NormanL
Mechanical
- Mar 22, 2012
- 1
thread108-242377
How to draw the free body diagram of a wheel subject to rolling resistance. Date 22 March 2012
To draw the free body diagram proceed as follows. The following assumptions apply.
1. Wheel is rolling to right of page at constant velocity.
2. The wheel is on a straight level road.
3. The weight of the wheel is ignored.
4. The wheel is driven by a pure couple ie no resultant force due to driving mechanism.
5. The wheel does not slip.
6. The vehicle is subject to aerodynamic drag and rolling resistance.
7. The air drag is 300 N
8. The rolling resistance on each wheel is 100N
Method.
1. Draw 2 concentric circles. The outer circel represents the
wheel and the inner circle represents the axle.
2. Mark four points as follows.
a) Point 1 at 12 oclock on inner circle.
b) Point 2 at 3 oclock on inner circle.
c) Point 3 at 6 oclock on outer circle
d) Point 4 at 4 to 5 oclock on outer circle.
3. Mark x y axes such that the x axis is positive to right.
and Y axis positive up.
4. Draw Arrow toward pt 1 in down direction.
5. Draw Arrow toward pt 2 in negative x direction.
6. Draw Arrow toward point 3 in positive x direction.
7. Draw Arrow toward pt 4 in radial direction toward centre.
8. Draw Moment around centre in clockwise direction.
9. Forces have following magnitudes.
F1=Vehicle weight supported by the wheel bearing.
F2=Half air drag + 1 front wheel roll resistance= 250N
F3=rolling resistance of 1 front wheel + rolling resistance
of 1 rear wheel + half vehicle air drag = 350N
F4x=Rolling resistance of rear wheel =100N
F4y=Weight supported by the wheel bearing.
Now if you do the sums
Sum forces in X direction =0
Sum forces in Y direction =0
Sum Moments centre =0
Notes.
The following forces do NOT do work F1,F3,F4y.
The following forces do work F2,F4x and the Moment M.
Now if assume that the wheel is at constant temp.
Work done on wheel = work done by wheel
Work-in=MdA
Work_out=(F2+F4x)dx
Hence
MdA=(F2+F4x)dx
where dx is a differential displacement of the wheel in x dirn
dA is a differential angular displacement of the wheel.
Noting that dx=rdA and substituting for dx
MdA=(F2+F4x)rdA
M=(F2+F4x)r
From statics
M=r(F3)
and F3=F2+F4x
M=r(F2+F4x) as before
Hence work-in equals the work-out and therefore we are in compliance with the law of conservation of energy.
The resultant force = 0 and resultant moment = 0 and we are in compliance with the laws of statics.
How to draw the free body diagram of a wheel subject to rolling resistance. Date 22 March 2012
To draw the free body diagram proceed as follows. The following assumptions apply.
1. Wheel is rolling to right of page at constant velocity.
2. The wheel is on a straight level road.
3. The weight of the wheel is ignored.
4. The wheel is driven by a pure couple ie no resultant force due to driving mechanism.
5. The wheel does not slip.
6. The vehicle is subject to aerodynamic drag and rolling resistance.
7. The air drag is 300 N
8. The rolling resistance on each wheel is 100N
Method.
1. Draw 2 concentric circles. The outer circel represents the
wheel and the inner circle represents the axle.
2. Mark four points as follows.
a) Point 1 at 12 oclock on inner circle.
b) Point 2 at 3 oclock on inner circle.
c) Point 3 at 6 oclock on outer circle
d) Point 4 at 4 to 5 oclock on outer circle.
3. Mark x y axes such that the x axis is positive to right.
and Y axis positive up.
4. Draw Arrow toward pt 1 in down direction.
5. Draw Arrow toward pt 2 in negative x direction.
6. Draw Arrow toward point 3 in positive x direction.
7. Draw Arrow toward pt 4 in radial direction toward centre.
8. Draw Moment around centre in clockwise direction.
9. Forces have following magnitudes.
F1=Vehicle weight supported by the wheel bearing.
F2=Half air drag + 1 front wheel roll resistance= 250N
F3=rolling resistance of 1 front wheel + rolling resistance
of 1 rear wheel + half vehicle air drag = 350N
F4x=Rolling resistance of rear wheel =100N
F4y=Weight supported by the wheel bearing.
Now if you do the sums
Sum forces in X direction =0
Sum forces in Y direction =0
Sum Moments centre =0
Notes.
The following forces do NOT do work F1,F3,F4y.
The following forces do work F2,F4x and the Moment M.
Now if assume that the wheel is at constant temp.
Work done on wheel = work done by wheel
Work-in=MdA
Work_out=(F2+F4x)dx
Hence
MdA=(F2+F4x)dx
where dx is a differential displacement of the wheel in x dirn
dA is a differential angular displacement of the wheel.
Noting that dx=rdA and substituting for dx
MdA=(F2+F4x)rdA
M=(F2+F4x)r
From statics
M=r(F3)
and F3=F2+F4x
M=r(F2+F4x) as before
Hence work-in equals the work-out and therefore we are in compliance with the law of conservation of energy.
The resultant force = 0 and resultant moment = 0 and we are in compliance with the laws of statics.
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