# An object with a mass of 12 kg is on a plane with an incline of  - pi/6 . If it takes 9 N to start pushing the object down the plane and 8 N to keep pushing it, what are the coefficients of static and kinetic friction?

Feb 9, 2016

Use a force vector diagram and trigonometry to find the forces parallel and perpendicular to the plane, generated by the object. Then by using:

$\textrm{C o e f f i c i e n t o f \mathfrak{i} c t i o n} = \frac{\textrm{p u s h \in g f \mathmr{and} c e} - \textrm{f \mathmr{and} c e p a r a l \le l \to p l a \ne}}{\textrm{f \mathmr{and} c e p e r p e n \mathrm{di} c \underline{a} r \to p l a \ne}}$
you should find :
${\mu}_{s t a t i c} = 0.66$ and ${\mu}_{\textrm{k \in e t i c}} = 0.65$

#### Explanation:

Use a force vector diagram and trigonometry to find the forces parallel and perpendicular to the plane, generated by the object.
The magnitude of the object's force is given by its mass multiplied by the acceleration due to gravity:
F = m a

If you're unsure what the force diagram looks like, there's a good description here:
http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html

Then use:

$\textrm{C o e f f i c i e n t o f \mathfrak{i} c t i o n} = \frac{\textrm{p u s h \in g f \mathmr{and} c e} - \textrm{f \mathmr{and} c e p a r a l \le l \to p l a \ne}}{\textrm{f \mathmr{and} c e p e r p e n \mathrm{di} c \underline{a} r \to p l a \ne}}$

The minus sign can be confusing. If the plane was an incline the object's force down the plane would be acting against our pushing force. Here we have a decline so our pushing force acts with the objects force down the plane. It works out because theta is negative, which will make sin(theta) negative, and as two negatives make a positive our pushing force will act with the objects force down the plane. Work this through mathematically and you will see how it all comes out.

The force perpendicular to the plane, which equals the normal force is:
${F}_{\textrm{p e r p e n \mathrm{di} c \underline{a} r}} = \left(m g\right) \cos \left(\theta\right)$

The force parallel to the plane, which is acting against friction is:
${F}_{\textrm{p a r a l \le l}} = \left(m g\right) \sin \left(\theta\right)$

Frictional force = $\mu \cdot \textrm{N \mathmr{and} m a l f \mathmr{and} c e} = \mu \cdot {F}_{\textrm{p e r p e n \mathrm{di} c \underline{a} r}}$
where $\mu$ is the coefficient if friction.

At the point where friction is overcome:
Frictional force = pushing force - ${F}_{\textrm{p a r a l \le l}}$

Now by equating the frictional forces:
$\mu \cdot {F}_{\textrm{p e r p e n \mathrm{di} c \underline{a} r}}$ = pushing force - ${F}_{\textrm{p a r a l \le l}}$

Dividing both sides by ${F}_{\textrm{p e r p e n \mathrm{di} c \underline{a} r}}$.

$\mu = \frac{\textrm{p u s h \in g f \mathmr{and} c e} - {F}_{\textrm{p a r a l \le l}}}{F} _ \left(\textrm{p e r p e n \mathrm{di} c \underline{a} r}\right)$

Using the trigonometric equations from earlier:

$\mu = \frac{\textrm{p u s h \in g f \mathmr{and} c e} - \left(m g\right) \sin \left(\theta\right)}{\left(m g\right) \cos \left(\theta\right)}$

Finally enter the values given in the question and you should find :
${\mu}_{s t a t i c} = 0.66$ and ${\mu}_{\textrm{k \in e t i c}} = 0.65$

Note that I used a value of $g = 10 m {s}^{-} 2$.

Best regards,

Rory.