Wheels Reaction Forces: Solving for Ra and Rb?

[BlueCB]

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.

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.

 

[Doc Al]

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?

 

[BlueCB]

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).

 

[Doc Al]

16 meters between the wheels (Ra to Rb).

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

 

[BlueCB]

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?

 

[Doc Al]

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.

 

[BlueCB]

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)?

 

[Doc Al]

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.

 

[BlueCB]

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"?

 

[SteamKing]

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.

 

[Doc Al]

And that would be "16 Rb"?

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

 

[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?

 

[Doc Al]

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.)

 

[BlueCB]

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)

 

[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)

 

[Naty1]

Is the aircraft in the question balanced or unbalanced?

What do you think? What criteria should be used?

 

[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?

 

[Doc Al]

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.