Push/Pull Force of wheeled cart on an incline

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To calculate the force required to push a 227 kg wheeled cart up a 15-degree incline, the relevant formula incorporates gravitational force and rolling resistance, expressed as F = mg sin(θ) + mgc cos(θ), where c is the rolling resistance coefficient. The distinction between using the radius of the wheel or not pertains to how rolling resistance is calculated, but for this scenario, the rolling resistance coefficient suffices. The coefficient of friction of the incline's surface is not directly relevant since the rolling resistance already accounts for the forces involved. It is suggested to consider pushing at a slight angle above the slope to reduce the normal force and, consequently, the rolling resistance. Testing the cart's actual resistance on a flat surface may provide further insights into its performance on the incline.
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Summary:: Looking for the formula to calculate force required to push a wheeled cart weighing 227 kg up a 15 degree incline.

I’m trying to find the formula for force required to push a 227kg cart with four wheels up an incline that is 15 degrees. From my physics classes I thought the formula is (F= m*g*Sin(theta) + m*g*c) where c is coefficient of the wheel material. I have also found a similar formula that uses the radius (r) of the wheel but I do not understand difference between ((m*g*c)/r) and m*g*c and where one applies over the other.

Also, if the coefficient of friction of the wheel as an example is 0.05, how do I include the coefficient of friction of the incline’s surface?

Again, just looking for a formula for minimum required force required to push the wheeled cart up the incline.

Thanks in advance.
 
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I assume you are attempting to denote by ##c## the rolling resistance coefficient. This is conventionally defined in two different (although similar) ways:\begin{align*}
f &= cN \\ \\
\mathrm{or} \ \ \ f &= \dfrac{\tilde{c}N}{r}
\end{align*}where ##f## is the force of rolling resistance, ##N## is the normal contact force and ##r## is the radius of the wheel.

On an incline, the total normal contact force is ##N = mg\cos{\theta}##. Therefore the total rolling resistance is ##f = mgc \cos{\theta} = \dfrac{mg\tilde{c}}{r} \cos{\theta}##. Add to this the gravitational contribution ##mg\sin{\theta}## to determine the result.
 
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Yes, rolling resistance coefficient. Thank you again for your help.
How do I take into account the friction coefficient of the ramp's surface into this equation?
 
The coefficient of friction ##\mu## applies to sliding, so isn't relevant here. All of the frictional force (in this idealisation, at least) is accounted for in the rolling resistance ##f = mgc \cos{\theta}##.

[However, the rolling resistance coefficient does probably depend on the coefficient of friction, in some complicated way. That's more a question of tribology, which I'm not qualified to comment on. In any case, you don't need this relationship.]
 
ergospherical said:
The coefficient of friction ##\mu## applies to sliding, so isn't relevant here. All of the frictional force (in this idealisation, at least) is accounted for in the rolling resistance ##f = mgc \cos{\theta}##.

[However, the rolling resistance coefficient does probably depend on the coefficient of friction, in some complicated way. That's more a question of tribology, which I'm not qualified to comment on. In any case, you don't need this relationship.]
No, rolling resistance is quite separate from the friction associated with relative tangential motion of surfaces. And to clarify, for rolling there will be static friction, usually denoted ##\mu_s##. But since there is no relative motion of the surfaces it does no net work.
For more on rolling resistance see 4.1 in https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/
 
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haruspex said:
No, rolling resistance is quite separate from the friction associated with relative tangential motion of surfaces. And to clarify, for rolling there will be static friction, usually denoted ##\mu_s##. But since there is no relative motion of the surfaces it does no net work.
For more on rolling resistance see 4.1 in https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/
Thanks, you're right. The rolling resistance arises as the sum of normal forces [which are greater at front than rear] over the curved contact area, whilst the tangential [static friction] forces over the contact area would indeed sum together to produce an additional static friction force.
 
Rusty_Shackleford said:
minimum required force required to push the wheeled cart up the incline.
You should consider the possibility of pushing in a direction a bit above the angle of the slope. That will reduce the normal force from the ground, and hence reduce rolling resistance. Maybe it will require a smaller force overall.
 
Could you field test the actual resistance to rolling of the cart on a horizontal surface of similar characteristics as of the ramp?
You may have surface imperfections, little valleys and crests, that may be more significative to actual rolling resistance of the wheels than the bearings coefficients of friction.
 
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