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

1. Oct 8, 2009

### johnpaul

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.

2. Oct 8, 2009

### tiny-tim

Welcome to PF!

Hi johnpaul! Welcome to PF!

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.

3. Oct 8, 2009

### Molydood

Re: Welcome to PF!

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

4. Oct 9, 2009

### johnpaul

hi tim, hi molydood,

please correct me if im 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

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

5. Oct 9, 2009

### johnpaul

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