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

• sachin
sachin
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

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.

sachin said:

phinds said:
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

I do not find that to be readable.

SammyS and BvU

• Introductory Physics Homework Help
Replies
7
Views
446
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
54
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
364
• Introductory Physics Homework Help
Replies
11
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
246
• Introductory Physics Homework Help
Replies
30
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
817
• Introductory Physics Homework Help
Replies
27
Views
3K