Question

A small mass m is set on the surface of a sphere, the figure.

If the coefficient of static friction is 0.75, at what angle ϕ would the mass start sliding?

Answer 1

This is the figure:
http://session.masteringphysics.com/problemAsset/1057220/4/GIANCOLI.ch05.p087.jpg

Answer 2

This is my free-body diagram.

Answer 3

I wanted to make sure that I got my angles on here right. I think I know how to solve it, but the cosine and sine always through me off. =/

Answer 4

its better to resolve mg, along the directions of friction and Normal..

Answer 5

So tilt the axis?

Answer 6

yup yup.. i think you can do it this way.. but then u need to resolve friction as well.. we ll try this method later
but m sure if u just resolve mg, like u said tilting the axis.. u ll get the answer pretty soon

Answer 7

Okay, let me try that then.

Answer 8

So, I kind of flipped the picture I drew in my notes so that the block was on top of the sphere so the normal force is opposite mg and not out at an angle. Is this right?

Answer 9

no need to flip the pic.. just consider the flipped axis.. :P

Answer 10

Oh, dang, I see what you mean.

Answer 11

Let me use the power of paint to see if I got this right.

Answer 12

ok.. i did it too.. u try.. and then i ll upload mine!

Answer 13

Here is what I think it is supposed to look like.

Answer 14

no thats wrong.. and when i say wrong.. i mean horribly wrong

why are you resolving normal force? its already along your chosen y direction, and friction along your negative x direction, the only thing you have to resolve is Mg..
concentrate bro/sis !!

Answer 15

1. Not appreciating the sarcasm as I am just as frustrated as you.
2. You said that I needed to resolve the normal force in your first response.

Answer 16

lol m not frustrated :D.. sorry about that.. i have a habit of doing that.. :-/..

and no.. read again. i did mention resolve mg..!!

Answer 17

@Hagopian13 can you try once again?!

Answer 18

I am in the process of working on it.

Answer 19

good good good.. ll wait..

Answer 20

Okay, makes enough sense.

Answer 21

as long as you get it :P

Answer 22

so did you get the answer @Hagopian13

Answer 23

Both angles are wrong.

Answer 24

So lets stick to the free body diagram shall we? come on finish that.. you can do it ;-)

Answer 25

I worked out the whole thing longhand and got 53.1 degrees off the y-axis for slide to begin. Does that sound right?

Answer 26

I used the equations and drawing from this link:
http://hyperphysics.phy-astr.gsu.edu/hbase/mincl2.html#c1

Answer 27

If you use the above link, remember that it is measuring the theta angle from the x-axis and that was 36.9 degrees off the x-axis

Answer 28

Hi!! Look at my attachment - hopefully I didn't make any mistakes in setting the diagram up.

I got the opposite of @biire, where I got \(36.9^\circ\) from the \(y\)-axis.

Answer 29

This is a neat physics trick.

Pretty much, you want to examine the case where friction force is less than the gravitational force that will make it slide.

So, I split up the forces in the picture.
I saw that you did similarly.

Anyway,
\(\left| F_f\right|\lt \left|mg\sin\phi\right|\)

We know what \(F_f\) is.
\(F_f=F_N\ \mu\)

And we know what \(F_N\) is.
\(F_N=mg\cos\phi\)

Going through those substitutions,
\(F_f=F_N\ \mu\)
\(F_f=mg\cos\phi\ \mu\)

\(\left| mg\cos\phi\ \mu\right|\lt \left|mg\sin\phi\right|\)

We can divide by \(\left| mg\right|\) to get
\(\left| \cos\phi\ \mu\right|\lt \left|\sin\phi\right|\)

Hmm.... \(\phi\) in two places... So... We fix that. \(\sin\) and \(\cos\) have a bunch of fun relations. One is \(\dfrac{\sin\theta}{\cos\theta}=\tan\theta\).
So we divide by \(\left|\cos\phi\right|\) so we wind up with \(\phi\) in only one spot. Take a minute and look at that cool trick, if you want.
So
\(\left|\mu\right|\lt\dfrac{\sin\phi}{\cos\phi}\)

\(\left|\mu\right|\lt\tan\phi\)

And then we take the inverse tangent of both sides. I honestly don't know how this effects the inequality. Sometimes things like this will create more inequalities. But we'll find out for this problem by common sense in the context. We are looking for the angle where gravity is more than friction.
\(\arctan\left|\mu\right|\ \ ?\ \ \phi\)
\(\arctan(0.75)=36.869 ....^\circ=36.9^\circ=\phi\)

So, if phi is less than this, normal force is greater (look at the diagram). So friction force will be greater than gravity.
At that angle we found, the friction and gravity are equal (as if we went through with equalities rather than inequalities).

So, when the angle is any greater than that... Which is pretty much just that angle... We see that the box will slide.

Feel free to ask questions for me or whoever is on! Take care!