# Physics Problem!

Find the relation b/w v1 and v2 in terms of v1,v2, and theta

Note by Ajitesh Mishra
4 years, 12 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Let $$L$$ be the length of the string and $$H$$ be the height difference between the pulleys.

Let $$x$$ be the length of the horizontal piece of string and $$y$$ be the length of the vertical piece of string.

The slanted piece of string has length $$L-x-y = \sqrt{x^2+H^2}$$. Thus, $$(L-x-y)^2 = x^2+H^2$$.

Differentiation yields: $$2(L-x-y)\left(-\dfrac{dx}{dt}-\dfrac{dy}{dt}\right) = 2x \dfrac{dx}{dt}$$.

By definition, $$\dfrac{dx}{dt} = -v_2$$ and $$\dfrac{dy}{dt} = v_1$$. Using trigonometry $$x = H\tan \theta$$ and $$L-x-y = H\sec \theta$$.

Therefore, $$2H\sec\theta \cdot (v_2 - v_1) = 2H\tan\theta \cdot (-v_2)$$. Solving yields $$v_1 = v_2(1+\sin \theta)$$.

- 4 years, 11 months ago

Awesome Solution Dude...!!!

- 4 years, 11 months ago

Wow! Good job Jimmy!

- 4 years, 11 months ago

using the virtual work method..-T(1+sin(thetha))x2 + Tx1=0. Differentiating this constraint equation we get the relation v2(1+sin(thetha))=v1. (x2 is wedge's displacement and x1 is the block's displacement)

- 4 years, 11 months ago

Another way to derive the relation is by using the fact that work done by tension is zero (by newton's third law). We get $$\sum$$T.s=0 . Differentiating we get, $$\sum$$T.v=0

- 4 years, 11 months ago

constraint relations

- 4 years, 11 months ago

xdx=ldL sinθ=dL/dx sinθ+1=(dL+dx)/dx sinθ+1=V1/V2

- 4 years, 11 months ago

concept behind the problem is use of constraint motion

- 4 years, 11 months ago

v1/(1+sinθ)=v2 I think so!!!

- 4 years, 11 months ago

$$v_2(1+\sin\theta)=v_1\sin\theta$$

- 4 years, 11 months ago

is the velocity v1 with respect to wedge or with respect to ground?

- 4 years, 11 months ago

won't it also be in terms of the masses of the two objects? (determining their acceleration from the tension in the thread and gravity)

- 4 years, 11 months ago

nope ..masses aren't needed..

- 4 years, 11 months ago

Does $$\theta$$ stay constant? My guess is no, but I can't tell for certain from the diagram.

- 4 years, 11 months ago

Yes , $$\theta$$ will definitely not remain the same . $$\theta$$ would gradually decrease .

- 4 years, 11 months ago

use the concept that length on the thread remains constant. :)

- 4 years, 12 months ago

i think $$v_1$$=$$v_2$$$$cos\theta$$ Not pretty sure but still..i think..

- 4 years, 12 months ago

no .. it isn't the right ans

- 4 years, 11 months ago

My guess... it's "v1 = v2 + v2*sin(theta)". Is that right?

- 4 years, 12 months ago

is it a guess or something else??

- 4 years, 11 months ago

More than a guess. I have used logic but might be wrong.

- 4 years, 11 months ago

by constrain relation,v2=v1tan theta

- 4 years, 12 months ago