[SOLVED] Mechanics problem

• Feb 25th 2009, 06:32 AM
jackiemoon
[SOLVED] Mechanics problem
Hi,

Can anyone help with solving this problem please:

Two rigid square plates, of different materials and with mass per unit area equal to λ1and λ2 respectively, are joined along one edge. The system is subject to gravity and can rotate freely around the pivot point A in the figure, which is fixed.

(see attached drawing)

Compute the angle formed by the segment AB with the vertical direction when the system is in equilibrium.

I'm having trouble understanding what exactly is being asked in this question. Can anybody shed any light on this please?

Thanks
• Feb 25th 2009, 08:19 AM
Equilibrium positio
Hello jackiemoon
Quote:

Originally Posted by jackiemoon
Hi,

Can anyone help with solving this problem please:

Two rigid square plates, of different materials and with mass per unit area equal to λ1and λ2 respectively, are joined along one edge. The system is subject to gravity and can rotate freely around the pivot point A in the figure, which is fixed.

(see attached drawing)

Compute the angle formed by the segment AB with the vertical direction when the system is in equilibrium.

I'm having trouble understanding what exactly is being asked in this question. Can anybody shed any light on this please?

Thanks

This question is all about finding the position of the centre of mass of the body, and saying that it lies directly beneath A.

So, if the squares have sides of length $\displaystyle 2a$, then their masses are $\displaystyle 4a^2\lambda_1$ and $\displaystyle 4a^2\lambda_2$.

In the attached diagram, then, take moments about G, the centre of mass:

$\displaystyle 4a^2\lambda_1PG = 4a^2\lambda_2GQ$, where P and Q are the centres of the two squares

$\displaystyle \Rightarrow PG = \frac{\lambda_2}{\lambda_1}GQ$

$\displaystyle = \frac{\lambda_2}{\lambda_1+\lambda_2}\times PQ$

$\displaystyle = \frac{\lambda_2}{\lambda_1+\lambda_2}\times 2a$

$\displaystyle \Rightarrow GO = a - PG = a - \frac{2\lambda_2}{\lambda_1+\lambda_2}a$

$\displaystyle = \frac{a(\lambda_1 - \lambda_2)}{\lambda_1 + \lambda_2}$

Now when the body is freely suspended from A, the line AG is vertical, and AB makes an angle $\displaystyle \theta$ with this line, where

$\displaystyle \tan\theta = \frac{GO}{AO}=\frac{GO}{a}$

$\displaystyle = \frac{\lambda_1 - \lambda_2}{\lambda_1 + \lambda_2}$