Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

enter image description here

Suppose we have a setup like this. Here $a_1,a_2,b_1,b_2$ are acceleration magnitudes($b_1,b_2$ being relative) and $P,Q,R,S$ are not pulley/blocks but are points on the rope. If I use a geometrical constraint then ($k$ is constant) $$PQ+QR+RS=k$$ $$\ddot {PQ}+\ddot{QR}+\ddot{RS}=0$$ And here since $\ddot {PQ}=b_1,\ddot {RS}=b_2, \ddot{QR}=-(a_1+a_2)$ we have $$b_1+b_2=a_1+a_2$$ This is fine, but how do I get a relation between $\vec{a_1},\vec{a_2},\vec{b_1},\vec{b_2}$? Can it be obtained in any geometrical way without listing out forces (assuming those triangular blocks are isosceles and have base angle $\theta$, base moving on ground)?

Thanks in advance.

share|improve this question
    
Hi Ashish Gaurav. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. –  Qmechanic Mar 28 '13 at 15:58
    
@Qmechanic: I apologize. This wasn't actually a homework question, and it was just that I wanted it to get some different tags. But if it resembles homework in some way, I'll soon put that back. Sorry for that. –  Ashish Gaurav Mar 28 '13 at 16:05
add comment

1 Answer 1

If you use direction vector $\hat{e}_1=(\sin\frac{\theta}{2},\cos\frac{\theta}{2})$ along PQ and $\hat{e}_2=(\sin\frac{\theta}{2},-\cos\frac{\theta}{2})$ along RS you get the positions:

$$ \vec{r}_Q = x_1 \hat{i} $$ $$ \vec{r}_P = x_1 \hat{i} - PQ\, \hat{e}_1 $$ $$ \vec{r}_R = -x_2 \hat{i} $$ $$ \vec{r}_S = -x_2 \hat{i} + RS\, \hat{e}_2 $$

Differentiating twice you get

$$ \ddot{\vec{r}}_Q = a_1 \hat{i} $$ $$ \ddot{\vec{r}}_P = a_1 \hat{i} - b_1\, \hat{e}_1 $$ $$ \ddot{\vec{r}}_R = -a_2 \hat{i} $$ $$ \ddot{\vec{r}}_S = -a_2 \hat{i} + b_2\, \hat{e}_2 $$

but you need the constraint of PQ + QR + RS = const to evaluate the componets. You can correctly shown that $(b_1+b_2)-(a_1+a_2)=0$ which allows you to find one acceleration in terms of the other three. So you will need 3 boundary conditions to fully solve the problem.

In this regard, the conservation of length is a fundamental principle and cannot be derived from other equations. Without it the physical cable connecting all the points cannot be described in this problem.

share|improve this answer
    
+1 for the fact that I cannot obtain a vector constraint, because length of the string is a scalar constraint. But I still wonder why you used the angle $\frac{\theta}{2}$ instead of $\theta$ in those direction vectors. I said the base angles were $\theta$ and not that their sum was $\theta$. Also, your choice of coordinate system is a little bit disturbing. I cannot happen to find a origin where the tails of position vectors lie. Please be kind enough to clarify that. –  Ashish Gaurav Mar 29 '13 at 8:44
    
The angle angle of the isosceles triangle is $\theta$ then the angle from vertical is $\frac{\theta}{2}$. –  ja72 Mar 31 '13 at 22:18
    
@AshishGaurav, they are just direction vectors along the incline where P and S slide. No common origin needed. –  ja72 Mar 31 '13 at 22:20
    
Agreed that no common origin is needed, but I said the base angle of the isosceles triangle is $\theta$, not the other angle(base angles are equal angles) –  Ashish Gaurav Apr 1 '13 at 6:43
1  
It would help to put the angle in the sketch as I see there is a communications issue here. Anyhow you have my intent and you can adjust the math for your situation. –  ja72 Apr 1 '13 at 13:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.