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

  • #1
sachin
60
7
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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  • #2
ladder problem.png
 
  • #3
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.
 
  • #5
phinds said:
Unreadable.
The diagram is readable and post #1 has all the text.
 
  • #6
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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