Addition of forces acting on object

In summary, The problem involves finding the magnitude of the total force acting on a hook in a 3-dimensional coordinate system. The hook is placed at point A=(0,0,3) and is connected to a hatch in the XY-plane at points B=(4,0,0) and C=(4,5,0) by two lines. The magnitudes of the forces in the lines are S[AB]=P and S[AC]=sqrt(2)*P. The attempt at a solution involves finding the unit vectors for each line force and then summing them to find the total force acting on the hook. However, there was an error in the calculation of the unit vectors, resulting in incorrect magnitude of the total force.
  • #1
kisengue
12
0
This should be simple, I've already solved several problems just like this one but it won't come out right however i try... apologies in advance for non-native english.

Homework Statement


In a 3-dimensional coordinate system, a hook is placed in A=(0,0,3). Two lines connect the hook down to a hatch that resides completely within the XY-plane. These lines keep the hatch horizontal. The lines attach to the corners of the hatch in B=(4,0,0) and C=(4,5,0). The magnitudes of the forces in the lines are S[AB]=P and S[AC]=sqrt(2)*P. (Did not know how to index S properly). What is the magnitude of the total force acting on the hook?

The Attempt at a Solution


First, i figure out the complete description of the line forces one by one by multiplying the magnitude of the force with the force vector divided by the length of the force vector, F=P*(AB)/|AB|;
F[AB]= P*(B-A)/|B-A| = P*(4,0,-3)/sqrt(4²+3²) = P*(4,0,-3)/5
F[AC]= sqrt(2)*P*(C-A)/|C-A| = sqrt(2)*P*(4,5,-3)/|4²+5²+3²| = sqrt(2)*P*(4,5,-3)/(5*sqrt(2)) = P*(4,5,-3)/5

Then, i sum the line forces= P*(4,0,-3)/5 + P*(4,5,-3)/5 = P*(8,5,-3)/5.

The magnitude of the total force acting upon the hook would then be 1/5*P. But it isn't. The answer says sqrt(5)*P. What did i miss?

Sorry if it's not optimally legible, i tried to tidy it up as best i could... thankful for any help!
 
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  • #2
I see some problems in your unit vectors. Your unit vectors should be

e1=(-4,0,3)/5 and e2=(-4,-5,3)/(5*sqrt(2))

Always draw a diagram! It will help ;).
 
  • #3
Hmm... the way I've done it thus far, and is described in my mechanics book, is to form the vector from A to B by subtracting A from B... so AB in the problem would be B-A=(4,0,0)-(0,0,3)=(4,0,-3), right? Or have i left something out?

Also, how would this change the magnitude of anything? I don't mean to argue, i just want to understand...
 
  • #4
It depends on which way the forces are acting. Think about it this way, the hook is at x=0, B is at x=4, so to since the hook is pulling UP on the hatch, you know that the x component of the force must be negative. The vector is from B to A!
 
  • #5
http://img222.imageshack.us/img222/1154/vectorsms2.jpg
 
Last edited by a moderator:
  • #6
Oh... you're right, of course! That's helpful. But that only reverses the direction of my final unit vector, it still has the same magnitude...

The problem is marked as "more difficult" in the book, and it doesn't seem all that hard so i guess there's something that i don't immediately consider.
 
  • #7
Aren't we trying to find the force on the hook?
AB = [4/5P, 0, -3/5P],
AC = [4P/5,5P/5,-3P/5]

Hope you guys don't mind me playing with this, too. I've never seen one like it.
I like the vector approach you took - much better than my trigonometry!

EDIT - simplified root(2)/root(50) to 1/5
 
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  • #8
oops, you know what, you were right with the unit vectors :$, sorry :(, but yea, no matter which way you do it you'll get the same answer! (opposite and equal forces) I am so sorry mate.

BUT I SEE THE PROBLEM!

When adding your vectors you had -3 + -3 = -3, should be -6.

sum the line forces= P*(4,0,-3)/5 + P*(4,5,-3)/5 = P*(8,5,-6)/5.If you do that and find the magnitude of the force you will find is [tex]\sqrt{5}[/tex]P

|F|=[tex]\sqrt{(8P/5)^2+(5P/5)^2+(6P/5)^2}[/tex]

You did the problem correctly, just a simple calculation error :).
 
  • #9
Do you have a short way to find the magnitude? I combined the first two with the Pythagorean theorem, then combined that result with the 3rd the same way. It does come out to root(5).

EDIT: Oh, I see it now - thanks!
I think I took a course on this in 1970 - Linear Algebra.
 

Related to Addition of forces acting on object

1. What is the definition of "addition of forces acting on an object"?

The addition of forces acting on an object refers to the mathematical process of combining multiple forces that are acting on the same object in order to determine the overall resultant force.

2. Why is it important to understand the concept of addition of forces?

Understanding the concept of addition of forces is important because it allows scientists to accurately predict the motion and behavior of an object based on the forces acting upon it. This is crucial in fields such as physics and engineering.

3. What is the difference between adding forces in the same direction versus in opposite directions?

When adding forces in the same direction, the resultant force will be equal to the sum of the individual forces. However, when adding forces in opposite directions, the resultant force will be equal to the difference between the two forces.

4. Can forces with different magnitudes and directions be added together?

Yes, forces with different magnitudes and directions can be added together using vector addition. This involves breaking down each force into its horizontal and vertical components and then adding them together separately to determine the overall resultant force.

5. How does the angle of a force affect its contribution to the overall resultant force?

The angle of a force affects its contribution to the overall resultant force by changing its horizontal and vertical components. A force acting at a greater angle will have a smaller horizontal and vertical component, resulting in a smaller overall contribution to the resultant force.

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