Does This Free-Body Diagram for a Forklift Make Sense?

[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.

 

[mcastillo356]

when it starts to climb the incline plane (whose angle we don't know; or is it also 40º?);

Hi haruspex!
The angle must not be 40º, much less instead.
I am trying to make sense of her words at the statement. "Along the handle", she mentions. The only way is to contextualize it. If I'm pushing to deal with a slope, everything tallies: only if I pretend to climb, teacher's statement makes sense, and the force exerted is downwards.
It doesn't mention (the statement) a inclined plane; but neither a coordinate system or any clue to determine the direction of the movement. Anyhow, I've $\color{red}aimed$ to explain what happens on the $x$ axis; I mean, without rise or fall... Well, I'm talking alone again.

I will quote the Physics Graduate:
1- He told me that the only way to represent the exerted force was that it was applied along the handle, this is, downwards.
2- He wrote this for the forum, talking about torques and the posibility to solve this problem fairly well, as it is used to be: with torques:
"The force of gravity generates a counterclockwise torque; then, we need another torque: this torque must be generated by the gardener, and it must have the same value as gravitational torque, but clockwise, because we need a total nule torque. This is not studied in this introductory problem; we only analyse the force (independent of the torque) that produce a linear acceleration".

 

[haruspex]

He wrote this for the forum, talking about torques and the posibility to solve this problem fairly well, as it is used to be: with torques:
"The force of gravity generates a counterclockwise torque; then, we need another torque: this torque must be generated by the gardener, and it must have the same value as gravitational torque, but clockwise, because we need a total nule torque.

Ok, so the Physics Graduate doesn't understand the question either. That is all utterly useless since we do not have enough information on the geometry of the wheelbarrow.

You have drawn it with the line of the handle passing through the centre of the wheel, but we are not told that, and neither do we know where the mass centre of the barrow+load is.
Call that mass centre O, the line along the handle L, and the point where the wheel contacts the ground C. Call the point on L directly above C the point A.
If we take moments about O, there are only two forces exerting a torque: the vertical reaction from C (anticlockwise) and the force applied along L. Since the latter must be clockwise, we can place O below L.
If the weight of barrow plus load is W, the reaction at C is W+F sin(40). We find that the line OA makes an angle $\alpha$ above the horizontal, where $W=F\tan(\alpha)\cos(40)$. It can be anywhere on that line.
This gives 86 degrees. That's a lot, but possible if the point A is high and the horizontal distance from O to A is short.

So, no need for a ramp. The question asks for the acceleration and you can calculate one, despite the bizarre requirement on how the gardener applies his force.

 

[mcastillo356]

Ok, so the Physics Graduate doesn't understand the question either. That is all utterly useless since we do not have enough information on the geometry of the wheelbarrow.

Yes, indeed

You have drawn it with the line of the handle passing through the centre of the wheel, but we are not told that,

Certain

and neither do we know where the mass centre of the barrow+load is.

Yes

Call that mass centre O, the line along the handle L, and the point where the wheel contacts the ground C. Call the point on L directly above C the point A.

Fine

If we take moments about O, there are only two forces exerting a torque: the vertical reaction from C (anticlockwise) and the force applied along L. Since the latter must be clockwise, we can place O below L.

Perfect

If the weight of barrow plus load is W, the reaction at C is W+F sin(40).

Fine

We find that the line OA makes an angle $\alpha$ above the horizontal, where $W=F\tan(\alpha)\cos(40)$. It can be anywhere on that line.

Brilliant

This gives 86 degrees. That's a lot, but possible if the point A is high and the horizontal distance from O to A is short.

Brilliant. Center of mass estimated

So, no need for a ramp. The question asks for the acceleration and you can calculate one, despite the bizarre requirement on how the gardener applies his force.

Fine, you've concluded that the center of mass is near to the wheel. That means Newton's second law (inertia) stands down there. Fine. The exerted force must be downwards. No ramp required at all.
I've waved my magic wand and a ramp has appeared.
Thanks!

 

[Lnewqban]

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.

Congratulations, señor Castillo!