Two candles making shadows of one another on opposite walls

AI Thread Summary
The discussion focuses on the dynamics of shadows cast by two candles burning at different rates. It establishes that the heights of the candles decrease over time, affecting the lengths of their shadows. The velocities of the shadows are derived using geometric relationships, leading to expressions for their speeds. A key point is the distinction between velocity and speed, with the latter being the absolute value of the former. The conversation concludes with the realization that both shadows could potentially move downward, especially if the candles burn at the same rate.
brotherbobby
Messages
749
Reaction score
169
Homework Statement
Two candles of equal height ##h## at the initial moment are at a distance ##a## from each other. The distance between each candle and the nearest wall is also ##a## (see figure below). With what speed will the shadows of the candles move along the walls if one candles burns down during a time ##t_1## and the other during a time ##t_2##?
Relevant Equations
1. (Uniform) Velocity of a point along a line : ##v = \dfrac{dx}{dt}##

2. For two similar right angled triangles of heights ##h_1## and ##h_2## and bases ##b_1## and ##b_2## : ##\quad\dfrac{h_2}{b_2}=\dfrac{h_2}{b_2}##
1715325726667.png


I copy and paste the problem as it appears in the text along with the diagram alongside. It is clear from the image that the base of the shadows don't move - only their tops would move either up or down. Let us assume the time ##t_1>t_2##, implying that candle 1 burns at a slower rate. How would things appear after a time of burning ##t## for both candles?

1715325865862.png

I draw my own diagram, showing the situation after a time ##t##. The rate at which the candles burn ##\small{r_i=\dfrac{h}{t_i}}##. The height of candle 1 after a time ##t## : ##\small{h_1 = h-r_1t=h-\dfrac{h}{t_1}t\Rightarrow h_1=h\left(1-\dfrac{t}{t_1}\right)\quad (1)}.## Likewise, ##\small{h_2= h\left(1-\dfrac{t}{t_2}\right)\quad (2)}.##

If the shadows have lengths ##s_1, s_2##, I have to find their velocities as the candles burn, i.e. ##\boxed{\dot{s_1}=?\;,\;\dot{s_2}=?}##.

In ##\triangle's## CDO and EFO, we have (by similarity) : ##\small{\dfrac{h_1-s_2}{2a}=\dfrac{h_2-s_2}{a}\Rightarrow h_1-s_2 = 2(h_2-s_2)\Rightarrow s_2=2h_2-h_1}=2h\left(1-\dfrac{t}{t_2}\right)-h\left(1-\dfrac{t}{t_1}\right)## upon using equations ##(1),(2)## above.
This simplifies to ##\small{s_2=h+h\left(\dfrac{1}{t_1}-\dfrac{2}{t_2}\right)t\Rightarrow \dot{s}_2=h\left(\dfrac{1}{t_1}-\dfrac{2}{t_2}\right)}\Rightarrow \boxed{\boldsymbol{\dot{s}_2=\dfrac{h}{t_1t_2}(t_2-2t_1)}}\quad (3)##
Since ##t_1>t_2,\; \dot{s}_2=\text{-ve}##.

From similar ##\triangle's## ABO and EFO, ##\small{\dfrac{s_1-s_2}{3a}=\dfrac{h_2-s_2}{a}\Rightarrow s_1-s_2=3(h_2-s_2)}##
##\small{\Rightarrow s_1=3h_2-2s_2=3\left(1-\dfrac{t}{t_2}\right)-2h-2h\left(\dfrac{1}{t_2}-\dfrac{2}{t_2}\right)t}## after using equations ##(2),(3)##.

This reduces to ##\small{s_1=h-\dfrac{2h}{t_1}t+\dfrac{h}{t_2}t\Rightarrow \dot{s}_1=-\dfrac{2h}{t_1}+\dfrac{h}{t_2}}\Rightarrow \boxed{\boldsymbol{\dot{s}_1=\dfrac{h}{t_1t_2}(t_1-2t_2)}}\quad (4)##
1715325971780.png


My answers miss those of the text by a change of sign.

Request : Where do you think am mistaken?
 
Last edited:
Physics news on Phys.org
You have defined the velocity in the opposite direction to the text. This will make it differ by a sign.

The problem also asks for speed, not velocity, so technically the correct answer is the absolute value of these expressions (which make them the same). The difference being that your expressions will always have (at least) one negative value.
 
Orodruin said:
You have defined the velocity in the opposite direction to the text. This will make it differ by a sign.
I think I should look into first myself and then post the text solution. I measured distances from the bottom up, taking the base as point 0.

1715327662116.png


Yes you were right. They are taking distances from the top down.

Orodruin said:
The difference being that your expressions will always have (at least) one negative value.

Yes that is what am not sure of. Let us have the expressions again, given my conventions.

Shadow ##s_1## moves with a speed ##\quad\small{\boxed{\boldsymbol{\dot{s}_1=\dfrac{h}{t_1t_2}(t_1-2t_2)}}}##

Shadow ##s_2## moves with a speed ##\quad\small{\boxed{\boldsymbol{\dot{s}_2=\dfrac{h}{t_1t_2}(t_2-2t_1)}}}##

I assumed ##t_1>t_2##. However, if ##t_1<2t_2\Rightarrow \dot{s}_1=-ve##. Either way, ##\dot{s}_2=-ve##.

So it could be that both shadows are moving down?
 
brotherbobby said:
So it could be that both shadows are moving down?
Of course, consider the case when both candles burn at the same speed. Then both shadows will move down at that speed.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Back
Top