Why Does Calculating Torque at the Ladder's Contact Point Not Work?

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SUMMARY

The discussion centers on the calculation of torque at the contact point of a ladder with the ground during its fall, specifically in the context of AP Physics rotational dynamics. The problem involves a uniform ladder of length L making an angle θ with the horizontal, with the correct movement of the bottom point (end B) identified as moving a distance 𝐿/2 (1 − cosθ) to the right. The participant struggles to apply the torque equation, torque = mg x l/2, effectively in their analysis, indicating a need for clarity on the forces involved.

PREREQUISITES
  • Understanding of rotational dynamics principles
  • Familiarity with torque calculations
  • Knowledge of gravitational force effects on rigid bodies
  • Ability to interpret and analyze physics problems involving angles and distances
NEXT STEPS
  • Study the concept of torque in detail, focusing on applications in rotational dynamics
  • Learn about the effects of gravity on rigid body motion
  • Explore the relationship between angles and distances in physics problems
  • Review examples of similar problems involving ladders and contact points in physics textbooks
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Students studying AP Physics, particularly those focusing on rotational dynamics, as well as educators looking for examples of torque applications in real-world scenarios.

sachin
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Homework Statement
while solving a problem on rotational dynamics for ap physics, m finding in the question given below,
I am unable to predict the form of movement of the bottom point as on the whole the ladder is not showing any particular rotational motion,

the question is A uniform ladder of length L is supported by a person holding end A, so the ladder makes an angle θ with the horizontal. There is no friction between the ladder and the floor. When the ladder is dropped it falls down because of gravity. Which of the following correctly describes the movement of end B as the ladder falls?
(A) It stays at rest.
(B) It moves a distance Lsinθ to the right.
(C) It moves a distance 𝐿/2 (1 − cosθ) to the right.
(D) It moves a distance 𝐿/2 (1 − sinθ) to the left.
(E) It moves a distance Lsinθ to the left.

this is question no 35 in the below link https://www.slideshare.net/slideshow/apphyscmrotationaldynamicsmultiplechoiceslides20170116-1pptx/260308461

the answer given is option (C) i.e It moves a distance 𝐿/2 (1 − cosθ) to the right.

I am trying to get the torque about the point of contact of the ladder on the ground = mg x l/2, but it is not leaving me anywhere, will request if any solution can be given, thanks.
Relevant Equations
torque about the point of contact of the ladder on the ground = mg x l/2
I am trying to get the torque about the point of contact of the ladder on the ground = mg x l/2, but it is not leaving me anywhere
 
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ladder problem.png
 
sachin said:
Homework Statement: while solving a problem on rotational dynamics for ap physics, m finding in the question given below,
I am unable to predict the form of movement of the bottom point as on the whole the ladder is not showing any particular rotational motion,

the question is A uniform ladder of length L is supported by a person holding end A, so the ladder makes an angle θ with the horizontal. There is no friction between the ladder and the floor. When the ladder is dropped it falls down because of gravity. Which of the following correctly describes the movement of end B as the ladder falls?
(A) It stays at rest.
(B) It moves a distance Lsinθ to the right.
(C) It moves a distance 𝐿/2 (1 − cosθ) to the right.
(D) It moves a distance 𝐿/2 (1 − sinθ) to the left.
(E) It moves a distance Lsinθ to the left.

this is question no 35 in the below link

the answer given is option (C) i.e It moves a distance 𝐿/2 (1 − cosθ) to the right.

I am trying to get the torque about the point of contact of the ladder on the ground = mg x l/2, but it is not leaving me anywhere, will request if any solution can be given, thanks.
Relevant Equations: torque about the point of contact of the ladder on the ground = mg x l/2

I am trying to get the torque about the point of contact of the ladder on the ground = mg x l/2, but it is not leaving me anywhere

Just think about the horizontal forces and the consequence.
 
phinds said:
Unreadable.
The diagram is readable and post #1 has all the text.
 
haruspex said:
The diagram is readable and post #1 has all the text.
? Here's what I see
1711064062852.png


I do not find that to be readable.
 
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