Help (required force to move a sliding door weighing 100 kg with two rollers)

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SUMMARY

The required force to initiate movement of a sliding door weighing 100 kg with two rollers is approximately 4 kg. This calculation considers the weight distribution of 50 kg per wheel and utilizes a coefficient of rolling resistance of 0.0005 for steel on steel. The primary resistance arises from friction between the wheel and the axle, rather than rolling friction, which is a misnomer in this context. Accurate computation of the frictional forces is essential for effective door operation.

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  • Understanding of basic physics concepts such as force, weight, and friction.
  • Familiarity with rolling resistance and its distinction from rolling friction.
  • Knowledge of basic mechanical engineering principles related to wheels and axles.
  • Ability to perform calculations involving coefficients of friction and weight distribution.
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  • Research the principles of rolling resistance in mechanical systems.
  • Learn about the different types of bearings and their impact on friction.
  • Explore the calculations for frictional forces in wheel and axle systems.
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Mechanical engineers, physics students, and anyone involved in the design or maintenance of sliding door systems will benefit from this discussion.

johnpaul
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what is the required force to start moving a sliding door weighing 100 kg with two rollers.
what are the forces acting on the system?

shall i used rolling friction on this?

what formula to use?

thanks a lot.
 
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Welcome to PF!

Hi johnpaul! Welcome to PF! :wink:

There's no such thing as "rolling friction", since a pair of rolling surfaces don't move with respect to each other.

There's rolling resistance, which is essentially energy lost due to deformation of the wheel, but I shouldn't think three's much of that with steel.

The main problem will be friction between the wheel and the axle though its centre. :smile:
 


tiny-tim said:
Hi johnpaul! Welcome to PF! :wink:

There's no such thing as "rolling friction", since a pair of rolling surfaces don't move with respect to each other.

There's rolling resistance, which is essentially energy lost due to deformation of the wheel, but I shouldn't think three's much of that with steel.

The main problem will be friction between the wheel and the axle though its centre. :smile:

But wouldn't the bearings be subject to rolling resistance as opposed to friction?
Maybe you assumed no bearings whereas I assumed bearings :-)
 
hi tim, hi molydood,

please correct me if I am wrong.

here's my computation:

Weight of door=100 kg
Number of wheels= 2
Weight per wheel =(100kg)/(2)= 50kg

F1= αW / r

F1- resistant force of a single wheel
α - coefficient of rolling friction
W - weight
r - radius (0.015m)

F1≈(0.0005 m)(50 kg) / (0.015 m)
F1≈ 1.67 kg; say 2kg

F''≈(F1)(2 wheels)
F''≈(2kg)(2 wheels)
F''≈4 kg

Force to move the 100 kg door horizontally:
F≈ 4 kg



Values for rolling friction from various sources are not consistent and the following values should only be used for approximate calculations.

Steel on Steel α = 0.0005m
 
hi,

can anyone advise me what is the proper way in computing the friction between wheel and the axle? its just a simple door carrier(please refer to the drawings above).

thanks
 

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