Gardener pushing a wheelbarrow

[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!