How High Can a Wheelbarrow Go Over a Step Relative to Its Wheel Radius?

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Homework Help Overview

The problem involves a uniformly-loaded wheelbarrow attempting to ascend a horizontal step. The scenario includes a person applying a maximum horizontal force to lift the wheelbarrow over the step, with the height of the step expressed as a fraction of the wheel's radius.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the wheelbarrow and the torque involved in lifting it over the step. There are attempts to derive relationships between the height of the step and the wheel's radius, with some questioning the relevance of torque in the context of the problem.

Discussion Status

There is an ongoing exploration of different diagrams and approaches to understand the mechanics involved. Some participants suggest focusing on the normal force and its relationship to the weight of the wheelbarrow, while others propose analyzing torques about the point of contact with the step. Multiple interpretations of the forces and angles involved are being examined.

Contextual Notes

Participants note the assumption of a horizontal force and the implications of the wheelbarrow's center of mass in relation to the forces acting on it. There is also mention of the need to clarify the direction of the normal force and the potential for frictional forces, which are not assumed in this scenario.

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Homework Statement


A person pushing horizontally a uniformly-loaded 45.2 kg wheelbarrow is attempting to get it over a horizontal step. The maximum horizontal force that person could apply is F = 410 N. What is the maximum height of the step, as a fraction of the wheel's radius, the person could get the wheelbarrow over? Assume that the height of the step is h=nR.

Homework Equations


τ=Fr
F = mg



The Attempt at a Solution


Calculating the wheelbarrow at the edge of step.

hence
Fr = mg

and I'm stuck here of how to apporach on to the next step
 
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This isn't a question about friction. Try drawing a diagram for a step height equal to about 1/2 the radius of the wheel (n=1/2). What forces act on the wheel?
 
To draw CWatters' diagram, you need to choose some point in the process of getting the wheel up the step. It should be obvious at what point the force needed is maximised; if not, put in an unknown for, say, the angle that the radius from wheel centre to point of contact with step makes to the horizontal.
 
diagram.png


This is the image I have came out with.
From this picture I can conclude that
τ = Frsinθ

The known values are:
F = 410

I know that Force of gravity is pushing down and the torque for wheel pushing upward. Hence,
mgR/2 = FRsinθ

(mg)/(2F) = sinθ

Is the approach correct? because I could not find the fraction for it
 
I don't think you have to worry about torque here. Just find the normal force upward as it depends on how high the step is and set it equal to the weight.
 
qqchan said:
diagram.png


This is the image I have came out with.
From this picture I can conclude that
τ = Frsinθ

The known values are:
F = 410

I know that Force of gravity is pushing down and the torque for wheel pushing upward. Hence,
mgR/2 = FRsinθ

(mg)/(2F) = sinθ

Is the approach correct? because I could not find the fraction for it
F is applied horizontally.
Which angle is theta? Is it what I suggested?
How did you get mgR/2?
 
haruspex said:
F is applied horizontally.
Which angle is theta? Is it what I suggested?
How did you get mgR/2?

Theta is the angle between the radius and the point of contact with the step. However, I am confused with "the angle that the radius from wheel centre to point of contact with step makes to the horizontal."

I got mgR/2 from the thought of finding the center of mass
 
Can I suggest a slightly different diagram. I would also look at the torques about the point where the wheel meets the step.
 

Attachments

  • barrow.jpg
    barrow.jpg
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Haruspex,

Wheel.png


Diagram.png


I have decided to remove my thought on torque and drew the body diagram again.

sin θ = (R-h)/R
θ = 1-h/R
θ = 1-n

Is this what you are suggesting?
 
  • #10
CWatters said:
Can I suggest a slightly different diagram. I would also look at the torques about the point where the wheel meets the step.

So according to your diagram
tanθ=Mg/2F

The torque where the points meet. The formula for torque is
τ=Frsinθ
 
  • #11
qqchan said:
Haruspex,

Wheel.png


Diagram.png


I have decided to remove my thought on torque and drew the body diagram again.

sin θ = (R-h)/R
θ = 1-h/R
θ = 1-n

Is this what you are suggesting?
I think you mean sin θ = 1 - n.
The new diagram is better, but you have the normal wrong. When a point contacts a smooth surface, the normal is the normal to the surface, in this case the wheel. Correct that, then write out the horizontal and vertical statics equations.
 
  • #12
qqchan said:
So according to your diagram
tanθ=Mg/2F

The torque where the points meet. The formula for torque is
τ=Frsinθ

I was thinking that to rotate about the point of contact with the step there must be a net clockwise torque.

Clockwise you have F(rSinθ)

Anticlockwise you have 0.5mg(rcosθ)

The brackets are just to highlight the distance term.

The factor of 0.5 comes from the fact that roughly half the weight is supported by the wheel on one end and half by the man on the handles at the other. I guess it depends what the expression "uniformly distributed" means but I reckon this is a reasonable approximation.

So I think the condition to climb the step is..

FrSinθ - 0.5mgrcosθ > 0

Edit: I'm happy to be convinced that another approach is better !
 
  • #13
CWatters said:
The factor of 0.5 comes from the fact that roughly half the weight is supported by the wheel on one end and half by the man on the handles at the other.
It says the force is horizontal. I think you have to take that literally or there's not enough information to answer the question. Of course, when pushing a wheelbarrow on the flat the person has to take some of the vertical load, but typically the handles are long enough that the wheel takes most of it. When trying to get up the step, there's no longer a need for the person to apply any vertical force. It can be purely horizontal now.
 
  • #14
So after changing the direction of the normal force. Here are the force vectors I get:

F = Fn
Fg = ma

τnet = 0

τnet = -Frsinθ + mgR/2sinθ


Fr = mgR/2sinθ
F = mg/2sinθ

2F/mg = sinθ

2F/mg = 1-n
Am I missing something? Or did I do write an equation wrong.
 
Last edited:
  • #15
qqchan said:
So after changing the direction of the normal force. Here are the force vectors I get:

F = Fn
Fg = ma

τnet = 0

τnet = -mgL/2 + mgRsinθ

Am I missing something? Or did I do write an equation wrong.
You seem to have made the normal force (Fn?) horizontal now. It should be radial to the wheel. F is horizontal, mg is vertical. There could also in principle be a frictional force from the step tangential to the wheel. You can show there is not, but you should not assume it.
 
  • #16
haruspex said:
You seem to have made the normal force (Fn?) horizontal now. It should be radial to the wheel. F is horizontal, mg is vertical. There could also in principle be a frictional force from the step tangential to the wheel. You can show there is not, but you should not assume it.

if Fn = normal force.
So, if it is tangential to the wheel, the best way to make it work would be the torque. There is no frictional force in this case.

τ=Frsinθ

F = Fn in such case.

is this what you are suggesting?
 
  • #17
qqchan said:
if Fn = normal force.
So, if it is tangential to the wheel, the best way to make it work would be the torque.
No, not tangential, radial. If there is a frictional force, that will be tangential.
If you take moments about the point of contact with the step, as CWatters suggested, you won't need to worry about this normal and frictional pair of forces. Neither will have any moment about that point. What torques do mg and F have about that point?
 
  • #18
haruspex said:
No, not tangential, radial. If there is a frictional force, that will be tangential.
If you take moments about the point of contact with the step, as CWatters suggested, you won't need to worry about this normal and frictional pair of forces. Neither will have any moment about that point. What torques do mg and F have about that point?
Iw2 would be the radial.

Iw2 = F

τnet = -Frsinθ + mgR/2sin(90-θ)

I guess these are the formulas
 
Last edited:
  • #19
qqchan said:
Iw2 would be the radial.
That's a formula for (twice) rotational kinetic energy!
τnet = -Frsinθ + mgR/2sin(90-θ)
Close, but lose the /2. That assumes the wheelbarrow's centre of mass is half way between the wheel and the handles. They're designed to have the centre of mass much further forward (when loaded), as close as possible to being directly over the wheel without risking tipping forwards. So you should take the weight as acting down through the centre of the wheel.
And r = R, I hope.
At equilibrium, what is the value of τnet?
 
  • #20
haruspex said:
That's a formula for (twice) rotational kinetic energy!

Close, but lose the /2. That assumes the wheelbarrow's centre of mass is half way between the wheel and the handles. They're designed to have the centre of mass much further forward (when loaded), as close as possible to being directly over the wheel without risking tipping forwards. So you should take the weight as acting down through the centre of the wheel.
And r = R, I hope.
At equilibrium, what is the value of τnet?

τnet = 0;

FRsinθ = mgRsin(90-θ)

F/(mg)=cotθ

(mg)/F = tanθ

Solve for θ

since sinθ = 1-n.

This way we could solve for n. I guess this is the way
 
  • #21
qqchan said:
τnet = 0;

FRsinθ = mgRsin(90-θ)

F/(mg)=cotθ

(mg)/F = tanθ

Solve for θ

since sinθ = 1-n.

This way we could solve for n. I guess this is the way

Loos right to me.
 
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  • #22
Hmm, that is right
 
Last edited:

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