Phatone Posts: 6, Reputation: 1 New Member #1 Nov 17, 2010, 12:51 AM
Magnitude and direction of the resultant and equilibrant of the forces acting on
Magnitude and direction of the resultant and equilibrant of the forces acting on the plate.

Thanks

Phatone
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #2 Nov 17, 2010, 07:12 AM

1. Make a sketch with the relevant angles.
2. Since the number of forces acting on the plate is not explicit, I'll advise finding the components of each force along an arbitrary horizontal and vertical axis.
3. Find the resultant force long the vertical and the horizontal.
4. Use vector addition to add the two forces to get the resultant magnitude of the forces.
5. Use appropriate trigonometric ratio such as the tangent ratio to get the angle of the resultant force.
 Phatone Posts: 6, Reputation: 1 New Member #3 Nov 17, 2010, 08:16 AM
Here is what I have got so far...

MAG HOR FOR VERT FOR HOR MOMENT VER MOMENT
F1 10 + 10 COS 45 +10 SIN 45 (7.07 X .5) (-7.07 X .5 )
F2 4 0 -4 0 (4 X .5 )
F3 2 -2 COS 60 -2 SIN 60 -2 X .5 -1.7 X .5
F4 5 -5 0 -(-5 X .5) 0

 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #4 Nov 17, 2010, 08:28 AM
Originally Posted by Phatone
Here is what i have got so far....

MAG HOR FOR VERT FOR HOR MOMENT VER MOMENT
F1 10 + 10 COS 45 +10 SIN 45 (7.07 X .5) (-7.07 X .5 )
F2 4 0 -4 0 (4 X .5 )
F3 2 -2 COS 60 -2 SIN 60 -2 X .5 -1.7 X .5
F4 5 -5 0 -(-5 X .5) 0
Um... are you dealing with moments?

By plate, I assumed that a plate was being pushed/pulled by forces, but now that you mention horizontal moment :confused:

If the situation is like a plate on a flat table being pulled by 4 forces, you add each force in the horizontal direction and each force in the vertical direction.

And I have absolutely no idea where the 0.5 comes from...
 Phatone Posts: 6, Reputation: 1 New Member #5 Nov 18, 2010, 12:29 AM
The 0.5 is the centre point on the plate. I attached a diagram on my first email not sure where it is now though??
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #6 Nov 18, 2010, 12:34 AM

No, the diagram is not there... =/

Do you have the link to the picture?
 Phatone Posts: 6, Reputation: 1 New Member #7 Nov 18, 2010, 12:36 AM
Magnitude and direction
Ok here is what I have so far can you please comment..

Thanks
 Phatone Posts: 6, Reputation: 1 New Member #8 Nov 18, 2010, 01:12 AM
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #9 Nov 18, 2010, 01:49 AM

Lol, okay, the link is this :)

And the question is sure clearer :)

Okay, use the table you were given.

The total moment is given by:

$Moment = (7.07\times 0.5) - (7.07\times 0.5) + (4\times 0.5) +(2\times 0.5) - (1.7 \times 0.5) - (5\times 0.5)$

I only followed the signs given to me that is clockwise is +ve and anticlockwise negative.

Now for the translational motion...

$Vertical\ Resultant = F_v = 10\sin45 -2 \sin60$

$Horizontal\ Resultant = F_h= 10\cos45 -2\cos60 -5$

$Resultant = \sqrt{F_v\ ^2 + F_h\ ^2}$

Direction is:

$Angle = \frac{F_v}{F_h}$ with the horizontal.
 Phatone Posts: 6, Reputation: 1 New Member #10 Nov 18, 2010, 02:10 AM
Brill thanks for the end working outs.
Please could you check the information in the table has I have had to work it out by just using the diagram above...

Thnaks..

Phats
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #11 Nov 18, 2010, 02:20 AM

You made the table? No wonder I had trouble with the moments... :o

Well, the forces are good, but not the moments. When it comes to moments, don't take the direction of the force you initially took for the Horizontal and Vertical force. Just follow the clockwise/anticlockwise direction set to put in the magnitudes.

In the table, I would rather put 'clockwise' or 'c/w' moment and 'anticlockwise' or 'a/w' instead of horizontal and vertical moments.

And you'll see that for F3, Horizontal moment is positive, F4, Horizontal moment is negative.

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