Refresher on pulling weight on a wheel

[Home Improvement]

<Moderator's note: Moved from a technical forum and thus no template.>

Hi all,

This is a simple system. Assume the weight is completely balanced on the wheel (not going to tilt to any direction). the weight (red block) weighs 200 kg. the wheel (2 of them) each weighs 4 kg.
how to calculate:
1. the minimum force needed to get the weight moving
2. total energy needed to get the system to move 2m from point of origin
3. as I understand it, the friction coefficient of 2 things in contact is not always constant. so, I;m not sure how to get this one metric.

Thanks

 

[BvU]

Hi,

You are posting in a physics forum, so you may expect a 'physics' answer. Since the question is 'how to calculate', the 'physics'answer is: not. At least, not with the information given.

You'll need extra information, namely on the friction. Without friction your minimum force is zero and the energy needed is zero too. (with a small addition: you need to borrow some energy to get it moving; but in principle you can get it back when you bring it to a stop). 'Physics'ally.

Real life brings in friction -- fortunately. A difficult beast, so we try to model it - but we really need to simplify considerably to get something workable:

We assume two categories: static friction and kinetic friction. The first relates to the 'get it moving' and the second to the 'keep it moving'.

So next next best answer (and the first useful one ) to the double zero is: A minimum force to get it moving is $\mu_s F_N$ and the minimum energy to displace the contraption over 2 m is $\mu_k F_N d$

$\mu$ are the friction coefficients -- in practice defined by $\mu \equiv { F_{\rm \ friction} \over F_N}\qquad$ (nicely circular ! )
$F_N$ is the normal force: that what's needed to keep the contraption from sinking into the floor. Usually $F_N = mg$, so here some 2080 N if $g$, the gravity acceleration, is 10 m/s²
$d$ is the displacement, so here 2 m.

And there you are -- all the help we can give you and no answer

 

[Home Improvement]

ok, so, is the friction coefficient the only thing missing from solving this?
if so, is there a table of static and kinetic coefficient table for different things? so like a table of density. is it guaranteed that if material A comes in contact with material B anytime, anywhere, the static and kinetic coefficient will be X and Y?
I remember in high school, the coefficient was provided in the problem, but I always wondered if those numbers were just arbitrary number between 0 and 1.

 

[CWatters]

I can only see the first of your three drawings and this shows a cart on wheels. So my answer is based on just that drawing...

The force required to start it moving will depend on the rolling resistance of the wheels plus any friction in the bearings. Normally friction in the bearings can be ignored (if they are good bearings).

https://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html

If you want it to accelerate you also need to include inertia.

Friction between the tyre and ground won't really come into it unless the wheel brakes are on causing the wheels to skid.

 

[Home Improvement]

The force required to start it moving will depend on the rolling resistance of the wheels plus any friction in the bearings. Normally friction in the bearings can be ignored (if they are good bearings).

https://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html

If you want it to accelerate you also need to include inertia.

Friction between the tyre and ground won't really come into it unless the wheel brakes are on causing the wheels to skid.

ok, so, assuming that they're good bearings (negligible friction), using the coefficient from the car on asphalt example (because the tire used in the diagram will be car tires with pressure 40 psi or 2.76 bar with speed of 0.2 km/h), the calculation becomes:
c = 0.005 + (1 / 2.76) x (0.01 + 0.0095 x (0.2/100)²) = 0.0087
so, the force needed to pull 200kg is
F = 200 x 9.8 x 0.0087 = 17.052 N ~ lifting a 1.7kg weight

is this correct? I mean, it would take so little effort to move 200kg on wheels with good bearings?
btw, no torque calculation involved in this?

thanks

 

[haruspex]

Assume the weight is completely balanced on the wheel (not going to tilt

no torque calculation involved in this?

If it was perfectly balanced when at rest, and stays balanced when the force is applied, the force will have to be applied at just the right height.

 

[BvU]

Now about this inertia: if you want to accelerate 208 kg to 200 m/h = 1/18 m/s you need some force too. Say you want to do this with a constant force over the full 2 m. Then Force * displacement = kinetic energy yields a puny extra required force of 0.16 N.

And note that with a 1 cm height difference from a loose tile your required force to overcome that jumps up considerably !

 

[CWatters]

If it was perfectly balanced when at rest, and stays balanced when the force is applied, the force will have to be applied at just the right height.

Or at the correct angle.

 

[CWatters]

ok, so, assuming that they're good bearings (negligible friction), using the coefficient from the car on asphalt example (because the tire used in the diagram will be car tires with pressure 40 psi or 2.76 bar with speed of 0.2 km/h), the calculation becomes:
c = 0.005 + (1 / 2.76) x (0.01 + 0.0095 x (0.2/100)2) = 0.0087
so, the force needed to pull 200kg is
F = 200 x 9.8 x 0.0087 = 17.052 N ~ lifting a 1.7kg weight

is this correct? I mean, it would take so little effort to move 200kg on wheels with good bearings?

From the link I posted above the rolling resistance or force required to pull 200kg is simply
F = c W
where c = 0.02 (from the table)
W= 200 * 9.81 = 1,962 N
so
F = 0.02 * 1962 = 39N

Not too far off your figure.

 

[BvU]

I mean, it would take so little effort to move 200kg on wheels

Experiment with a Dinky Toy (or any other) truck and e.g. an 0.1 kg load -- see if it needs 0.1/208 x 17 Newton (a small weight over a pulley) to get going

(from http://teacher.nsrl.rochester.edu/PhyInq/Experiments/P09/P09_A_of_Cart_1.html)

 

[CWatters]

Just for info I googled to see what force would be required to drag 200Kg on steel skids (eg without wheels). Found this old paper from 1962 in which NASA measures the coefficient of friction between a steel skid and asphalt (in the context of rocket landing skids)..

http://www.dtic.mil/dtic/tr/fulltext/u2/270810.pdf

Fig 9 give the coefficient of friction for steel on asphalt as about 0.25. So it would take about

200 * 9,81 * 0.25
= 490N

Although that would be the force to keep it moving not start it moving.

 

[Home Improvement]

And note that with a 1 cm height difference from a loose tile your required force to overcome that jumps up considerably !

what do you mean with loose tile?

Just for info I googled to see what force would be required to drag 200Kg on steel skids (eg without wheels). Found this old paper from 1962 in which NASA measures the coefficient of friction between a steel skid and asphalt (in the context of rocket landing skids)..

http://www.dtic.mil/dtic/tr/fulltext/u2/270810.pdf

Fig 9 give the coefficient of friction for steel on asphalt as about 0.25. So it would take about

200 * 9,81 * 0.25
= 490N

Although that would be the force to keep it moving not start it moving.

so, for this case, there's no static or kinetic coefficient? just 1 rolling coefficient

 

[haruspex]

Or at the correct angle.

True, but the diagram shows it as horizontal.