How Do You Calculate the Coefficient of Kinetic Friction on a Ramp?

[ughschool]

A box slides down a 30° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box and the ramp


ok so the only equations I know are μk= Fk(kinetic friction)/Fn(normal) and that Fn=mass x gravity
so i don't know how to get Fk or mass... so help on solving this please?


im not really sure how to start other than maybe using the angle to find the x and y components of the ramp but again I don't know what I would do with that information

 

[rock.freak667]

Well using the angle, what are the forces acting parallel to the ramp?

Then if you add up those forces they will give you a resultant force, what does Newton's 2nd Law say about these forces?

 

[ughschool]

forces acting parallel to the ramp: gravity and friction?

 

[rock.freak667]

forces acting parallel to the ramp: gravity and friction?

No, the force due to gravity acts vertically downwards, you can split this force into 2 components, one perpendicular to the ramp and another parallel to the ramp. Can you find these two forces? (given that the force due to gravity = weight = mg)

Also yes the force due to friction is parallel to the ramp.

 

[ughschool]

Newton's 2nd law: Fnet=acceleration x mass

 

[ughschool]

how would I find the force due to gravity if I don't have mass?

 

[rock.freak667]

Newton's 2nd law: Fnet=acceleration x mass

That's correct, so that is one part down, you just need to figure out the forces now in terms of mass and acceleration due to gravity.

how would I find the force due to gravity if I don't have mass?

For now, don't put in any numbers, just leave mass as 'm' and acceleration due to gravity as 'g' and angle as 'θ'.

 

[ughschool]

ok so
Fnet= a x m
g in the x direction= m X 9.81 cosθ
g in the y direction= m x 9.81 sinθ
is that right??

 

[ughschool]

wait do i switch those? because my teacher has us rotate our axis so that the ramp is on the X axis

 

[rock.freak667]

ok so
Fnet= a x m
g in the x direction= m X 9.81 cosθ
g in the y direction= m x 9.81 sinθ
is that right??

If you mean x-direction as parallel to the ramp and y as perpendicular, then yes.

How would you find the force due to friction using your equation with μ_k?

(Hint: the normal force = force perpendicular to the ramp)

 

[ughschool]

ok so
Fn= m x 9.81 sinθ
so uk=Fk/m x 9.81 sinθ

im still not sure how that gives me Fk

 

[rock.freak667]

ok so
Fn= m x 9.81 sinθ
so uk=Fk/m x 9.81 sinθ

im still not sure how that gives me Fk

Write F_k as the subject.


Now consider your forces in the x-direction i.e. parallel to the ramp. In what direction are the forces acting? (Which one will point down the ramp and which one up the ramp?)

After that, the question says that the box slides down the ramp, so is F_net pointing down the ramp or up the ramp?

 

[ughschool]

Fk=uk x (m x 9.81 sinθ)

left?

Fnet is pointing down the ramp?

 

[rock.freak667]

Fk=uk x (m x 9.81 sinθ)

Good

left?

You have two forces parallel to the ramp, component of the weight and the friction, which one acts down the ramp and which up? When you figure this one, what would be the resultant of these two forces? (Use symbols as in A+B or A-B and so on. Not numbers)

Fnet is pointing down the ramp?

This is correct.

 

[ughschool]

weight is down the ramp and friction up right?

and the resultant is sqrt(a^2+b^2)?

 

[rock.freak667]

weight is down the ramp and friction up right?

Yes, so if you had two forces pulling in opposite directions and you know the direction in which the resultant is in, what would you do to the two forces to get that resultant?

and the resultant is sqrt(a^2+b^2)?

No.

 

[ughschool]

add them together?

 

[rock.freak667]

add them together?

Right but if one is opposing the other would it be positive or negative?


Using the forces you found above, add these up and what do you get?

 

[ughschool]

a negative resultant

 

[rock.freak667]

a negative resultant

In terms of the forces, one will be positive and the other negative. Using the forces you found about, put one as negative and then add them up.

What do you get?

Also remember that this would be the resultant force which is the same as mass x acceleration or 'ma'.

 

[ughschool]

which forces are you talking about?

 

[rock.freak667]

which forces are you talking about?

The forces parallel to the ramp, the x-direction.

 

[ughschool]

so weight and friction
so -weight + friction= -resultant

 

[rock.freak667]

so weight and friction
so -weight + friction= -resultant

Right!

So putting in all of the expressions you found before, what would you have and what cancels out?

 

[ughschool]

Fk=Uk x (m x 9.81 sinθ)
and -weight + friction= - resultant

im not sure what combines those?

 

[rock.freak667]

Fk=Uk x (m x 9.81 sinθ)
and -weight + friction= - resultant

im not sure what combines those?

You told me that the frictional force is F_k and the component of the weight is mgsinθ and that the resultant is 'ma'. So put those into the equation.

 

[ughschool]

-(m x 9.81 sinθ) + ((m x 9.81 sinθ) x Uk)=-(m X a)
sooooo Uk=-ma?

 

[rock.freak667]

-(m x 9.81 sinθ) + ((m x 9.81 sinθ) x Uk)=-(m X a)
sooooo Uk=-ma?

No, the term in magenta is not the x-component you previously gave. Put the other one in.

 

[ughschool]

-(m x 9.81 cosθ)?

 

[rock.freak667]

-(m x 9.81 cosθ)?

Yes.

 

[ughschool]

-cosθ x m x a / sinθ = Uk?

 

[ughschool]

wait no negative

 

[ughschool]

maybe?

 

[rock.freak667]

-cosθ x m x a / sinθ = Uk?

No.


This is your equation

-mgsinθ + mμ_kgcosθ = -ma

Solve for 'a'.

 

[ughschool]

gsinθ + μkgcosθ = a?