Mastering Physics: Kinetic and Static Friction

[rwcollings]

Homework Statement

A box of textbooks of mass 24.5 rests on a loading ramp that makes an angle with the horizontal. The coefficient of kinetic friction is 0.260 and the coefficient of static friction is 0.360.

As the angle is increased, find the minimum angle at which the box starts to slip.

Homework Equations

N+f_s+W=0

N=mg cos$$\vartheta$$
f_s=mg sin$$\vartheta$$

The Attempt at a Solution

In my notes, it gives an example of an object at rest on an inclined plane and sets up the two above equations with the angle and weight of the object both given. The problem is that when I add up the three forces N, f_s, and W I do not get "0". This is leaving me quite confused.

I have tried substituting the equations above into the top equation equal to zero, but I cannot figure out how to solve for theta since Weight is involved.

Can anyone give me a push in the right direction. I feel like I am forgetting something.

 

[LowlyPion]

Welcome to PF.

You must remember that the frictional force is really only a maximum number that needs to be overcome in order for there to be motion along the incline.

Hence what you need to consider is the point at which your

m*g*sinθ = μ*m*g*cosθ

 

[rwcollings]

Thank you for the quick response (sorry for my delay). So what does the symbol "μ" represent in the equation? Will I be able to set the equation equal to zero and factor out the common angle?

Thanks,

Rob

 

[Kruum]

Thank you for the quick response (sorry for my delay). So what does the symbol "μ" represent in the equation? Will I be able to set the equation equal to zero and factor out the common angle?

$$\mu}$$ is the coefficient of friction. And the rest you got right.

 

[rwcollings]

Hence what you need to consider is the point at which your

m*g*sinθ = μ*m*g*cosθ

In this equation both mg are the same correct? So that would make μcosθ=sinθ

I realize looking at this that I can't just factor out that angle. Is there a way to determine at what angle the book is no longer stationary short of plugging in random angles?

 

[LowlyPion]

In this equation both mg are the same correct? So that would make μcosθ=sinθ

I realize looking at this that I can't just factor out that angle. Is there a way to determine at what angle the book is no longer stationary short of plugging in random angles?

Have you run across the trig identity that Tanθ = Sinθ/Cosθ before?

 

[rwcollings]

Yes, thanks for the reminder. That's one of those things that you end up smacking yourself in the head.

 

[LowlyPion]

So ... for θ just take the arctan

Tan^-1(X) = θ

 

[rwcollings]

Yep, got the answer and it was correct.

Thank you for the help