Wheels Reaction Forces: Solving for Ra and Rb?

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Homework Help Overview

The discussion revolves around calculating the reaction forces at two wheels (Ra and Rb) in a mechanical system, with specific attention to the forces and distances involved in the setup. Participants are analyzing the equations derived from the forces acting on the system and questioning the assumptions made in the calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the distances between the wheels and the forces acting on them. There is a focus on the correct formulation of torque equations and the implications of using different axes for calculations. Questions arise regarding the interpretation of terms like "5 Rb" and whether they should be adjusted based on the distances from the pivot point.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem and clarifying the relationships between forces and torques. Some participants have proposed revised equations and calculations, while others are still uncertain about the assumptions regarding balance and equilibrium in the system.

Contextual Notes

There is mention of specific forces (25 kN, 5 kN, and 8 kN) and their respective distances from the pivot point, which are critical to the calculations. Participants are also reflecting on the nature of the system, questioning whether it is balanced or unbalanced, which influences their approach to solving for the reaction forces.

BlueCB
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Homework Statement



In the diagram below, work out the reaction force in each of the wheels.
Note there are two wheels that make up Ra.

15dkmq0.png




The Attempt at a Solution



0 = 25,000*2 + 5,000*15 -5 Rb - 8,000*2
0 = 50,000 + 75,000 - 5 Rb - 16,000
0 = 109,000 - 5 Rb
5 Rb = 109,000
Rb = 109,000 / 5 = 21,800 (5,600)

8,000 + 25,000 + 5,000 = 38,000 N
38,000 - 21,800 = 16,200 N
Ra = 16,200 N
Ra*2 = 32,400 N
38,000 - 32,400 = 5,600 N
Rb therefore = 5,600 N

Ra*2 = 32,400. Ra = 16,200 N
Rb = 5,600 N

This has been marked incorrect, the '5 Rb' reference especially.
And Ra and Rb are wrong.
 
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BlueCB said:
0 = 25,000*2 + 5,000*15 -5 Rb - 8,000*2
Looks to me like you are using the A wheel as your axis. So what's the distance between the wheels?
 
Doc Al said:
Looks to me like you are using the A wheel as your axis. So what's the distance between the wheels?
16 meters between the wheels (Ra to Rb).
 
BlueCB said:
16 meters between the wheels (Ra to Rb).
Right. So why did you write "5 Rb"? Correct that.
 
Doc Al said:
Right. So why did you write "5 Rb"? Correct that.
Would that make it "16 Rb" instead then?

I thought it was 5 Rb due to the 5 kN force being 1 meter away from Rb = 5*1 = 5?
 
Last edited:
BlueCB said:
Would that make it "16 Rb" instead then?
Right.

I thought it was 5 Rb due to the 5 kN force being 1 meter away from Rb = 5*1 = 5?
What matters is the distance of each force from wheel A, which is your axis of rotation. You already accounted for the torque from the 5 kN force, which is 15 m from wheel A.
 
Doc Al said:
Right.What matters is the distance of each force from wheel A, which is your axis of rotation. You already accounted for the torque from the 5 kN force, which is 15 m from wheel A.
So would it just be the case of completely removing "5 Rb" from the equation if I've already accounted for it via (5,000*15)?
 
BlueCB said:
So would it just be the case of completely removing "5 Rb" from the equation if I've already accounted for it via (5,000*15)?
5,000*15 is the torque due to the 5 kN force, not the torque due to the unknown force Rb. You still need a torque term due to Rb.
 
Doc Al said:
5,000*15 is the torque due to the 5 kN force, not the torque due to the unknown force Rb. You still need a torque term due to Rb.
And that would be "16 Rb"?
 
  • #10
You must also be careful in finding the reaction at each of the front wheels once you have solved for RA.
You should write a 'sum of the forces' equation to go along with your 'sum of the moments about A' equation.
 
  • #11
BlueCB said:
And that would be "16 Rb"?
Exactly. (Making sure to give it the proper sign.)
 
  • #12
0 = 25,000*2 + 5,000*15 -16 Rb - 8,000*2
0 = 50,000 + 75,000 - 16 Rb - 16,000
0 = 109,000 - 16 Rb
16 Rb = 109,000
Rb = 109,000 / 16 = 6812.5 N (round it to 6800)

8,000 + 25,000 + 5,000 = 38,000 N
38,000 - 6800 = 31200 N
Ra = 31200 / 2 = 15600 N
38,000 - 31,200 = 6,800 N
Rb therefore = 6,800 N

Is this any better?
 
  • #13
BlueCB said:
0 = 25,000*2 + 5,000*15 -16 Rb - 8,000*2
0 = 50,000 + 75,000 - 16 Rb - 16,000
0 = 109,000 - 16 Rb
16 Rb = 109,000
Rb = 109,000 / 16 = 6812.5 N (round it to 6800)

8,000 + 25,000 + 5,000 = 38,000 N
38,000 - 6800 = 31200 N
Ra = 31200 / 2 = 15600 N
38,000 - 31,200 = 6,800 N
Rb therefore = 6,800 N

Is this any better?
Yes, looks good. (Assuming that Ra is the force on each front wheel.)
 
  • #14
Doc Al said:
Yes, looks good. (Assuming that Ra is the force on each front wheel.)
Yep, 15,600 N on each front wheel (15,600*2 = 31,200)
 
  • #15
Right, still a bit unsure.
Just dug up some old notes related to moments;

Moment of a force (Nm) = force (N)*perpendicular distance from pivot to line of action of the force (M)
M = F*X

Perpendicular distances = moment = force*distance from pivot

Balanced object = sum of clockwise moments = sum of anti-clockwise moments

• Decide where the pivot is.
• Decide whether the object is balanced or unbalanced.
• For each force, decide whether the force is tending to turn the object clockwise or anti-clockwise.
• Decide whether any distance given in the question is the perpendicular distance from the pivot to the line of action.Is the aircraft in the question balanced or unbalanced? (just want to clarify before I re-submit for marking)
 
  • #16
Is the aircraft in the question balanced or unbalanced?

What do you think? What criteria should be used?
 
  • #17
I'd say unbalanced.

The pivot is Ra.
25 kN and 5 kN forces are tending to turn it clockwise and the 8 kN force anti-clockwise.
So a total of 30 kN clockwise force and 8 kN anti-clockwise, would tend to make one assume it's unbalanced?
 
  • #18
BlueCB said:
I'd say unbalanced.
Well is the plane sitting in equilibrium or not? Any reason to think it is rotating?

Note that you had to assume equilibrium (and balanced torques and forces) in order to solve for the reaction forces.
 

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