# Finding largest angle - mechanics

1. Feb 26, 2012

### blackandyello

1. The problem statement, all variables and given/known data
The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.4, determine the largest angle θ for which the assembly will remain at rest.

http://screensnapr.com/e/vU9M3X.jpg [Broken]

2. Relevant equations

summation of forces among horizontal
summation of forces among vertical
summation of forces - point
3. The attempt at a solution

so i use the fbd of the whole body, and take summation of forces on x axis, and i got Na = Nc (normal at point a is equal to point c). then, i take summation of forces on y axis, so thats

Fa - 6N -8N +Fb = 0 (where fa and fb are friction forces on a and b), and since fa = fb (their normals are equal), i use f = uN and yielded Na = 17.5 N.

I then take summation of moments at point B( bar of A to B) and heres eqn

(6)(300)(cos(x)) - 17.5 (600)(cos(x)) + 7(600)(sin(x)) = 0 and

the value of angle is 64 degress. the book says 10 degress. Where did I go wrong?

Last edited by a moderator: May 5, 2017
2. Feb 26, 2012

### ehild

The bars are not fixed together, but connected with a pin, they can turn around. Write out the conditions for equilibrium for both.
The force of friction is static, so you can not say that fa=fb. It is only true that neither of them can exceed 0.4 times the normal force.

ehild

3. Feb 28, 2012

### ehild

I would not like to waste this problem...

The equation is correct for the resultant of all vertical components, but the forces of static friction are not the same: Fa is not equal to Fb. The force of static friction can not exceed the maximum value, so Fa, Fb ≤μN and the greater one is equal to it.

The bars can turn around B with respect to each other, so the torques must be balanced separately for both. If L is the length of the rods, the sum of torques on the left rod is

NLsinθ+6L/2 cosθ-Fa Lcosθ=0 => Ntan(θ)=Fa-3, Fa=Ntan(θ)+3

The equation for the right rod is

NLsinθ+8L/2 cosθ-Fb Lcosθ=0 => Ntan(θ)=Fb-4, Fb=Ntan(θ)+4

That means Fa<Fb≤0.4N. Taking Fb equal to 0.4N, Ntan(θ)+4=0.4N,

N(0.4-tan(θ))=4

The equation for vertical force components yields Fa+Fb=14. Plugging in Fa and Fb:

Ntan(θ)+3+Ntan(θ)+4=14 =>2N tan(θ)=7

Solve the equations in bold for θ.

ehild

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