# trapezoid

• November 28th 2007, 12:25 AM
trapezoid
Hi, I need some help on this question.

A unifom lamina ABCEF is obtained from a rectangle ABCD with AB = CD = 8cm and BC = AD = 6cm, by removing the triangle EDF, where E and F lie on CD, AD respectively, with CE = 2cm and AF = 3cm.

1. Find the distance of the centre of mass of the lamina ABCEF from AB and AD.

2. The lamina is suspended freely from F and hangs in equilibrium under gravity. Find the angle which AF makes with the vertical.

I'm stuck and don't know how to start.
• November 28th 2007, 05:00 AM
TKHunny
It's a rectangle with uniform density.

I would situate it on a set of coordinate axes, with A at the Origin and B on the Positive x-axis. This puts D on the Positive y-axis.

Area of FDE = ??
Area of AFECB = ??

Center of Mass of FDE = (??,??)
Center of Mass of AFECB = (??,??)

Fill in the blanks and let's see where that leads us?
• November 28th 2007, 05:08 AM
CaptainBlack
Quote:

Hi, I need some help on this question.

A unifom lamina ABCEF is obtained from a rectangle ABCD with AB = CD = 8cm and BC = AD = 6cm, by removing the triangle EDF, where E and F lie on CD, AD respectively, with CE = 2cm and AF = 3cm.

1. Find the distance of the centre of mass of the lamina ABCEF from AB and AD.

2. The lamina is suspended freely from F and hangs in equilibrium under gravity. Find the angle which AF makes with the vertical.

I'm stuck and don't know how to start.

You start by drawing a diagram.

RonL
• November 28th 2007, 07:08 AM
Hi, guys Im still stuck on this question cause I got answer's that is completely wrong like distance from AB as 6 and AD as 2 so can anyone show me a step to step cause im hopeless on this. :(
• November 28th 2007, 07:48 AM
TKHunny
Use the very clear drawing and fill in the blanks I suggested.

Let's see what you get.
• November 28th 2007, 07:54 AM
earboth
Quote:

Hi, I need some help on this question.

A unifom lamina ABCEF is obtained from a rectangle ABCD with AB = CD = 8cm and BC = AD = 6cm, by removing the triangle EDF, where E and F lie on CD, AD respectively, with CE = 2cm and AF = 3cm.

1. Find the distance of the centre of mass of the lamina ABCEF from AB and AD.

2. The lamina is suspended freely from F and hangs in equilibrium under gravity. Find the angle which AF makes with the vertical.

I'm stuck and don't know how to start.

Hello,

to #1:

I take the drawing of CaptainBlack.

The area of the pentagon ABCEF can be calculated by Area of rectangle - area of right triangle $= 6 \cdot 8 - \frac12 \cdot 6 \cdot 3 = 39$

Now choose 2 lines of equilibrum that means if you support the pentagon along the line the pentagon is in equilibrum. Thats only possible if the areas on both sides of such a line of equilibrum are equal.

First I was looking for a right triangle with an area of 19.5 cm² (in red)

Next I was looking for a trapezium with an area of 19.5 cm² (in blue)

The intersection of these 2 lines is the centre of mass.

As you see you get 2 congruent triangles thus the centre of mass is the midpoint of the lines of equilibrum.

I'll leave the rest for you.
• November 28th 2007, 11:38 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
You start by drawing a diagram.

RonL

Note deliberate error in attachment AF should be 3cm not 2cm marked.
• November 28th 2007, 12:17 PM