Rough Inclined Plane - Find Coefficient of friction

[FaraDazed]

Homework Statement

a block of mass 2kg is in at rest on a rough plane inclined at 30°, it is in limiting equilibrium with with a force of 20N applied horizontally (to the ground).

Find the coefficient of static friction between the block and the plane.

The Attempt at a Solution

[itex]
R=mgcos30+20sin30\\
.\\
mgsin30-μR=20cos30\\
mgsin30-μmgcos30+μ20sin30=20cos30\\
9.8-16.97μ+10μ=17.32\\
9.8-26.97μ=17.32\\
-26.97μ=7.52\\
μ=-0.28\\
[/itex]
Now I now my answer cannot be minus something as 0<μ<1, however, I am unsure as to whether it is my understanding of the question or the algebra I have got wrong. I am sure its a silly mistake, it usually is with me but any help or advice is greatly appreciated.

 

[Doc Al]

[itex]
R=mgcos30+20sin30\\

mgsin30-μR=20cos30\\
mgsin30-μmgcos30+μ20sin30=20cos30\\[/itex]

When you substituted the two terms of R, only one of them got the minus sign.

 

[Doc Al]

It looks like that was just a typo on your part and that your calculation did carry the minus sign throughout.

But you made an assumption about the direction of the friction force that may not be justified. Try interpreting the problem like so: If that 20 N force were any greater, the block would start to slide up the incline.

 

[FaraDazed]

Thanks for your help I just realized my silly mistake. Yeah I carried the minus sign through but it wasnt a typo it was a stupid mistake lol, I was saying to myself (as I was doing the problem) that if its going to be pushed up then the frictional force is in the opposite direction but still put -uR instead of + and even after looking at it a hundred times couldn't think what was wrong, as I was getting wrong answers (positive ones) to begin with so when I finally get the correct figure (only by luck) but minus it, I couldn't see where I went wrong.

 

[Nickie]

Your understanding of the question and the algebra are both correct. However, the negative value for the coefficient of friction indicates that the block is actually sliding down the inclined plane rather than being at rest. This is not possible in a situation of limiting equilibrium.

To find the correct coefficient of friction, you can rearrange the equation to solve for μ:

μ = (mgsin30 - 20cos30) / (mgcos30 + 20sin30)

Substituting in the given values, we get μ = (2*9.8*0.5 - 20*0.87) / (2*9.8*0.87 + 20*0.5) = 0.14

This is a positive value within the range of 0<μ<1, indicating that the block is at rest and in limiting equilibrium on the inclined plane with a coefficient of friction of 0.14.