Finding the Angle of a 2kg Block on a Slippery Plane

[contlee]

The coefficient of friction between a surface and a block in .41. Determine the angle needed for a 2kg block to slide down the plane with a constant speed.

Well after drawing a free body diagram I really couldn't see any way to go about solving. So i decided to name what i knew. I know the massc in 2kg or 20N. I know the coefficient of friction is .41. I remember a concept saying tan(angle)= coefficient of friction. But that seems to direct. If you did use the tangent method the angle would be 23. But then why would the information about the weight be included?

 

[tiny-tim]

Welcome to PF!

Hi contlee! Welcome to PF!

… I remember a concept saying tan(angle)= coefficient of friction. But that seems to direct. If you did use the tangent method the angle would be 23.

You'll never remember all those formulas for the exam, so you need to be able to derive them as you go.

You drew a free-body diagram, so it had weight, normal force, and reaction force.

Now either use components, or draw a vector triangle … and find the normal force first (that's always easiest, because the normal acceleration is always zero! ).

But then why would the information about the weight be included?

Sometimes they give you unncecessary information just to test you.

 

[inutard]

Hi contlee. Technically speaking, your formula is correct. However, if you were to do this problem from scratch, the problem could also be solved by breaking the forces down to their components as tiny-tim suggests. If you solve for the angle in general terms, you'd see that the weight term in the equation would always cancel out, thus giving rise to
co-eff of friction = tan (x)

 

[contlee]

but would the tan method work

 

[inutard]

yes.