Calculating Force and Torque for Moving Mass on Wheels

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    Friction Motion
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Homework Help Overview

The discussion revolves around calculating the force and torque required to move a 25 kg mass on wheels along aluminum alloy rails, aiming for a speed of 12 m/s over a distance of 15 meters. Participants are exploring the necessary formulas and calculations involved in this physics problem, particularly focusing on the effects of friction and the dynamics of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss finding acceleration and force using Newton's laws, with one poster attempting calculations based on given parameters. Others suggest drawing free-body diagrams and considering rotational dynamics. Questions arise about the role of friction and the complexity of the calculations involved.

Discussion Status

The discussion is ongoing, with participants sharing different approaches to the problem. Some have provided insights into the independence of acceleration from friction under certain conditions, while others express confusion about the calculations presented. There is no clear consensus yet, as participants are still exploring various aspects of the problem.

Contextual Notes

Participants note the absence of a known friction coefficient, which complicates the calculations. There is also mention of the need for simplifying assumptions in the analysis.

fox2005eng
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Homework Statement



Hi for all, basically I have a mass of 25Kg, want to move this mass with speed upto 12m/s for 15 meter long, where the mass held on 4 wheel (R= 50mm) and moving on aluminum alloy rails. I want to know the force needed and torque required, so I can be able to choose the suitable DC motor required.
actually I have tried most formulas but with no luck, as I don't know the friction coefficient.
However,
what is the suitable formulas are required here according to my question?
steps of calculations if possible?
Thanks in Advance.

Homework Equations


The Attempt at a Solution



1 – Find the time:
Given:
Velocity = 12 (m/s).
V: Velocity (m/s).
x: Distance (m).
t: Time (sec).
According to the rail length, which is = 14.9728 (m)
Time taken from the beginning of the rail until the end at speed of 12 (m/s) is:
t=V/x
= 1.2477 (seconds) taken from start of the rail until the end.
2- Find the acceleration:
V: Final velocity = 12(m/s)
V0: Initial velocity = 0 (m/s)
a: Acceleration (m/s2).
t: Time (seconds).
V=V0+2at
a = 9.62 (m/s2).
3- Find the Force:
Mass of the machine is 25 Kg , acceleration is 9.62 (m/s2), by using equation of motion can obtain the required force, by the following formula:
Newton second law:
F=ma
F: Force (N).
m: Mass = 25 Kg.
a: Acceleration = (m/s2).
Assume no friction "Ignore Friction":
F= 240.5 (N).
4- Find required torque:
Torque depends on force:
T = F x r sin theta ,
F= 240.5 (N)
r: wheel radius = 0.05 (m).
theta : Angle = 90.
T= 12 (N.m).
 
Last edited:
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The first thing to do is to find the acceleration of the cart given a force F. To do this, draw free-body diagrams on the wheels as well as the body, write out Newton's second law and the rotational second law, and solve. This will get quite complicated, and you'll need to make some simplifying assumptions.
 
You will, by the way, find that acceleration is independent of the coefficient of friction as long as the wheels don't slip.
 
ideasrule said:
You will, by the way, find that acceleration is independent of the coefficient of friction as long as the wheels don't slip.

it seems to complicated for me.

what about calculations that I made above ?

is it wrong ?

thanks for advance.
 
ideasrule said:
The first thing to do is to find the acceleration of the cart given a force F. To do this, draw free-body diagrams on the wheels as well as the body, write out Newton's second law and the rotational second law, and solve. This will get quite complicated, and you'll need to make some simplifying assumptions.

how to find the rotational inertia ?

thanks
 

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