Gardener pushing a wheelbarrow

[mcastillo356]

Homework Statement

A gardener pushes a wheelbarrow that has a handle that forms a 40º angle over the horizontal. if the wheelbarrow mass is 32 kg and the gardener pushes the forklift with a force of 65 N along the handle, calculate:
a) The weight of the wheelbarrow.
b) If the forklift is loaded with a 50 kg sack of earth, what will be the normal force the forklift will experience?.
c) What will be the acceletation of the forklift, supposed there is no friction?.
d) Draw a sketch with all the vectors that take place in the exercise, and designate them properly.

Relevant Equations

Newton's second law

It's the first exercise of the exam I've had today.
d) No images found. As the level required is lowly (access to university) I've drawn the forklift like a free-body diagram, with three vectors: normal, weight, and forty degrees upwards force... does it make sense?
a) w=mg=802 N.
b)##N=w-F\cdot{sin{(\theta)}}=762\;N##
c)##F_x=m\cdot{a_x}##; ##65\;N\cdot{\cos{(40)}}=82\;kg\cdot{a_x}##; ##a_x=0,61\;m/s^2##
Regards

 

[Doc Al]

A few preliminary questions.

" A gardener pushes a wheelbarrow that has a handle that forms a 40º angle over the horizontal."
Is the handle pointing down or up?

" if the wheelbarrow mass is 32 kg and the gardener pushes the forklift with a force of 65 N along the handle"
Forklift? Do the terms "wheelbarrow" and "forklift" refer to the same thing?

 

[mcastillo356]

Hi Doc Al!
The handle is pointing up; wheelbarrow and forklift refer to the same things. I've translated this first exercise from spanish. It's a "carretilla" in spanish.

 

[Doc Al]

I've drawn the forklift like a free-body diagram, with three vectors: normal, weight, and forty degrees upwards force... does it make sense?

Yes.

a) w=mg=802 N.

Did they want the weight of the wheelbarrow or wheelbarrow plus load?

b)

Looks good.

c)

Looks good.

 

[mcastillo356]

Hi Doc Al!
It looks nonsense, but I think the first question is annoying: only the wheelbarrow's weight. They also give an amount of points (1,5 over 10) for the answer that looks strange.

 

[haruspex]

b)##N=w-F\cdot{sin{(\theta)}}=762\;N##

A gardener pushes a wheelbarrow that has a handle that forms a 40º angle over the horizontal
The handle is pointing up

If the gardener pushes on a handle that points up, what does that do to the normal force from the ground?

 

[Doc Al]

If the gardener pushes on a handle that points up, what does that do to the normal force from the ground?

It's true that "pointing up" is ambiguous. I don't think I phrased my earlier "clarifying question" very clearly!

A better version: When the gardener pushes along the handle, is he pushing up or down? (At an angle, of course.)

 

[mcastillo356]

Hi Doc Al, haruspex, thanks!

 

[haruspex]

Hi Doc Al, haruspex, thanks!

Nice of you to thank me, but I would really like you to answer my question.
The way I interpret "handle pointing up" is that the gardener is pushing down on the barrow. That makes your answer wrong.

 

[mcastillo356]

I've read all the thread again, and I think I am not being clear at all. My intention is to put things in common: that is the target of the forum. I'll be back today. I am going to obtain an image.

 

[Doc Al]

I am going to obtain an image.

Excellent.

 

[mcastillo356]

The first post is well translated, except for the word "forklift", which is completely wrong: I thought they were synonymous. That means the gardener pushes, that is, uses physical force, in order to move the wheelbarrow forward. Also, there is an angle between the barrow's handle and the ground, that measures forty degrees.
This changes my point of view about the exercise, so I propose another solution:
a)##\vec{w}=m_1\vec{g}##; the same result given in my first post. ##m_1## is the wheelbarrow's mass;
b)##\vec{N}=\vec{F}\sin{40}+(m_1+m_2)\vec{g}##. ##m_2## is the sack of earth;
c)##\sum{F_x}=\vec{F}\cos{40}=ma_x##

 

[Doc Al]

This changes my point of view about the exercise, so I propose another solution:
a)##\vec{w}=m_1\vec{g}##; the same result given in my first post. ##m_1## is the wheelbarrow's mass;
b)##\vec{N}=\vec{F}\sin{40}+(m_1+m_2)\vec{g}##. ##m_2## is the sack of earth;
c)##\sum{F_x}=\vec{F}\cos{40}=ma_x##

Good, this makes much more sense. (Thanks to @haruspex for raising the flag.) Most wheelbarrows that I've seen had the person pushing downward --- and this one is no different.

For (a): in your first post, you used the total mass, not just ##m_1##.

Your diagram seems a bit not-to-scale. The force ##F## is only 65 N, but it appears much larger than the weight.

 

[jbriggs444]

Most wheelbarrows that I've seen had the person pushing downward --- and this one is no different.

No wheelbarrow I have ever seen has the person pushing downward. If you find yourself pushing downward, you are using the wheelbarrow incorrectly.

In a properly used wheel barrow, one is lifting on the two handles with a force that is nominally vertical. This is nominally at right angles to the handles which are more or less horizontal.

The load sits balanced more toward the wheel and less toward the handles. This is adjustable based on how one loads the barrow. One does not want the load too far forward because the wheelbarrow becomes difficult to manage and can even spontaneously tip forward.

The effect of this to make the lifting effort easier than simply carrying the load in one's hands. The wheel supports the bulk of the load. The user supports less than half the weight of the load.

While standing in place and before setting the load in motion, this pattern of support means that the handles which are sloping slightly upward from barrow to user are under a slight tension. The strength of the user's grip keeps the handles from slipping out from the user's gloves. [Or keeps the gloves from sliding up and off the handles]

In practice, slope and surface conditions matter greatly. In order to set the load in motion, the user leans forward and applies a force that is angled forward from the vertical. This makes the force more nearly at right angles to the handles. If the ground is soggy or upward sloping, the user may have to lean forward further still and exert a force with an even farther forward-leaning angle. This can result in putting the handles into compression.

The only circumstances in which the user's applied force on the handles would be purely pointing down the long axis of the handles would be with a badly balanced load centered directly over the wheel and with the wheel up against an obstruction such that maximum effort is needed to get past said obstruction.

A 40 degree angle from the horizontal is pretty aggressive. One would normally keep the handles more nearly horizontal -- closer to 30 degrees.

Edit: For reference, the sort of wheelbarrow that is described above is pictured here.

 

[haruspex]

No wheelbarrow I have ever seen has the person pushing downward. If you find yourself pushing downward, you are using the wheelbarrow incorrectly.

In a properly used wheel barrow, one is lifting on the two handles with a force that is nominally vertical. This is nominally at right angles to the handles which are more or less horizontal.

The load sits balanced more toward the wheel and less toward the handles. This is adjustable based on how one loads the barrow. One does not want the load too far forward because the wheelbarrow becomes difficult to manage and can even spontaneously tip forward.

The effect of this to make the lifting effort easier than simply carrying the load in one's hands. The wheel supports the bulk of the load. The user supports less than half the weight of the load.

While standing in place and before setting the load in motion, this pattern of support means that the handles which are sloping slightly upward from barrow to user are under a slight tension. The strength of the user's grip keeps the handles from slipping out from the user's gloves. [Or keeps the gloves from sliding up and off the handles]

In practice, slope and surface conditions matter greatly. In order to set the load in motion, the user leans forward and applies a force that is angled forward from the vertical. This makes the force more nearly at right angles to the handles. If the ground is soggy or upward sloping, the user may have to lean forward further still and exert a force with an even farther forward-leaning angle. This can result in putting the handles into compression.

The only circumstances in which the user's applied force on the handles would be purely pointing down the long axis of the handles would be with a badly balanced load centered directly over the wheel and with the wheel up against an obstruction such that maximum effort is needed to get past said obstruction.

A 40 degree angle from the horizontal is pretty aggressive. One would normally keep the handles more nearly horizontal -- closer to 30 degrees.

Agreed, but more information would be needed to solve the realistic version. The angle the handles make is irrelevant - we could change that just by altering where they attach to the barrow. We'd need the fulcrum geometry: point of contact with ground, centre of mass of barrow and load, point where handles are held.

 

[Doc Al]

No wheelbarrow I have ever seen has the person pushing downward. If you find yourself pushing downward, you are using the wheelbarrow incorrectly.

In a properly used wheel barrow, one is lifting on the two handles with a force that is nominally vertical. This is nominally at right angles to the handles which are more or less horizontal.

The load sits balanced more toward the wheel and less toward the handles. This is adjustable based on how one loads the barrow. One does not want the load too far forward because the wheelbarrow becomes difficult to manage and can even spontaneously tip forward.

The effect of this to make the lifting effort easier than simply carrying the load in one's hands. The wheel supports the bulk of the load. The user supports less than half the weight of the load.

While standing in place and before setting the load in motion, this pattern of support means that the handles which are sloping slightly upward from barrow to user are under a slight tension. The strength of the user's grip keeps the handles from slipping out from the user's gloves. [Or keeps the gloves from sliding up and off the handles]

In practice, slope and surface conditions matter greatly. In order to set the load in motion, the user leans forward and applies a force that is angled forward from the vertical. This makes the force more nearly at right angles to the handles. If the ground is soggy or upward sloping, the user may have to lean forward further still and exert a force with an even farther forward-leaning angle. This can result in putting the handles into compression.

The only circumstances in which the user's applied force on the handles would be purely pointing down the long axis of the handles would be with a badly balanced load centered directly over the wheel and with the wheel up against an obstruction such that maximum effort is needed to get past said obstruction.

A 40 degree angle from the horizontal is pretty aggressive. One would normally keep the handles more nearly horizontal -- closer to 30 degrees.

All good and agreed. I should have been clearer that I was addressing just the artificial problem as stated.

(I was having similar thoughts after I had made my earlier comments -- that this problem is quite unrealistic as to how wheelbarrows actually work. Thanks for your comments.)

 

[mcastillo356]

Well, I think I must post the teacher. I'll be back
Thanks for your comments. Truly indeed

 

[Lnewqban]

Well, I think I must post the teacher. I'll be back
Thanks for your comments. Truly indeed

Could you post a link to the original problem?

 

[mcastillo356]

Ok. It's spanish.

 

[Lnewqban]

Thank you.
It seems to me that the problem can't refer to a wheelbarrow, because we run into the problem of the lever effect and it does not give you any information about distances among hands, load and wheel.
Let's assume that the word "carretilla" refers to a push cart dolly instead.
The word "asa", previously translated as forklift, should become handlebar.

 

[haruspex]

Thank you.
It seems to me that the problem can't refer to a wheelbarrow, because we run into the problem of the lever effect and it does not give you any information about distances among hands, load and wheel.
Let's assume that the word "carretilla" refers to a push cart dolly instead.
The word forklift should become handlebar.

View attachment 268843

That makes more sense.

 

[Lnewqban]

That makes more sense.

In this particular problem, the handlebar is not perpendicular to the platform, but it forms an angle of 40° with respect to it (sorry, I could not find a suitable picture).
The gardener pushes with 65 N along that handlebar (same 40° respect to horizontal).

 

[haruspex]

In this particular problem, the handlebar is not perpendicular to the platform, but it forms an angle of 40° with respect to it (sorry, I could not find a suitable picture).
The gardener pushes with 65 N along that handlebar (same 40° respect to horizontal).

But in reality, the angle of the handle is only loosely related to the direction the force will be applied unless the handle swivels where it meets the cart.

 

[mcastillo356]

Anyhow, I've already posted the teacher. Let's see what she says. It's my personal opinion. No gardener in Spain uses the thing you mention.

 

[Lnewqban]

But in reality, the angle of the handle is only loosely related to the direction the force will be applied unless the handle swivels where it meets the cart.

I agree.
Just as stated in the original wording.
Perhaps to give a clear direction of the exerted force.

 

[Lnewqban]

Anyhow, I've already posted the teacher. Let's see what she says. It's my personal opinion. No gardener in Spain uses the thing you mention.

Ask her about these missing dimmensions:

 

[mcastillo356]

Fine, its an exercise I must solve. I need time. She will help me. I'm sure at 99%. Regards.
Marcos

 

[mcastillo356]

Lnewqban, It is not needed to understand spanish to understand this exercise. It's the one stated by you in your picture. It's statics. The link seems endless: a lot of statics exercises; and it's full of advertising, but very good
Until monday my teacher is not going to answer, as she always do

 

[Lnewqban]

Lnewqban, It is not needed to understand spanish to understand this exercise. It's the one stated by you in your picture. It's statics. The link seems endless: a lot of statics exercises; and it's full of advertising, but very good
Until monday my teacher is not going to answer, as she always do

Hola, Marcos.
In that video, you can see that the problem is just to resolve the distribution of forces along a lever, having a balance of moments respect to the fulcrum.
Note that all forces have a vertical direction of application, reason for which the distances that matter to solve the balance of moments are the horizontal ones.

For that reason, if your problem is similar, we need the distances between the point of application of each force and the common fulcrum.

Please, note that the force of the hands of the gardener are vertically up only.
That is because she is not trying to accelerate the wheelbarrow forward, just keeping a steady speed (not fighting any rolling resistance in theory).
Bassically, she is only counter-acting gravity (not lifting up or lowering down the handlebar) and walking.

If she was trying to accelerate the wheelbarrow forward, like before going up a ramp for example, the force of her hands would be pointing up and forward, having horizontal and vertical components, both positive.
That vertical component would have to be equal in magnitude to the previous vertical force she was exerting while walking, so balance of moments remains.

The force exerted by the hands of the gardener of your problem points in the wrong direction if he is actually driving a wheelbarrow.
He simply can't push forward and down 40° respect to the ground and still make the loaded wheelbarrow accelerate forward.

That, and the absence of the distances between forces and fulcrum, is what made me think of a different type of "carretilla" (a wheeled flat platform with a handlebar).
My apologies if I am wrong.

 

[mcastillo356]

It's me who must apologize. I've misunderstood, the statement, and that is because I have not made use of the common sense. Simply I've read word by word, not trying to conclude the physical terms in which is written. Not trying to extract the main message. Not portraying, not describing the statement as what it is written for: to describe a physical fact.
I strongly agree with your words, but let's not advance any hypotesis about the solution. We might be loosing more precious time: she is good, very good: the statement of the exam's exercise must be self-contained. I am sure. And not despite there is a lack of information, as you pretty well say: probably because it's written for beginners.

 

[Lnewqban]

... she is good, very good: the statement of the exam's exercise must be self-contained. I am sure. And not despite there is a lack of information, as you pretty well say: probably because it's written for beginners.

No apologies needed.
As you learn the details of the problem from your teacher, please come back, we should be able to help with either case.
Saludos

 

[mcastillo356]

Still waiting for news from her.
Lnewqban, Doc Al, haruspex, jbriggs444, I will post as soon as she posts me. Meanwhile I am working on it, but nothing I can rely on enough to publish.

 

[mcastillo356]

Hi, PF

On next saturday I will post. I will have the help of a Physics Graduate on friday. Anyhow, I would estimate PF's advice, Lnewqban, Al Doc, haruspex... No news from the teaching staff, so I've decided to work it out with or whithout them... If they happen to post before friday, I would be thankful, and their words should be definitive to make up my mind about the exercise.
Greetings. Meet you this weekend

 

[mcastillo356]

<iframe src="https://www.geogebra.org/classic/tfjcvjjb?embed" width="800" height="600" allowfullscreen style="border: 1px solid #e4e4e4;border-radius: 4px;" frameborder="0"></iframe>

Well, my personal opinion: as you can see in the link above, the force exerted is downwards. This way:
1- The statement is well, the 40º grades are over the horizontal, the wheelbarrow is pushed along the handle;
2- The wheelbarrow is at the very moment when it starts to climb the incline plane (whose angle we don't know; or is it also 40º?);
3- The wheel experiences a torque, steps into the incline.
Now, does it make sense all said in relation with the problem?
No news from teachers' staff. The PG has gave me this idea.
I've passed the exam: 6,30 over 10. I've passed all the subject. Next week, or in two weeks, I will pay the registration.
Every opinion welcome. Greetings

 

[haruspex]

when it starts to climb the incline plane

What inclined pane? I see no prior mention of that.
If the gardener is pushing down at 40 degrees and the slope goes up at 40 degrees he's going nowhere unless his mass is at least 85 kg x cos(50) x cos(50) / cos(80) = 195 kg. And he will certainly need to exert more than 65N.