1. ## mechanics question

can some explain to me how they get tana=r/2r-x as i do not see how triangle with angle a is right angled which it has to be to use tan. just because the three forces are going through o would not make it right angled

2. ## Re: mechanics question

I'm going to do this problem in static equilibrium a little differently than the way the answer does. See attached diagram ...

$\displaystyle \sum F_x = 0$ ... force right = force left

$N = T\sin{\alpha}$

$\displaystyle \sum F_y = 0$ ... force up = forces down

$T\cos{\alpha} + f = W$

since the sphere is about to slip, the force of static friction is a maximum $\implies f = \mu \cdot N = \mu \cdot T\sin{\alpha}$

$T\cos{\alpha} + \mu \cdot T\sin{\alpha} = W$

$\displaystyle \sum \tau = 0$ ... torque CCW = torque CW

about the point where the sphere touches the wall ...

$T \cdot 2r\sin{\alpha} = W \cdot r \implies W = 2T \sin{\alpha}$

returning to the equilibrium equation in the y-direction & substituting for $W$ ...

$T\cos{\alpha} + \mu \cdot T\sin{\alpha} = 2T \sin{\alpha}$

divide every term by $T\cos{\alpha}$ ...

$1 + \mu \cdot \tan{\alpha} = 2\tan{\alpha}$

$1 = 2\tan{\alpha} - \mu \cdot \tan{\alpha}$

$1 = \tan{\alpha}(2 - \mu)$

$\dfrac{1}{2-\mu} = \tan{\alpha}$

3. ## Re: mechanics question

Thanks . The diagram is really helpful. It will take me a day or a few to take this all in.