Engineering statics equilibrium problem

[Alison A.]

Homework Statement

Here is the prompt/picture

The hint given is a FBD of the bead is recommended to being this problem. Find the coordinates of B so that both the magnitude and orientation of the elastic cord force can be properly represented. Also, two mutually orthogonal normal force directions (to bar AC) need to be included to permit a general representation of normal forces acting on the bead. A shortcut to solving for the elastic cord force can be obtained by writing the vector equilibrium equation, then taking the dot product of that equation with a unit vector pointing in the direction of bar AC. Since the normal forces are by definition perpendicular to the bar, their contribution is zero, and a single scalar equations remains for the for P.

Homework Equations

ΣF_x=0
ΣF_y=0
ΣF_z=0
Basic trigonometry

The Attempt at a Solution

Alright so I started off by finding the coordinates of all the points
A (124, 0, 0)mm
B (?, 31, 21)mm
C (0, 62, 42)mm
D (62, 0, 62)mm

To find B I made a right triangle with A and C to find the magnitude. Since B is in the middle of AC as stated, I divided the magnitude I found and got 65.5. So B (65.5, 31, 21)mm.

I found the unit vector of AC like the hint suggested and found that to be (-0.856, 0.428, 0.290).

I am stuck here, I don't know how to go about finding the vector equilibrium.

Thank you for any help!

 

[SteamKing]

The Attempt at a Solution

Alright so I started off by finding the coordinates of all the points
A (124, 0, 0)mm
B (?, 31, 21)mm
C (0, 62, 42)mm
D (62, 0, 62)mm

To find B I made a right triangle with A and C to find the magnitude. Since B is in the middle of AC as stated, I divided the magnitude I found and got 65.5. So B (65.5, 31, 21)mm.

Why did you do this? If the y coordinate and z coordinate of point B are both halfway between the y and z coordinates of points A and C, it stands to reason that the x coordinate of B will also be halfway between the x coordinates for these two points.

I found the unit vector of AC like the hint suggested and found that to be (-0.856, 0.428, 0.290).

I am stuck here, I don't know how to go about finding the vector equilibrium.

Thank you for any help!

Find the correct location for point B.

Once you know this, you can then find the tension force in the cord BD, given the spring constant and its unstretched length.

 

[Alison A.]

Why did you do this? If the y coordinate and z coordinate of point B are both halfway between the y and z coordinates of points A and C, it stands to reason that the x coordinate of B will also be halfway between the x coordinates for these two points.Find the correct location for point B.

Once you know this, you can then find the tension force in the cord BD, given the spring constant and its unstretched length.

A (124, 0, 0)mm
B (62, 31, 21)mm
C (0, 62, 42)mm
D (62, 0, 62)mm

T_BD=k(l-l₀) = 4(20-51.4) = -125.6

 

[SteamKing]

A (124, 0, 0)mm
B (62, 31, 21)mm
C (0, 62, 42)mm
D (62, 0, 62)mm

T_BD=k(l-l₀) = 4(20-51.4) = -125.6

Tensions are usually taken to be positive. The stretch in the spring would be calculated as stretched length - unstretched length = 51.4 - 20 = 31.4 mm
The tension would therefore be 4 N/mm * 31.4 mm = 125.6 N.

 

[Alison A.]

So I add this value with the unit vector in the direction of AC?

 

[SteamKing]

So I add this value with the unit vector in the direction of AC?

The tension in the cord is pulling the bead at B against the bar AC. You want to convert the magnitude of this tension, 125.6 N, into a force vector F_BD parallel to BD.

Then, according to the directions in the problem statement, you want to take the dot product of F_BD and the unit vector u_AC to find out what magnitude P must be.

 

[Alison A.]

So the unit vector of BD is (0, -0.6031, 0.7977), then multiplied by the tension is (0, -75.75, 100.2)

The dot product of that and u_AC is -3.367 which doesn't seem right...

 

[SteamKing]

So the unit vector of BD is (0, -0.6031, 0.7977), then multiplied by the tension is (0, -75.75, 100.2)

This looks OK.

The dot product of that and u_AC is -3.367 which doesn't seem right...

I think it's correct. I think the fact that F_BD ⋅ u_AC is negative means that P is in the direction of point C.

 

[Alison A.]

The answer was positive, hm, anyway I got it right. So for finding the reaction between bead B and rod AC, it would be just the force that is perpendicular to P right? I know the hint says nothing about it but wouldn't the cross product between them?

 

[Alison A.]

I know there are infinitely number of vectors perpendicular to bar AC, but I don't know how to find it specifically where bead B is. Thank you so much for your help, I need to figure out these problems within the next hour.