# Static Friction of hollow sphere rolling up incline

For hollow sphere rolling up incline:

I know that the kinetic friction will equal 2/3 acceleration since:

μgcosθ=2/3a

acceleration = 3/5gsinθ

so...kinetic friction = 2/5gsinθ

But how I do calculate the force of static friction??

Would the static friction be the max static friction?

Max static friction = tan θ times mgcosθ

Would the static friction exerted by incline on the hollow sphere be the max static friction?

tiny-tim
Homework Helper
hi sweetpete28! acceleration = 3/5gsinθ

so...kinetic friction = 2/5gsinθ

yes, the acceleration is 3/5gsinθ,

but there's no kinetic friction for a rolling object (beacuse there's no relative motion at the point of contact) …

that 3/5gsinθ is the static friction! Hi tiny-tim,

So the magnitude of the acceleration is equal to the force of the static friction??

tiny-tim
Homework Helper
oops!

Hi tiny-tim,

So the magnitude of the acceleration is equal to the force of the static friction??

oops! i copied the wrong line! i meant that 2/5gsinθ But that is what I did and the answer is wrong:

For a I got 3.77 --> (3/5)gsin θ = (3/5)(9.81)(sin39.8degrees) = 3.77 (this is right)

Force of static friction = (2/5)gsinθ = (2/5)(9.81)(sin39.8degrees) = 2.51 (this is wrong)

tiny-tim
Homework Helper

Mass of hollow sphere = 6.24kg but how does this factor in if:

μgcosθ = (2/3)a

solve for a: a = (3/5)gsinθ = 3.77 m/s^2

so Force of friction should = (2/3)(3.77) = 2.51 N....still dont understand why this is wrong...

tiny-tim
Homework Helper
that's an acceleration!

multiply by the mass ok...

so Force of static friction = (m)(2/5)gsinθ = (6.24kg)(2/5)(9.81m/s^2)(sin39.8) = 15.7N...??

Right...?

tiny-tim
should be 