there is no net change or movement.

What are the conditions for equilibrium?

The conditions for equilibrium are that the net force acting on an object must be zero, and the net torque acting on an object must also be zero.

How do you determine if a system is in equilibrium?

To determine if a system is in equilibrium, you must first calculate the net force and net torque acting on the system. If both values are equal to zero, then the system is in equilibrium.

What happens to a system if it is not in equilibrium?

If a system is not in equilibrium, it will experience a net force and/or net torque, causing it to accelerate or rotate.

Can a system be in equilibrium if it is moving?

Yes, a system can be in equilibrium even if it is moving. This is known as dynamic equilibrium, where the object is moving at a constant velocity and the net force and net torque are both zero.

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