Static Equilibrium: Understanding Changes in Tension and Vertical Force

In summary, the tension force T and the magnitude of the vertical force Fv exerted by the wall on the hinge change when the load on the uniform beam is moved to the left. The distance and angle decrease, causing an increase in the vertical force exerted by the wall. For the second question regarding a ladder, the answer is B because the distance and angle decrease, which affects the torque equation and the force exerted by the wall. The notation used in the conversation may be confusing, but considering moments is necessary to properly answer the question.
  • #1
jakeginobi

Homework Statement



If the load on the uniform beam shown below is moved to the left, how do the tension force T and the magnitude of the vertical force Fv exerted by the wall on the hinge change?

The answer is A

Homework Equations


torque = Fdsin(theta)

The Attempt at a Solution


My equation was Tsinx = fd. I understand when the object is moved to the left the distance decreases, and the angle decreases, but I don't understand how the vertical force increases exerted by the wallAlso for the 2nd question ( the ladder one) why is the answer B? What I thought was Fgdcos theta = FN2(d)(cos theta) + FN1(d)(cos theta), so when the distance for FN1 increases wouldn't the force increase too?
 

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  • #2
jakeginobi said:
and the angle decreases,
What angle decreases? As I read the question, only the load is moving. The cable stays put.
 
  • #3
jakeginobi said:
What I thought was Fgdcos theta = FN2(d)(cos theta) + FN1(d)(cos theta),
I cannot understand your notation. Is this a force equation or a moment equation? What is Fgd? Is FN2(d) the force FN2 as a function of d, the distance up the ladder, or is it a force multiplied by a distance?
(You will need to consider moments to answer the question.)
 
  • #4
haruspex said:
I cannot understand your notation. Is this a force equation or a moment equation? What is Fgd? Is FN2(d) the force FN2 as a function of d, the distance up the ladder, or is it a force multiplied by a distance?
(You will need to consider moments to answer the question.)
Oh I meant Fg(d) and it's for the torque equation
 
  • #5
jakeginobi said:
Oh I meant Fg(d) and it's for the torque equation
Ok. What point are you taking moments about?
You seem to be using the same d for different distances and the same θ for different angles.
 

1. What is static equilibrium?

Static equilibrium refers to the state of an object or system where all forces acting on it are balanced, resulting in zero net force and zero acceleration. In other words, the object or system is at rest or moving at a constant velocity.

2. How is static equilibrium different from dynamic equilibrium?

Static equilibrium is a state of balance where all forces are equal and opposite, resulting in no overall motion. Dynamic equilibrium, on the other hand, refers to a state of balance where there is constant motion but the net force is still zero.

3. What are the conditions for static equilibrium?

The conditions for static equilibrium are that the net force acting on the object or system must be zero, and the net torque (or rotational force) must also be zero. This means that all forces must be balanced and there can be no unbalanced moments acting on the object or system.

4. How is static equilibrium used in real life?

Static equilibrium is used in many everyday situations, such as when building structures like bridges and buildings. Engineers must ensure that all forces acting on these structures are balanced to prevent collapse. Additionally, our bodies are constantly maintaining static equilibrium to keep us upright and balanced.

5. What happens if an object is not in static equilibrium?

If an object is not in static equilibrium, it will experience a net force and/or net torque, causing it to accelerate or rotate. This can result in the object falling, tipping over, or moving in an unintended direction. In order to reach a state of static equilibrium, the object may need to be repositioned or additional forces may need to be applied.

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