# Thread: [SOLVED] Find coefficient of friction?

1. ## [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?

2. Hello fardeen_gen
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:

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

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

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

Then, the Sine Rule again to find AB:

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

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

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 $N, \mu N$
• the tension in the string is $T$

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

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

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

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

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

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

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

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

Then, the Sine Rule again to find AB:

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

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

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 $N, \mu N$
• the tension in the string is $T$
Resolve vertically: $mg = \mu N + T\cos\alpha$

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

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

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

Shouldn't it be AB = d.sin C/sin α?