Question about Momentum and torque

[kozmosnak]

Homework Statement

The force on the rope

Relevant Equations

Momentum and Troqhe balance equations

TL;DR Summary: I have this Question, and I have no idea, how am I supposed to set the Forces to create a balance since they are not perpendicular to the moving arm.

I have this question I have spend a lot of time on. It basically shows two sticks which are connected in the middle to create a Hourglass shape and on the top side there is a ball which has the mass of 1G (it is known just not a number)

Each of the sticks has their own mass which is G2. Right at the top there is a rope that holds two ends of the stick together so it won’t collapse. All the measurements that was given is on the attachment file.

What I need to figure out is how much Force the rope has. Even though one can see the answers right below it doesn’t make it understandable. Thank you everyone for helping <3

 

[renormalize]

Fixing your diagram:

 

[haruspex]

how am I supposed to set the Forces to create a balance since they are not perpendicular to the moving arm.

So take the components which are perpendicular to the arm.
You can assume there is no friction and that the system is symmetric.
Draw the Free Body Diagram for one stick and post it.

 

[kozmosnak]

So take the components which are perpendicular to the arm.
You can assume there is no friction and that the system is symmetric.
Draw the Free Body Diagram for one stick and post it

I have drawn it but cant get any further

 

[haruspex]

I have drawn it but cant get any further

Use the FBD of the ball to determine $F_k$.
Use the symmetry to determine $A_y$.

 

[kozmosnak]

Use the FBD of the ball to determine $F_k$.
Use the symmetry to determine $A_y$.

So the force Fk is Fg1 which is to see G

G1/sin(30)*2=Fk

 

[Steve4Physics]

@kozmosnak, if you take moments (torques) about the centre (where the sticks cross) you don't need $A_x$ or $A_y$.

 

[kozmosnak]

@kozmosnak, if you take moments (torques) about the centre (where the sticks cross) you don't need $A_x$ or $A_y$.

Yes but the problem I’m facing is that the forces are not perpendicular to the moving arm. They have angles that’s why I am confused. For example the rope force is horizontal but the axis is there with a 60 degree angle how can I add them up

 

[kozmosnak]

@kozmosnak, if you take moments (torques) about the centre (where the sticks cross) you don't need $A_x$ or $A_y$.

Yes but the problem I’m facing is that the forces are not perpendicular to the moving arm. They have angles that’s why I am confused. For example the rope force is horizontal but the axis is there with a 60 degree angle how can I add them hp

 

[Steve4Physics]

Yes but the problem I’m facing is that the forces are not perpendicular to the moving arm. They have angles that’s why I am confused. For example the rope force is horizontal but the axis is there with a 60 degree angle how can I add them hp

Look at the diagram carefully:

If you are unfamiliar with this try watching the video:

 

[haruspex]

So the force Fk is Fg1 which is to see G

G1/sin(30)*2=Fk

Yes.

Yes but the problem I’m facing is that the forces are not perpendicular to the moving arm. They have angles that’s why I am confused. For example the rope force is horizontal but the axis is there with a 60 degree angle how can I add them hp

See if https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/ helps.