Is there a geometrical way to obtain a relationship between these vectors? 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)?

• 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

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.

• +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
• 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