# [SOLVED] Find coefficient of friction?

• Apr 13th 2009, 08:55 PM
fardeen_gen
[SOLVED] Find coefficient of friction?
A picture is attached to a vertical wall by means of string AC with length l forming an angle α with the wall. The height of the picture BC = d. The bottom of the picture is not fastened. At what coefficient of friction between the picture and the wall will the picture be in equilibrium?

For diagram, Mechanics on Flickr - Photo Sharing!

I keep getting the value of μ in terms of weight of the body but that is not a known variable. How to find μ in terms of the lengths given?
• Apr 13th 2009, 10:29 PM
Hello fardeen_gen
Quote:

Originally Posted by fardeen_gen
A picture is attached to a vertical wall by means of string AC with length l forming an angle α with the wall. The height of the picture BC = d. The bottom of the picture is not fastened. At what coefficient of friction between the picture and the wall will the picture be in equilibrium?

For diagram, Mechanics on Flickr - Photo Sharing!

I keep getting the value of μ in terms of weight of the body but that is not a known variable. How to find μ in terms of the lengths given?

First, use the Sine Rule to find angle B:

$\displaystyle \frac{\sin B}{l}=\frac{\sin \alpha}{d}$

$\displaystyle \Rightarrow \sin B = \frac{l\sin\alpha}{d}$

and $\displaystyle \cos B = \sqrt{1-\frac{l^2\sin^2\alpha}{d^2}}$

Then, the Sine Rule again to find AB:

$\displaystyle AB=\frac{d\sin\alpha}{\sin C} = \frac{d\sin\alpha}{\sin(B+\alpha)}$, noting that $\displaystyle \sin C = \sin(B + \alpha)$

And if the perpendicular distance from B to AC produced = $\displaystyle h$, then $\displaystyle h = AB\sin\alpha =\frac{d\sin^2\alpha}{\sin B\cos\alpha +\cos B\sin\alpha}$

Then, we assume that

• the picture is on the point of slipping at B
• the picture has mass m, and is uniform, so the centre of mass is at the mid-point of BC
• the forces at B are $\displaystyle N, \mu N$
• the tension in the string is $\displaystyle T$

Resolve vertically: $\displaystyle mg = \mu N + T\cos\alpha$

Resolve horizontally: $\displaystyle N = T\sin\alpha$

Take moments about B: $\displaystyle \tfrac12mgd\sin B = Th$

You now have all you need to eliminate $\displaystyle mg, N$ and $\displaystyle T$ and obtain an equation for $\displaystyle \mu$.

• Apr 15th 2009, 05:47 AM
fardeen_gen
Quote:

Hello fardeen_genFirst, use the Sine Rule to find angle B:

$\displaystyle \frac{\sin B}{l}=\frac{\sin \alpha}{d}$

$\displaystyle \Rightarrow \sin B = \frac{l\sin\alpha}{d}$

and $\displaystyle \cos B = \sqrt{1-\frac{l^2\sin^2\alpha}{d^2}}$

Then, the Sine Rule again to find AB:

$\displaystyle AB=\frac{d\sin\alpha}{\sin C} = \frac{d\sin\alpha}{\sin(B+\alpha)}$, noting that $\displaystyle \sin C = \sin(B + \alpha)$

And if the perpendicular distance from B to AC produced = $\displaystyle h$, then $\displaystyle h = AB\sin\alpha =\frac{d\sin^2\alpha}{\sin B\cos\alpha +\cos B\sin\alpha}$

Then, we assume that

• the picture is on the point of slipping at B
• the picture has mass m, and is uniform, so the centre of mass is at the mid-point of BC
• the forces at B are $\displaystyle N, \mu N$
• the tension in the string is $\displaystyle T$
Resolve vertically: $\displaystyle mg = \mu N + T\cos\alpha$

Resolve horizontally: $\displaystyle N = T\sin\alpha$

Take moments about B: $\displaystyle \tfrac12mgd\sin B = Th$

You now have all you need to eliminate $\displaystyle mg, N$ and $\displaystyle T$ and obtain an equation for $\displaystyle \mu$.

Shouldn't it be AB = d.sin C/sin α?

After eliminating mg, T and N,
μ = {l cos α + 2√(d^2 - l^2 sin^2 α)}/l sin α, which is the required answer.

Sine Rule For the Win! :)
• Apr 15th 2009, 06:46 AM