- #1

- 321

- 20

## Homework Statement

A wooden block with mass of .0507 kg is placed at the midpoint of a 1m long wooden board, where the coefficient of kinetic friction (mk) = .275, at what angle should the board be placed so that the block takes time t = .387 s to slide to the lower end of the board? You may find a spread sheat program helpful in answer this question

## Homework Equations

f=ma

## The Attempt at a Solution

So I know:

[itex]Fg = .0507 kg * 9.81[/itex]

[itex]Fn = Fgcos(\Theta)[/itex]

mk = .275

That means Fk (force of kinetic friction) [itex] Fk = mkFgcos(\Theta) [/itex]

here I use f = ma

[itex]Fgsin(\Theta) - Fgcos(\Theta)mk = m * a [/itex]

[itex]Fg(sin(\Theta) - mkcos(\Theta)) = m * a [/itex]

since Fg = mass* 9.81 the two masses cancel leaving

[itex] 9.81(sin(\Theta) - mkcos(\Theta)) = a [/itex]

Now I use:

[itex] \Delta x = V_{0x}t + (1/2) a_x t^2 [/itex]

[itex] 1/ (.378^2 * 9.81) = (sin(\Theta) - mkcos(\Theta)) [/itex]

1/(.378^2 * 9.81)= .6806268265 which I will just call "c."

So now I end up with

[itex] Sin(\Theta) = c + mkcos(\Theta) [/itex]

[itex] Cos(\Theta) = ( c-sin(\Theta) )/( -mk ) [/itex]

[itex] \Theta = arcsin(c+ mkcos(\Theta)) [/itex]

[itex] \Theta = arccos( (c-sin(\Theta) )/( -mk )) [/itex]

Now this is as far as I've reached.

[itex] \Theta = arcsin(c+ mkcos(\Theta)) = arccos( (c-sin(\Theta) )/ (-mk)) [/itex]

but I see no possible way of solving this for Theta...

My biggest hint was:

https://www.thestudentroom.co.uk/showthread.php?t=1873196

Which said "x^2 + y^2 = 1"

But I don't understand and don't think I can use this because x and y are sides, correct? While I'm dealing with angles...

Last edited: