Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum 
Oct 26th 2015, 08:29 AM

#1  Senior Member
Join Date: Jun 2014
Posts: 306
 Determine the load
I am asked to find the load in AD, BD, and DC....
I have AD(2/surd38) +BD(2/surd29)=800
AD(3/surd38)=DC
I can only formed two equation.... But , so many unknown... How to solve this???

 
Oct 26th 2015, 01:50 PM

#2  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,326

You have three axes here (x, y, and z), so you can set up three equations based on balancing of forces in all three dimensions.
In the xdirection: AD (5/sqrt(50)) + BD(5/sqrt(29)) =0
In the ydirection: AD(3/sqrt(50)) + CD = 0
In the zdirection: BD (2/sqrt(29) + AD (4/sqrt(50)) = 0
Now you can solve the three equations in three unknowns.
I see that you have surd(38) in the denominator for AD in both of your equations  I think you may have thought the z dimension is 2, but it's actually 4.

 
Oct 26th 2015, 05:10 PM

#3  Senior Member
Join Date: Jun 2014
Posts: 306

Originally Posted by ChipB You have three axes here (x, y, and z), so you can set up three equations based on balancing of forces in all three dimensions.
In the xdirection: AD (5/sqrt(50)) + BD(5/sqrt(29)) =0
In the ydirection: AD(3/sqrt(50)) + CD = 0
In the zdirection: BD (2/sqrt(29) + AD (4/sqrt(50)) = 0
Now you can solve the three equations in three unknowns.
I see that you have surd(38) in the denominator for AD in both of your equations  I think you may have thought the z dimension is 2, but it's actually 4. 
why it is actually 4?
is it ok if i assume the BD is directed outside of the page at first ? In your calculation , it's clearly that you have assumed it to be directed towards inside the page at first.... For CD , i have gt it = negative value , so it should point downwards , am i right ? can i assume it to be acting downwards at first ?

 
Oct 27th 2015, 05:50 AM

#4  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,326

Point A's zdimension is 4 meters  note that it is 2 meters from the origin to point B, then 2 more meters along the zaxis to point A. See attached.
As for signs of things  I used the positive x, y and z directions as defined by the diagram, and positive forces for tension. You should find that AD is a positive value, BD is negative because it is in compression, and CD is negative because it is tension n the negative y direction.
Last edited by ChipB; Oct 27th 2015 at 07:28 AM.

 
Oct 28th 2015, 09:01 AM

#5  Senior Member
Join Date: Jun 2014
Posts: 306

Originally Posted by ChipB Point A's zdimension is 4 meters  note that it is 2 meters from the origin to point B, then 2 more meters along the zaxis to point A. See attached.
As for signs of things  I used the positive x, y and z directions as defined by the diagram, and positive forces for tension. You should find that AD is a positive value, BD is negative because it is in compression, and CD is negative because it is tension n the negative y direction. 
how do u know that BD is compression at first sight ?

 
Oct 28th 2015, 10:48 AM

#6  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
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Originally Posted by ling233 how do u know that BD is compression at first sight ? 
In general you may not know that ahead of time  but the math should work out to tell you it is so. However in this case it's pretty obvious  if you replaced the BD rigid bar with a rope, you can see that the structure would collapse. A rope cannot withstand compression, only tension.

 
Oct 28th 2015, 09:18 PM

#7  Senior Member
Join Date: Jun 2014
Posts: 306

Originally Posted by ChipB In general you may not know that ahead of time  but the math should work out to tell you it is so. However in this case it's pretty obvious  if you replaced the BD rigid bar with a rope, you can see that the structure would collapse. A rope cannot withstand compression, only tension. 
Ya, from the ans, we .know that the BD will collapse if it is a rope...
If there is no other forces involve , how do we knw that the. BD is compression at first sight?

 
Oct 29th 2015, 05:27 AM

#8  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,326

I'm not sure I understand your question, but if there are no forces being applied anywhere then the force in BD is 0, so it would be neither in compression nor tension.

 
Oct 29th 2015, 09:24 AM

#9  Senior Member
Join Date: Jun 2014
Posts: 306

Originally Posted by ChipB I'm not sure I understand your question, but if there are no forces being applied anywhere then the force in BD is 0, so it would be neither in compression nor tension. 
well , do you mean the rope can only take tension , but not compression ? the bar can be either compression or tension ?

 
Oct 29th 2015, 09:39 AM

#10  Physics Team
Join Date: Jun 2010 Location: Morristown, NJ USA
Posts: 2,326

Originally Posted by ling233 well , do you mean the rope can only take tension , but not compression ? the bar can be either compression or tension ? 
Ever tried to push on a rope (put it in compression)? What happens?
As for a rigid bar: you can push on it, or pull on it, and it transmits that force to whatever is at the other end  correct?

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