# Homework Help: Reaction force in wheels?

1. Jul 28, 2013

### BlueCB

1. The problem statement, all variables and given/known data

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

3. 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.

2. Jul 28, 2013

### Staff: Mentor

Looks to me like you are using the A wheel as your axis. So what's the distance between the wheels?

3. Jul 28, 2013

### BlueCB

16 meters between the wheels (Ra to Rb).

4. Jul 28, 2013

### Staff: Mentor

Right. So why did you write "5 Rb"? Correct that.

5. Jul 28, 2013

### BlueCB

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: Jul 28, 2013
6. Jul 28, 2013

### Staff: Mentor

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.

7. Jul 28, 2013

### BlueCB

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)?

8. Jul 28, 2013

### Staff: Mentor

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.

9. Jul 28, 2013

### BlueCB

And that would be "16 Rb"?

10. Jul 28, 2013

### SteamKing

Staff Emeritus
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. Jul 28, 2013

### Staff: Mentor

Exactly. (Making sure to give it the proper sign.)

12. Jul 28, 2013

### BlueCB

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. Jul 28, 2013

### Staff: Mentor

Yes, looks good. (Assuming that Ra is the force on each front wheel.)

14. Jul 28, 2013

### BlueCB

Yep, 15,600 N on each front wheel (15,600*2 = 31,200)

15. Jul 28, 2013

### BlueCB

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. Jul 28, 2013

### Naty1

What do you think? What criteria should be used?

17. Jul 28, 2013

### BlueCB

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. Jul 28, 2013

### Staff: Mentor

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.