How Do You Calculate Wall Forces in Beam Equilibrium Problems?

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In summary, the wall exerts a horizontal force of 2202 N on the left end of the beam and a vertical force of 0 N.
  • #1
shaka23h
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A 1090-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. Using the data shown in the figure, find (a) the magnitude of the tension in the wire and the magnitudes of the (b) horizontal and (c) vertical components of the force that the wall exerts on the left end of the beam.

http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c09/ch09p_20.gif


I was able to get part A but I'm having trouble on b and c, I just don't know how to draw this free body diagram?

Thanks for all ur help
 
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  • #2
Okay -- to work you through the free body diagram of the beam...

First can you tell us all the forces that will have to be on the diagram?
Where, and in what direction, are all of these forces?

Maybe attach a picture of this so we can check it out.
 
  • #3
well at tutoring today the guy didn't explain it too sell. But I know there is the weight of the ladder and the weight of the box going down oh and the tension force of course of the top string. which I solved to equal 2202 N

I can't attach a picture of my hand drawing cause I don't have a scanner :( I really just need a equation for this problem for I'm scared that it'll pop up on the exam tomorrow :(

Thanks a lot
 
Last edited:
  • #4
The problem also told you there were two forces at place it joins the wall... those need to go in there. Since you don't know the magnitudes of them (that's what you are looking for) you need to just use them as variables...

Then -- to find two unknowns, you'll need to use two equations.
Use -- net force is zero, net torque is zero.

Does that help?
 
  • #5
shaka23h said:
A 1090-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. Using the data shown in the figure, find (a) the magnitude of the tension in the wire and the magnitudes of the (b) horizontal and (c) vertical components of the force that the wall exerts on the left end of the beam.

http://edugen.wiley.com/edugen/courses/crs1507/art/qb/qu/c09/ch09p_20.gif


I was able to get part A but I'm having trouble on b and c, I just don't know how to draw this free body diagram?

Thanks for all ur help
Since you already have determined the tension in part A, and since the reaction at the top of the wall must be equal and opposite to (in line with) that tension force, it just becomes a matter of solving for the horizontal and vertical forces at the lower part of the wall by Newton 1 : sum of x component of forces = 0, and sum of y component of forces = 0.
 
  • #6
PhanthomJay said:
by Newton 1 : sum of x component of forces = 0, and sum of y component of forces = 0.

Yeah! that's even easier -- you get two equations just by the net force equals zero. No need for torque. Must be getting late here for me. :yuck:
 
  • #7
you guys//lady are awsome :)

Thanks a lot
 

1. What is rigid body equilibrium?

Rigid body equilibrium refers to a state in which a body is at rest or moving at a constant velocity without any external forces acting on it. This means that the net force and net torque on the body are both equal to zero.

2. How is equilibrium achieved in a rigid body?

Equilibrium is achieved in a rigid body when the sum of all external forces acting on the body is equal to zero, and the sum of all external torques acting on the body is also equal to zero. This can be achieved through the balancing of forces and torques on the body.

3. What is the difference between static equilibrium and dynamic equilibrium?

In static equilibrium, the body is at rest and all forces and torques acting on it are balanced. In dynamic equilibrium, the body is moving at a constant velocity, but the forces and torques acting on it are still balanced.

4. How is the center of mass related to rigid body equilibrium?

The center of mass is a point within a body where the entire mass of the body can be considered to be concentrated. In rigid body equilibrium, the center of mass remains at a constant velocity or at rest, as there are no external forces or torques acting on it.

5. What are some real-life applications of rigid body equilibrium?

Rigid body equilibrium is used in many everyday objects, such as buildings, bridges, and vehicles. It is also important in fields such as engineering, physics, and biomechanics, as it helps to analyze and design structures and machines for stability and balance.

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