Statics of rigid bodies

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February 10th 2008, 05:49 AM
free_to_fly

Statics of rigid bodies

I'm stuck on the following question:

ABCD is a uniform square lamina with side 4 and mass 8kg. It's hinged at A so that it is free to move in the vertical plane. It is mantained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons. Find the value of F.

I don't know how to resolve the forces when taking moments at A or if they even need to be resolved since a square has 90 degrees so the forces are penpendicular to the side AB. Can someone help please?
February 10th 2008, 06:31 AM
earboth

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Quote:

Originally Posted by free_to_fly

I'm stuck on the following question:

ABCD is a uniform square lamina with side 4 and mass 8kg. It's hinged at A so that it is free to move in the vertical plane. It is mantained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons. Find the value of F.

I don't know how to resolve the forces when taking moments at A or if they even need to be resolved since a square has 90 degrees so the forces are penpendicular to the side AB. Can someone help please?

The plate ABCD revolves around point A.

I've attached a sketch of the situation as I understand it.
February 10th 2008, 06:47 AM
free_to_fly

So taking moments at A:
2*8*9.8cos45=2*4F

Is that correct?
February 10th 2008, 08:01 AM
TrevorP

You should have more moments than that.

Ok here's how I would do it:

$\mbox{Assume CCW +}$

$\sum M_A = 0$

$-(8kg)(9.81m/s^2)(\cos(45))(2m) + (F)(4m) + (F)(4m) = 0$

$2(F)(4m) = (8kg)(9.81m/s^2)(\cos(45))(2m)$

$F = \frac{(8kg)(9.81m/s^2)(\cos(45))(2m)}{2\cdot4m} = \boxed{13.9 N}$

That would be the answer for BOTH forces assuming that both have to be the same.
February 10th 2008, 08:44 PM
earboth

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Quote:

Originally Posted by free_to_fly

So taking moments at A:
2*8*9.8cos45=2*4F

Is that correct?

Yes, it is the right turning momentum.

The left turning momentum is caused by the forces $F_1$ and $F_2$ and the effective lever arm l.

The length of $l = 4 \cdot \sqrt{2}$
and since $|\overrightarrow{F_1}| = |\overrightarrow{F_2}|$ the resulting force of $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ is $\vec R$ :

$|\vec R| = |\overrightarrow{F_1}| \cdot \sqrt{2}$

Because $\vec R$ is perpendicular to l you don't have to take any angle into account. The solid is in equilibrium if both momenti are equal:

$2 m \cdot 8 kg \cdot 9.81 \frac{m}{s^2} = 4 \cdot \sqrt{2}m \cdot |\vec R| = 4 \cdot \sqrt{2}m \cdot |\overrightarrow{F_1}| \cdot \sqrt{2}$ will yield:

$|\overrightarrow{F_1}| = 2 \cdot 9.81 \ N = 19.62 N$