Solving Torque Problem: Unknown Forces & Equations

[Ivan Antunovic]

I used symbol M for torque by my mistake in equations,(because M is used in my language),If I am right americans use symbol T for torque.

At equation (1) I get two unknowns : force T and force Rbx
tried more equations but it gets more complicated,what am I doing wrong...

note:I've choosen point A for calculating torque!

 

[Ivan Antunovic]

I used symbol M for torque by my mistake in equations,(because M is used in my language),If I am right americans use symbol T for torque.

At equation (1) I get two unknowns : force T and force Rbx
tried more equations but it gets more complicated,what am I doing wrong...

note:I've choosen point A for calculating torque and also clockwise direction is chosen to be NEGATIVE,counter clockwise is chosen POSITIVE !

 

[SteamKing]

I used symbol M for torque by my mistake in equations,(because M is used in my language),If I am right americans use symbol T for torque.

At equation (1) I get two unknowns : force T and force Rbx
tried more equations but it gets more complicated,what am I doing wrong...

note:I've choosen point A for calculating torque!

Speaking as an American, I use M for 'moment', rather than torque. Using T for 'torque' gets confusing, since T is often used for 'Tension'.

In your equation (1), you have written (√3 / 3) * R_Bx as one of the components of the moment about A. Since you have already included the forces in the wire in calculating the moment about A, this is an unnecessary and incorrect inclusion in your moment summation.

The correct moment summation about A allows you to calculate the magnitude of the tension T in the wire, which can then be used to determine the reaction at A. You are also including the vertical component of the tension in the wire twice in equation (2), once as R_By and again as T_y.

A proper free body drawing of the beam ADC would include the beam and the components of the tension in the wire at C. Since the beam is in equilibrium and there is only the wire between C and B, once you determine the tension in the wire, the reactions at point B are automatically known. There is no need to include the components of R_B in your force summation.

 

[Ivan Antunovic]

Speaking as an American, I use M for 'moment', rather than torque. Using T for 'torque' gets confusing, since T is often used for 'Tension'.

In your equation (1), you have written (√3 / 3) * R_Bx as one of the components of the moment about A. Since you have already included the forces in the wire in calculating the moment about A, this is an unnecessary and incorrect inclusion in your moment summation.

The correct moment summation about A allows you to calculate the magnitude of the tension T in the wire, which can then be used to determine the reaction at A. You are also including the vertical component of the tension in the wire twice in equation (2), once as R_By and again as T_y.

A proper free body drawing of the beam ADC would include the beam and the components of the tension in the wire at C. Since the beam is in equilibrium and there is only the wire between C and B, once you determine the tension in the wire, the reactions at point B are automatically known. There is no need to include the components of R_B in your force summation.

So R_By and T_y are basically the same forces? I thought R_By is the force of the reaction that wall produces on the rod,but when I think of it must be the same thing because Tension is actually produced by gravitational force that pulls down the rod and they are obviously both in the same direction.(hope you understand my thought)

So I get from equation (1) that T= (2.58)/(3*10^4) =7.75 kN ?

 

[SteamKing]

So R_By and T_y are basically the same forces? I thought R_By is the force of the reaction that wall produces on the rod,but when I think of it must be the same thing because Tension is actually produced by gravitational force that pulls down the rod and they are obviously both in the same direction.(hope you understand my thought)

That's why it's important to learn how to draw a proper free body diagram: it allows you to isolate only the forces acting directly on the member of interest.
The reactions at point A directly influence what happens to the beam, from the standpoint of its equilibrium. The same cannot be said of the reactions at B, because their influence has already been taken into account by including the tension in the wire, which acts to keep the beam in equilibrium under the action of the load G.

So I get from equation (1) that T= (2.58)/(3*10^4) =7.75 kN ?

You've made a slight mistake in writing your moments from one line to the next.

The moment due to load G = 2 m × 10 kN = 20 kN-m = 2 × 10⁴ N-m

T = 2 × 10⁴ N-m / 2.58 m = 7.75 × 10³ N = 7.75 kN

 

[Ivan Antunovic]

That's why it's important to learn how to draw a proper free body diagram: it allows you to isolate only the forces acting directly on the member of interest.
The reactions at point A directly influence what happens to the beam, from the standpoint of its equilibrium. The same cannot be said of the reactions at B, because their influence has already been taken into account by including the tension in the wire, which acts to keep the beam in equilibrium under the action of the load G.You've made a slight mistake in writing your moments from one line to the next.

The moment due to load G = 2 m × 10 kN = 20 kN-m = 2 × 10⁴ N-m

T = 2 × 10⁴ N-m / 2.58 m = 7.75 × 10³ N = 7.75 kN

Yes ,I've noticed that mistake in equation (1),that force R^B only confused me,
and thank you for your help.
Ivan.

 

[Ivan Antunovic]

Yes ,I've noticed that mistake in equation (1),that force Rb at point B only confused me(notice it was already drawn in the task),
and thank you for your help.
Ivan.