Can Vectors in a Box be Solved Without Given Coordinates?

In summary, the problem asks for the coordinates of the vectors F1, F2, and F3 in a box. There are no coordinates given, and the problem is asking for the displacement vector from corner A to corner B.
  • #1
EternityMech
95
0

Homework Statement



3 Forces/vectors in a box going from a corner.

F1=260N F2=75N F3=60N

then it says calculate F0 which is F0=F1+F2+F3
problem is that the answer is not the sum of the vectors its 347.3N

its also wants the angles of the vectors to the corner

problem is no cordinates are given for any of the vectors. and there's a picture of the box with measurements but does not specify the cordinates of any of the vectors. they all seem to be half way almost.

Homework Equations



is it possible to solve, don't you need the cordinates?

picture:

Vbox.jpg

The Attempt at a Solution



non
 
Last edited:
Physics news on Phys.org
  • #2
The way you have it drawn, the *angle of the* vector on the right hand side would have to be ##θ=tan^{-1}{\frac{3}{4}}##. However, that is assuming that your interpertation of the vector splitting the side down the center is accurate.
 
Last edited:
  • #3
EternityMech said:

Homework Statement



3 Forces/vectors in a box going from a corner.

F1=260N F2=75N F3=60N

then it says calculate F0 which is F0=F1+F2+F3
problem is that the answer is not the sum of the vectors its 347.3N

its also wants the angles of the vectors to the corner

problem is no coordinates are given for any of the vectors. and there's a picture of the box with measurements but does not specify the coordinates of any of the vectors. they all seem to be half way almost.

Homework Equations



is it possible to solve, don't you need the coordinates?

picture:
[ img]http://s21.postimg.org/yhypu214n/Vbox.jpg

The Attempt at a Solution



non

It would help us help you, if you would give us the exact wording of the problem as it was given to you.

Your figure does not label which of the red vectors is F1, which is F2, or which is F3 .

One of those vectors is "anti-parallel" to the x-axis, i.e. it's in the negative x direction.

Another is parallel to a vector going from corner A to the opposite corner, corner B.

The remaining vector is parallel to a vector from corner A on the right face of the block, to the opposite corner of that face.

From the dimensions of the box, you should be able to construct a unit vector in the direction each of the three force vectors.
 
Last edited by a moderator:
  • #4
SammyS said:
It would help us help you, if you would give us the exact wording of the problem as it was given to you.

Your figure does not label which of the red vectors is F1, which is F2, or which is F3 .

One of those vectors is "anti-parallel" to the x-axis, i.e. it's in the negative x direction.

Another is parallel to a vector going from corner A to the opposite corner, corner B.

The remaining vector is parallel to a vector from corner A on the right face of the block, to the opposite corner of that face.

From the dimensions of the box, you should be able to construct a unit vector in the direction each of the three force vectors.

ya sorry changed it. this is the exact wording. no cordinates are given. and it wants exact answers it has given me the F0 and the angles precisely they are 107.6 149.7 and 66.2
 
  • #5
Hello! If you have the dimensions of the box, you can find the coordinates by setting up your own coordinate system.
 
  • #6
There are x, y, and z axes drawn in the box. As I look at this perspective drawing, it looks to me like it's drawn so that the coordinate system is drawn inside the box, with the origin being in the lower left-hand corner to the rear of the box. If that is the intent, then this is a left-handed coordinate system. -- That's very strange !

As SteamKing and I hinted at earlier, what is the displacement vector from corner A to corner B ? F1 is parallel to this vector.


By the way, this problem is very straight forward -- except for that left-handed system.
 
  • #7
SammyS said:
There are x, y, and z axes drawn in the box. As I look at this perspective drawing, it looks to me like it's drawn so that the coordinate system is drawn inside the box, with the origin being in the lower left-hand corner to the rear of the box. If that is the intent, then this is a left-handed coordinate system. -- That's very strange !

As SteamKing and I hinted at earlier, what is the displacement vector from corner A to corner B ? F1 is parallel to this vector.


By the way, this problem is very straight forward -- except for that left-handed system.

care to explain or give tips? i mean the cordinates of the vectors F1 F2 F3 are not give only their magnitudes.
 
  • #8
EternityMech said:
care to explain or give tips? i mean the cordinates of the vectors F1 F2 F3 are not give only their magnitudes.
Start by answering this question:

What is the displacement vector from corner A to corner B ?
 
  • #9
SammyS said:
Start by answering this question:

What is the displacement vector from corner A to corner B ?

its not given and no cordinates are given that's the point of the assignment did you read my initial post?
 
  • #10
EternityMech said:
its not given and no cordinates are given that's the point of the assignment did you read my initial post?
In the figure that you posted in your OP, I see a box that's 12 meters wide, 4 meters high and 3 meters deep. I also has a coordinate system sketched in a corner.
 
Last edited:

Related to Can Vectors in a Box be Solved Without Given Coordinates?

1. What are vectors in a box?

Vectors in a box refer to a set of vectors that are contained within a specific region or boundary. They can be represented graphically as arrows pointing in a specific direction and magnitude.

2. What is the importance of using vectors in a box?

Vectors in a box are important in many scientific fields, such as physics and engineering, as they allow us to represent and analyze complex systems and movements in a simplified manner. They also help us understand the forces and motions involved in these systems.

3. How do you calculate the magnitude of a vector in a box?

The magnitude of a vector in a box can be calculated using the Pythagorean theorem, where the square of the vector's magnitude is equal to the sum of the squares of its components in each direction.

4. Can vectors in a box have negative components?

Yes, vectors in a box can have negative components, as they represent both direction and magnitude. A negative component indicates a direction opposite to the positive direction of the axis it is on.

5. How are vectors in a box used in real-world applications?

Vectors in a box are used in various real-world applications, such as navigation systems, computer graphics, and modeling of physical systems. They are also used in physics and engineering calculations, such as calculating forces and velocities.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Linear and Abstract Algebra
Replies
9
Views
411
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
8K
  • Introductory Physics Homework Help
Replies
5
Views
430
Back
Top