- #1
Rijad Hadzic
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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...
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