Torque needed to accelerate rail-mounted vehicle

In summary: Fnet is the total external longitudinal force. It includes the needed traction force (m*a = 5*9 = 45 N ) from the wheels, and I also need to compensate for the counterforces in the other direction to still achieve the acceleration of 9 m/s2. In summary, the torque needed to reach the desired acceleration is 6.10 Nm.
  • #1
Ole Forsell
10
0
<Moved from a technical forum. Therefore no template.>I am currently analysing the forces acting on a moving vehicle. The vehicle is mounted on a rail with wheels on both upper and lower side of the rail with a spring increasing the reaction force, and thereby the friction. The vehicle should be designed to reach an acceleration, a, of 9 m/s2.

The vehicle have a mass, m, of 5 kg, and a cross-section area, A, of 0,04 m2. The wheel have a radius, r, of 0,106 m.

Until now I have used the following equations:

Fnet = Ft + Fd + Fr
Where
Fnet: Total longitudinal force required to reach the acceleration
Ft = Traction force from the driving wheel
Fd = drag force
Fr = rolling resistance

From this I get Fnet = 57,56 N
Do I need to include the friction force in this equation?

From this I have used two different formulas to calculate the Torque required to reach the needed acceleration: T = Fnet*r and T = I*alpha.
Where
I: moment of inertia of the wheel
alpha: angular velocity

I know the second equation is wrong, since only parameters for the driving wheel is included, but i have no idea how to use this for the whole vehicle.
With the first equation I get a required torque of approximately 6,10 Nm. I do not believe this is correct, and I think I am missing out on something.

I've been searching the forum for this problem, but I have not yet found an answer to this specific problem.
 
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  • #2
Ole Forsell said:
Fnet = Ft + Fd + Fr
Where
Fnet: Total longitudinal force required to reach the acceleration
Ft = Traction force from the driving wheel
Fd = drag force
Fr = rolling resistance

From this I get Fnet = 57,56 N
The only figures I see for inputs are 5 kg, 9 m/s2, 0,04 m2 and 0,106 m. Starting from those inputs, how did you deduce 57,56 N?
 
  • #3
jbriggs444 said:
The only figures I see for inputs are 5 kg, 9 m/s2, 0,04 m2 and 0,106 m. Starting from those inputs, how did you deduce 57,56 N?

Im sorry for the limited explanation. I used the different equation for drag force, rolling force and F = ma for the traction force. The reason why I didnt't explain it more thouroughly is that it is a lot of parameters and coefficients. This is not really the part I am questioning. What I want to know is which different forces i need to take into account, and how to calculate the required torque.
Hope this clarify your question in some way.
 
  • #4
Ole Forsell said:
Im sorry for the limited explanation. I used the different equation for drag force, rolling force and F = ma for the traction force. The reason why I didnt't explain it more thouroughly is that it is a lot of parameters and coefficients. This is not really the part I am questioning. What I want to know is which different forces i need to take into account, and how to calculate the required torque.
Hope this clarify your question in some way.
If Fnet is 57.56 N and the cart's mass is 5 kg then the resulting acceleration will be 11.5 m/s2. So there is definitely something amiss.
 
  • #5
jbriggs444 said:
If Fnet is 57.56 N and the cart's mass is 5 kg then the resulting acceleration will be 11.5 m/s2. So there is definitely something amiss.
I think you are mixing up Fnet and Ft. Ft = ma = 5*9 = 45 N. From this we also need to include the drag force due to air resistance which is given by Fd = 1/2*p*v2*Cd*A and rolling resistance F = Crr*N. Then all the forces is summed together to get Fnet. But these forces are not the problem. I'm trying to figure out which other forces that will have an impact on the system.
 
  • #6
Ole Forsell said:
I think you are mixing up Fnet and Ft. Ft = ma = 5*9 = 45 N.
Perhaps you can explain your sign conventions and the source of each force.

Fnet would ordinarily be the total external force on the cart. Fnet = ma
Ft would naturally be the traction force on the cart.
Fd would naturally be the drag force on the cart.
Fr being rolling resistance would normally manifest as a torque rather than as a force.
 
  • #7
jbriggs444 said:
Perhaps you can explain your sign conventions and the source of each force.

Fnet would ordinarily be the total external force on the cart. Fnet = ma
Ft would naturally be the traction force on the cart.
Fd would naturally be the drag force on the cart.
Fr being rolling resistance would normally manifest as a torque rather than as a force.

Fnet is the total external longitudinal force. It includes the needed traction force (m*a = 5*9 = 45 N ) from the wheels, and I also need to compensate for the counterforces in the other direction to still achieve the acceleration of 9 m/s2.

In my understanding, torque and force goes hand in hand. To find the needed torque from the driving wheel, all forces need to be added toghether. That means that the torque would need to be represented as a force. This is no problem as I have the diameter of the wheel.

I think my question might have been a bit missleading. If we forget the numbers for now, which forces/torques would have an impact on the movement of the car?
 
  • #8
Ole Forsell said:
Fnet is the total external longitudinal force. It includes the needed traction force (m*a = 5*9 = 45 N ) from the wheels, and I also need to compensate for the counterforces in the other direction to still achieve the acceleration of 9 m/s2.
You compensate for the counterforces by increasing Ft. The result vector sum of all forces is an Fnet that is exactly large enough to produce the required acceleration.
Ole Forsell said:
In my understanding, torque and force goes hand in hand. To find the needed torque from the driving wheel, all forces need to be added toghether. That means that the torque would need to be represented as a force. This is no problem as I have the diameter of the wheel.
The problem is that rolling resistance manifests as a torque. If you are already tracking Ft as a force, there is no place left to put Fr as a force. It is better modeled as if it were friction at the axle (as if the bearings were sticky).
 

Related to Torque needed to accelerate rail-mounted vehicle

What is torque?

Torque is a measure of the rotational force applied to an object. In the context of a rail-mounted vehicle, it is the force needed to rotate the wheels and accelerate the vehicle.

How is torque calculated?

Torque is calculated by multiplying the force applied to an object by the distance from the point of rotation. In the case of a rail-mounted vehicle, it is the force applied by the motor and the diameter of the wheels.

Why is torque important in accelerating a rail-mounted vehicle?

Torque is important because it is what allows the wheels to rotate and propel the vehicle forward. The greater the torque, the faster the acceleration of the vehicle.

What factors affect the torque needed to accelerate a rail-mounted vehicle?

The torque needed to accelerate a rail-mounted vehicle depends on several factors such as the weight of the vehicle, the diameter of the wheels, the friction between the wheels and the rails, and the incline of the track.

How can the torque needed to accelerate a rail-mounted vehicle be increased?

The torque needed to accelerate a rail-mounted vehicle can be increased by increasing the force applied by the motor, increasing the diameter of the wheels, reducing friction between the wheels and the rails, and decreasing the incline of the track.

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