Calculate the tension force of 3 supporting cables that hold a crate

[plazprestige]

Homework Statement

A 500 N crate is at position A = (45, 70, -115) in a standard xyz coordinate system. Three cables support the crate and have terminal points B, C, D, whose coordinates are (0, 70, 0), (-60, 0, 0), and (45, -65, 0). Calculate the tension force of EACH of the three cables.

Homework Equations

Vector properties

The Attempt at a Solution

My first course of action was to find the component form of vectors AB, AC, and AD (representing the three cables). Respectively, my results were <-45, 0, 115>, <-105, -70, 115>, and <0, -135, 115>.

Beyond this point, however, I don't know how to proceed. In the textbook, there is an example problem involving a camera and a tripod. This example showed the procedure for calculating the supporting force of the tripod legs. However, F1 = F2 = F3 for the example because the legs were of equal length and arranged symmetrically. In this problem, neither conditions apply.

I also tried breaking the problem down by recognizing the net force on the system is zero, but had no luck. This problem is given in the context of a calculus textbook which is just introducing vectors in three dimensions (so no dot-products, cross-products, or calculus is needed to solve the problem).

I would like to know how to solve the problem. I already have the answers.

Thanks in advance!

 

[barryj]

I think you have an error in one of your AB, AC, or AD vectors, Check them.

 

[barryj]

Please double check all of your coordinates. They seem funny to me.

 

[gneill]

@barryj: I can't see anything wrong with the given coordinates.

@plazprestige: Consider shifting the coordinate system to be centered on the crate where the cables meet. Write the cable vectors from that point of view. Now, the forces that the cable tensions apply at this "origin", along with the crate's weight, should sum to zero if the crate is static. You should be able to form the appropriate three equations in three unknowns. (HINT: convert the three "cable vectors" to unit vectors first).

 

[barryj]

Well, for one thing, (-65) - (70) is not 135.

 

[barryj]

I plotted the AB, AC, and AD vectors on an xy grid with z pointing up. They are all off to one side. I don't know how this could be stable.

 

[gneill]

Well, for one thing, (-65) - (70) is not 135.

Can you explain? What's the significance?

Granted, with the given coordinates at least one of the "tensions" will have to be negative (a compression), but the problem is solvable as-is.

 

[plazprestige]

@barryj: I can't see anything wrong with the given coordinates.

@plazprestige: Consider shifting the coordinate system to be centered on the crate where the cables meet. Write the cable vectors from that point of view. Now, the forces that the cable tensions apply at this "origin", along with the crate's weight, should sum to zero if the crate is static. You should be able to form the appropriate three equations in three unknowns. (HINT: convert the three "cable vectors" to unit vectors first).

Perhaps it's been too long (took physics last year), but I do not know the three equations to set up the system you're describing. If you could provide a little bit more detail, that would be great.

 

[plazprestige]

Well, for one thing, (-65) - (70) is not 135.

True. Forgot the negative sign. Apologies.

 

[gneill]

Perhaps it's been too long (took physics last year), but I do not know the three equations to set up the system you're describing. If you could provide a little bit more detail, that would be great.

The forces applied by the cables to the crate act along the same direction as the cable vectors. You can find unit vectors that lie along those directions. Each unit vector multiplied by the (unknown) tension in that cable represents the force applied by that cable. So the unit vectors split those forces into components... There are three axes and you've got components for each from each cable as well as the crate weight.

 

[barryj]

By calculating the vectors AB, AC, and AD you are essentially shifting the coordinate system so the crate is at the origin. Notice that the components are all 115. Now plot the x and Y components and see if there is a way to have a stable system if the crate is at the origin. I don't think so.

 

[plazprestige]

The forces applied by the cables to the crate act along the same direction as the cable vectors. You can find unit vectors that lie along those directions. Each unit vector multiplied by the (unknown) tension in that cable represents the force applied by that cable. So the unit vectors split those forces into components... There are three axes and you've got components for each from each cable as well as the crate weight.

What I am unsure about is how to relate the position vectors of the cables to the force vectors. For the tripod problem I described in the OP, the tension vector for each tripod leg was some scalar "c" multiplied by the position vector. This scalar was the same for all three tripod legs, and it was known that the vertical force component of the three tripods was the weight of the camera divided by 3 since the force was equally distributed. With that information, "c" was determined and then obviously the force vectors since they were just the scalar "c" multiplied by the position vectors.

In this case, the scalar c is not the same (or is it?) for the three position vectors.

 

[gneill]

What I am unsure about is how to relate the position vectors of the cables to the force vectors. For the tripod problem I described in the OP, the tension vector for each tripod leg was some scalar "c" multiplied by the position vector. This scalar was the same for all three tripod legs, and it was known that the vertical force component of the three tripods was the weight of the camera divided by 3 since the force was equally distributed. With that information, "c" was determined and then obviously the force vectors since they were just the scalar "c" multiplied by the position vectors.

In this case, the scalar c is not the same (or is it?) for the three position vectors.

In this case you have three different scalars representing the tensions. Multiply each scalar by a unit vector in the direction of the corresponding cable. Those are your force vectors for the cables. You want to solve for the scalars. Don't forget the crate weight...

 

[plazprestige]

In this case you have three different scalars representing the tensions. Multiply each scalar by a unit vector in the direction of the corresponding cable. Those are your force vectors for the cables. You want to solve for the scalars. Don't forget the crate weight...

unit vector AB = <-0.364, 0, 0.931>
unit vector AC = <-0.615, -0.410, 0.674>
unit vector AD = <0, -0.761, 0.648>

Force vector AB = a<-0.364, 0, 0.931>
Force vector AC = b<-0.615, -0.410, 0.674>
Force vector AD = c<0, -0.761, 0.648>

where a, b, and c are scalars.

This is how I interpreted your directions. I have three unknowns (the scalars). What are my equations for this system though? Also, why did I need to make my original vectors unit vectors?

 

[gneill]

unit vector AB = <-0.364, 0, 0.931>
unit vector AC = <-0.615, -0.410, 0.674>
unit vector AD = <0, -0.761, 0.648>

Force vector AB = a<-0.364, 0, 0.931>
Force vector AC = b<-0.615, -0.410, 0.674>
Force vector AD = c<0, -0.761, 0.648>

where a, b, and c are scalars.

This is how I interpreted your directions. I have three unknowns (the scalars). What are my equations for this system though? Also, why did I need to make my original vectors unit vectors?

Yes, good so far.

You want unit vectors so that your scalars are the magnitudes of the forces acting along the unit vector directions. The unit vectors just provide directions, and "carve" the forces into their components.

You want to form a set of equations such that the like components of the forces sum to the required values. It's particularly easy to set up in matrix form when you have unit vectors to work with... place your scalars in a vector F, the required sums in vector S. Then let M be a matrix containing the unit vectors (you figure it out!), so that ##M\cdot F = S##.

 

[plazprestige]

Yes, good so far.

You want unit vectors so that your scalars are the magnitudes of the forces acting along the unit vector directions. The unit vectors just provide directions, and "carve" the forces into their components.

You want to form a set of equations such that the like components of the forces sum to the required values. It's particularly easy to set up in matrix form when you have unit vectors to work with... place your scalars in a vector F, the required sums in vector S. Then let M be a matrix containing the unit vectors (you figure it out!), so that ##M\cdot F = S##.

Ah, that makes a lot of sense (regarding the use of unit vectors).

However, I am not very familiar with the use of matrices with vectors. Could you please explain it in terms of just an algebraic system of equations?

 

[barryj]

You can just write out the three simultaneous equations and then solve any way you like. Matrices do all the grunt work for you.

I think one of them should be .931B + .67C + .64 D = 500

 

[gneill]

You can set up your equations any way you want, and solve them any way you wish. I suggested matrix form because it would be expedient here. But if you're not familiar with it, use what you know.

You have three equations in three unknowns. The unknowns are the scalar cable force magnitudes. The equations are written to satisfy the constraints of the problem. In this case the constraint is static equilibrium for all three directions of motion for the crate.

What conditions need to hold for the system to be in static equilibrium?

 

[plazprestige]

Ah, I see. The net force of the entire system must equal 0 because there is no acceleration. This means the net force in the positive z direction must be 500 N to counter the 500 N downward weight. The force in the +x direction plus the force in the -x direction = 0, and the same holds true for the y directions.

Am I correct?

 

[gneill]

Ah, I see. The net force of the entire system must equal 0 because there is no acceleration. This means the net force in the positive z direction must be 500 N to counter the 500 N downward weight. The force in the +x direction plus the force in the -x direction = 0, and the same holds true for the y directions.

Am I correct?

You are correct.

 

[plazprestige]

You are correct.

Thank you very much for your assistance. I appreciate that you took the time to walk me through the concept of the problem rather than just spitting out the answer. I also now see why the matrices are useful (for solving the system of equations). I know that much about matrices, at least.

 

[gneill]

Thank you very much for your assistance. I appreciate that you took the time to walk me through the concept of the problem rather than just spitting out the answer. I also now see why the matrices are useful (for solving the system of equations). I know that much about matrices, at least.

You're very welcome

You may find that the results are peculiar (as pointed out by barryj) for a set of cables. They would be less peculiar if one or more of the cables happened to be a rigid rod...

 

[barryj]

Wait, let's see the answers. Everything that has been said above is correct but I still think when solved you will get a negative value for a tension and that is not possible. I would like to know the answers and then be able to figure out where my bad assumption is.