# An object with a mass of #7 kg# is on a plane with an incline of # - pi/6 #. If it takes #15 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?

If angle of inclination is

(see explanation regarding angle)

We're asked to find the coefficient of static friction

We'll call the positive *up* the incline (the direction of

There is no net vertical force, so we'll look at the horizontal forces (we WILL use the normal force magnitude

We're given that the object's mass is

Since the angle is

#-(pi)/6# , this would be the angle goingdownthe incline (the topmost angle in the image above). Therefore, the actual angle ofinclinationis

#pi/2 - (pi)/6 = ul((pi)/3#

The formula for the coefficient of static friction

Since the object in this problem "breaks loose" and the static friction eventually gives way, this equation is simply

Since the two vertical quantities

Since

The coefficient of static friction is thus

The coefficient of kinetic friction

It takes

The coefficient of kinetic friction is thus

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The coefficient of static friction is ( \mu_s = 0.36 ) and the coefficient of kinetic friction is ( \mu_k = 0.2 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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