Push/Pull Force of wheeled cart on an incline

In summary: Testing on a horizontal surface may give a more accurate representation of the resistance to rolling.
  • #1
Rusty_Shackleford
2
0
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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  • #2
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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  • #3
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?
 
  • #4
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.]
 
  • #5
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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  • #6
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.
 
  • #7
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.
 
  • #8
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.
 

Related to Push/Pull Force of wheeled cart on an incline

1. What is the definition of "Push/Pull Force"?

Push/Pull Force refers to the amount of force required to move an object, either by pushing or pulling, in a specific direction.

2. How does the incline of the surface affect the Push/Pull Force of a wheeled cart?

The incline of the surface affects the Push/Pull Force of a wheeled cart by increasing or decreasing the amount of force required to move the cart. A steeper incline will require more force, while a flatter incline will require less force.

3. What factors influence the Push/Pull Force of a wheeled cart on an incline?

The factors that influence the Push/Pull Force of a wheeled cart on an incline include the weight of the cart, the angle of the incline, the surface friction, and the force applied by the person pushing or pulling the cart.

4. How can the Push/Pull Force of a wheeled cart on an incline be calculated?

The Push/Pull Force of a wheeled cart on an incline can be calculated by using the formula F = mg sinθ, where F is the force, m is the mass of the cart, g is the acceleration due to gravity, and θ is the angle of the incline.

5. Is it easier to push or pull a wheeled cart on an incline?

It depends on the direction of the incline and the force applied. Generally, it is easier to push a wheeled cart on a downward incline, as gravity helps to move the cart. However, on an upward incline, it may be easier to pull the cart as the force applied is in the same direction as the movement of the cart.

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