Calculate Force AB & Angle Support System

[Dellis]

Hi, I been trying to do this exercise, visually it looks very basic but I just need some guidance, thanks in advance :).

Homework Statement

Calculate the Force AB in Cable and the angle(theta symbol) for support system shown.

Point A pinned and Point B connected as a cable at an angle( not given ), Point B with a Force of 100 going to the right and a Weight of 200, like displayed below.


A
\_________>F=100
|B
|
|
|
[W]=200

Homework Equations

\SigmaFx=0
\SigmaFy=0
\SigmaM=0

The Attempt at a Solution

1st step- The free Body Diagram showing all the forces

2nd step- Guidance needed at this point, is this the right direction?

\SigmaFxy= 100-200=0

 

[PhanthomJay]

Hi, I been trying to do this exercise, visually it looks very basic but I just need some guidance, thanks in advance :).

Homework Statement

Calculate the Force AB in Cable and the angle(theta symbol) for support system shown.

Point A pinned and Point B connected as a cable at an angle( not given ), Point B with a Force of 100 going to the right and a Weight of 200, like displayed below.


A
\_________>F=100
|B
|
|
|
[W]=200

Homework Equations

\SigmaFx=0
\SigmaFy=0
\SigmaM=0

The Attempt at a Solution

1st step- The free Body Diagram showing all the forces

Great start. What are the forces acting, both known and unknown?

2nd step- Guidance needed at this point, is this the right direction?

\SigmaFxy= 100-200=0

No, what are you doing here? It's sum of F_x = 0, and sum of F_y =0. What are the x components of all the forces; what are the y components of all the forces? You'll get 2 equations with 2 unknowns (T and theta), which you can solve. Alternatively, it may be simpler to compute the resultant (magnitude and direction) of the applied loads, and the Tension must be equal and opposite to that resultant.

 

[Dellis]

Great start. What are the forces acting, both known and unknown?

No, what are you doing here? It's sum of F_x = 0, and sum of F_y =0. What are the x components of all the forces; what are the y components of all the forces? You'll get 2 equations with 2 unknowns (T and theta), which you can solve. Alternatively, it may be simpler to compute the resultant (magnitude and direction) of the applied loads, and the Tension must be equal and opposite to that resultant.

1. Forces are on a STABLE situation, Force POINT A meets with POINT B which then its Force shows going in a Horizontal direction, then the TENSION of the Weight literally comes down from POINT B2. So its like this?...

F_x = 0
F_x= 100lb

F_y= 0
F_y = -200 lb3. I am confused, is it possible to even use the Triangle Method for this? or that would be making things to complicated?

Show me a route I can follow, this problem visually seems to be basic but I just don't know how to approach it.

 

[PhanthomJay]

1. Forces are on a STABLE situation, Force POINT A meets with POINT B which then its Force shows going in a Horizontal direction, then the TENSION of the Weight literally comes down from POINT B


2. So its like this?...

F_x = 0
F_x= 100lb

F_x can't simultaneously be 0 and 100. The equation at joint B is sum of all forces in the x direction is 0. You have 100 units of force acting to the right, and T* cos theta units of force acting to the left.

F_y= 0
F_y = -200 lb

Same comment; 200 acting down and T*sin theta acting up, sum of which are 0.

3. I am confused, is it possible to even use the Triangle Method for this? or that would be making things to complicated?

Yes, that's the easiest way. Use Pythagorus' theorem to calculate the hypotenuse of a triangle with legs of 200 and 100, and calculate the proper angle. The tension is then equal and opposite to this result.

 

[Dellis]

Yes, that's the easiest way. Use Pythagorus' theorem to calculate the hypotenuse of a triangle with legs of 200 and 100, and calculate the proper angle. The tension is then equal and opposite to this result.

Right direction?C(2)= A(2)+B(2)

A(2)= 100(2) + 200(2)

\sqrt{}A(2)= 3000

A= 173.2 lb

 

[PhanthomJay]

Right direction?


C(2)= A(2)+B(2)

A(2)= 100(2) + 200(2)

\sqrt{}A(2)= 3000

A= 173.2 lb

Ahh, your math is not so good. C is the hypotenuse of the triangle, which you are trying to find. A and B are the legs, given. C must be greater than A or B. It's C^2 = A^2 + B^2. Solve for C.

 

[Dellis]

Ahh, your math is not so good. C is the hypotenuse of the triangle, which you are trying to find. A and B are the legs, given. C must be greater than A or B. It's C^2 = A^2 + B^2. Solve for C.

Yeah that's why I need help with it , but anyways I did it again and Its the same thing, so the answer will be now

C^2 = A^2 + B^2

C^2= 200^2 + 100^2

\sqrt[]{}C^2= 30000

C= 173.2 lbWhat is the next step?, help me out here, I know you probably did the problem in your head already, guide me.

 

[PhanthomJay]

Yeah that's why I need help with it , but anyways I did it again and Its the same thing, so the answer will be now

C^2 = A^2 + B^2

C^2= 200^2 + 100^2

yes.

\sqrt[]{}C^2= 30000

C= 173.2 lb

Your math is in error.

C^2 = 40000 + 10000 =50000
\sqrt C} = \sqrt 50000 = ?