IsaacPhysics - Climbing up a corner -

[fernihough]

Homework Statement

Struggling with a tricky (level 6) IsaacPhysics problem and need help

Relevant Equations

F = uN
Newton's LoM

Hi,

I'm stuck on this problem and have tried several different ways of approaching it
https://isaacphysics.org/questions/corner_climbing?board=7156e89c-b9a6-4e4e-886f-b8e0deed28c9


The thread on this site seemed helpful, but was never finished and I still can't get the answer.
https://www.physicsforums.com/threads/wall-climbing-force-problem.62044/

Can anyone help please?

Regards,
Mark

 

[Delta2]

Hi Mark and welcome to Physics Forums.

Unfortunately the rules here at PF say that we can't give you the full blown solution, but instead you must type here your best try and we can see what wrong steps you might have got and give you hints towards the right direction.
This problem is not so hard, it just requires a bit of thinking in three dimensions, while the majority of high school problems are usually in at most two dimensions.
Just carefully find all the forces in the x,y,z directions and then use the conditions that the total force in each x,y,z component is zero that is $$\sum F_x=0,\sum F_y=0,\sum F_z=0$$.

 

[fernihough]

Ok.

So in the Vertical direction, (y in the video), then there are two frictional forces (one on each wall)
So Sum(Fy) = 2 x Friction-y - mg = 0
So Friction-y = mg/2
(But does that assume there is no vertical component of the force from the person on the wall?!)

Taking one wall from the top (assuming the person exerts a force, F (without any vertical component)
can we say that F^2 = N^2 + Friction-x^2

Then I assumed F = mu * N

and that's where I get stuck!

 

[Delta2]

The video (and you) are using y-direction where I would use z-direction and using the same x- direction for the horizontal friction forces from the walls, while I would use $F_y$ and $F_x$ for those two friction forces. But anyway let's focus on one wall so maybe we can reduce the problem to two dimensions.

You can use the inequality $F\leq\mu N$ only for the horizontal and vertical friction forces. You are right that $F_y=\frac{mg}{2}$. Use that $F_y\leq \mu N$ to find a lower bound for the friction coefficient $\mu$.

Now here comes the bit tricky part. What balances the friction force $F_x\leq \mu N$?

EDIT : @haruspex has corrected me, we can use that $F\leq\mu N$ only for the total friction force $F=\sqrt{F_x^2+F_y^2}$. This complicates things just a bit more.

 

[haruspex]

Friction-y = mg/2

Yes.

does that assume there is no vertical component of the force from the person on the wall

No, and we do not care about forces on the wall, since those include whatever the wall is standing on. We are analysing the forces on the person.
The second diagram is misleading because it shows Fy acting down instead of up.

assuming the person exerts a force, F (without any vertical component)

F is defined in the question as the whole force parallel to the surface. It has horizontal and vertical components.

can we say that F^2 = N^2 + Friction-x^2

No. How do you get that?
F is the resultant of Fx and Fy.

F = mu * N

Yes.
Try to answer @Delta2's question about balancing Fx.

You can use the equation F=μN only for the horizontal and vertical friction forces.

Not sure what you mean by that. We can use it only for the total frictional force, F.

 

[Delta2]

We can use it only for the total frictional force, F.

Yes you are very right on that.

 

[fernihough]

In terms of balancing Fx, looking at the forces on the body as a whole from the top, then doesn't Fx = N?

 

[Delta2]

In terms of balancing Fx, looking at the forces on the body as a whole from the top, then doesn't Fx = N?

Yes $F_x$ from one wall is balanced by the normal force from the other wall.

 

[Delta2]

So now you know $F_x$ and $F_y$ calculate the total friction force $F$ and set it $F\leq\mu N$ to calculate a lower bound for $\mu$

 

[fernihough]

The question initially asks for minimum force, F first, so my calc's were thus:
(I'll use Fr for friction..)

Fr-total^2 = Fr-x^2 + Fr-y^2

substitute Fr-x = N and Fr-y = mg / 2

Fr-total^2 = N^2 + (mg/2)^2

subs Fr-total = μN

Fr-total^2 = (Fr-total / μ)^2 + (mg/2)^2

which simplifies to

Fr-total = sqrt( (mg)^2 / ( 4( μ^2 - 1 ) )I tried this as an answer and the reply was that this was the Normal force, not the frictional force.
(Just to see where this went, I then multiplied the answer above by μ and it told me this was the frictional force, which makes sense if I had calculated N and not Fr-total).

I have the answer to the total force on the body now F^2 = N^2 + Fr-total ^2
BUT, I don't understand where I went wrong in calculating N and not Fr-total...

Ideas?

Thanks for your continued help :-)

 

[fernihough]

Yes, that's what I started with - see below...

But the IsaacPhysics response tells me that I have just calculated the Normal force, not the total frictional force...

I can get to the correct answer for the total force on the body if I know why the above is the Normal force, not the total frictional force...

 

[Delta2]

You did a simple algebra mistake $$1-\frac{1}{\mu^2}=\frac{\mu^2-1}{\mu^2}$$ hence at the very last step the correct expression for $F_{r-total}$ is $$\sqrt{\mu^2\frac{(mg)^2}{4(\mu^2-1)}}$$

 

[fernihough]

Ah - now I feel dumb! ;-)
(I obviously missed the μ ^ 2 - trying to rush the simplification!)

Thank you, that all makes sense now.

The second part of the question is easy in terms of minimum μ - anything less than 1 gives imaginary roots.
This segues to quite a nice comment about the coefficient of friction as well:
https://simple.wikipedia.org/wiki/Coefficient_of_friction
:-)

 

[Delta2]

My way of proving that $\mu\geq 1$ which I think provides better physical intuition is that since $$\sqrt{N^2+(\frac{mg}{2})^2}\leq\mu N$$ we can get by dividing both sides by $N\geq 0$ that $$\mu\geq \sqrt{1+(\frac{mg}{2N})^2}$$. The term $(\frac{mg}{2N})^2$ becomes zero when the normal force N becomes infinite so the minimum value of $\mu=\sqrt{1+0}=1$.

 

[fernihough]

Cool.
Thanks again :-)

 

[haruspex]

It might be interesting to generalise it to any angle between the walls.