newtonian mechanics

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June 3rd 2010, 06:09 AM
mathaddict

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newtonian mechanics

THe diagram shows particles A and B connected by a light inextensible string that passes over a small smooth pulley C. Pulley C is fixed at the top of a wedge which is fixed on a horizontal plane. Particle A is on the face of the wedge which makes an angle of alpha with the horizontal with the portion of the string AC parallel to the line of the greatest slope of the face. Particle B is on the face of the wedge which makes an angle of beta with the horizontal with the portion of the string BC prarallel to the line of the greatest slope of the face. Weight of each particle A and B is W and the coefficient of friction of each particle with the faces of the wedge is $\mu$ . If $\alpha>\beta$ and particles A and B are about to slide, show that

$ \mu=\tan \frac{1}{2}(\alpha-\beta) $
June 3rd 2010, 07:28 AM
skeeter

Quote:

Originally Posted by mathaddict

THe diagram shows particles A and B connected by a light inextensible string that passes over a small smooth pulley C. Pulley C is fixed at the top of a wedge which is fixed on a horizontal plane. Particle A is on the face of the wedge which makes an angle of alpha with the horizontal with the portion of the string AC parallel to the line of the greatest slope of the face. Particle B is on the face of the wedge which makes an angle of beta with the horizontal with the portion of the string BC prarallel to the line of the greatest slope of the face. Weight of each particle A and B is W and the coefficient of friction of each particle with the faces of the wedge is $\mu$ . If $\alpha>\beta$ and particles A and B are about to slide, show that

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

since A and B are about to slide, the static friction force is at a maximum, and the system is in equilibrium.

forces on A ...

$ W\sin{\alpha} - T - \mu W\cos{\alpha} = 0 $

forces on B ...

$ T - \mu W\cos{\beta} - W\sin{\beta} = 0 $

combine the equations ...

$ W(\sin{\alpha} - \sin{\beta}) - \mu W(\cos{\alpha} + \cos{\beta}) = 0 $

solving for $\mu$ ...

$ \mu = \frac{\sin{\alpha} - \sin{\beta}}{\cos{\alpha} + \cos{\beta}} $

I'll leave the identity work for you.
June 4th 2010, 03:28 AM
mathaddict

Quote:

Originally Posted by skeeter

since A and B are about to slide, the static friction force is at a maximum, and the system is in equilibrium.

forces on A ...

$ W\sin{\alpha} - T - \mu W\cos{\alpha} = 0 $

forces on B ...

$ T - \mu W\cos{\beta} - W\sin{\beta} = 0 $

combine the equations ...

$ W(\sin{\alpha} - \sin{\beta}) - \mu W(\cos{\alpha} + \cos{\beta}) = 0 $

solving for $\mu$ ...

$ \mu = \frac{\sin{\alpha} - \sin{\beta}}{\cos{\alpha} + \cos{\beta}} $

I'll leave the identity work for you.

thanks Skeeter , this is the continuation of the problem.

If alpha is 60 degrees and beta is 30 degrees , show that the tension of the string is $(\sqrt{3}-1)W$

so i started by evaluating $\mu=2-\sqrt{3}$

Then calculate the reaction forces on each particles and using $F_A+F_B=0$

i tried to solve the simultaneous equation but ended up cancelling everything . Could you check if i am on the right track ?
June 4th 2010, 04:14 AM
skeeter

calculate $\mu$ ...

$\mu = \tan\left(\frac{\alpha-\beta}{2}\right)$

$\mu = \tan(15) = 2-\sqrt{3}$

use the equation for forces on A ...

$ W\sin{\alpha} - T - \mu W\cos{\alpha} = 0 $

$ T = W\sin(60) - (2-\sqrt{3})W\cos(60) $

$T = W(\sqrt{3}-1)$

or B ...

$ T - \mu W\cos{\beta} - W\sin{\beta} = 0 $

$T = W[\mu\cos(30) + \sin(30)]$

$ T = W(\sqrt{3}-1) $