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Prove torque equation 2sin^2(theta/2) * cos(theta + beta) = sin (beta)

  • Thread starter UCD2
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  • #1
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Homework Statement



A uniform smooth plank weight W and lenght 2a is hinged to the bottom horizontal edge of a smooth, fixed plane inclined at angle (beta) to the horizontal. A sphere of radius (1/2)a and weight 2W is placed between the plank and the plane. Assume no friction. Prove that, in the position of equilibrium, the angle (theta) between the plank and the plane will be given by the equation

2sin^2(theta/2) * cos(theta + beta) = sin (beta)


Homework Equations



sum of torque =0
sum of force along(x and y) = 0

The Attempt at a Solution


I try to draw a FBD, but i dont know where to start to get to that equation :x.

(the pic is drawn not to scale)
 

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Answers and Replies

  • #2
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this is what i try so far T.T *writing equation with blood*

sum of force along the x-axis of the ball = N(plank)(sin[theta]) - 2w(sin[beta]) = 0

N(plank) = 2w(sin[beta]) / (sin[theta])

take torque of the plank ( draw another seperate diagram of the plank{with blood :P} ) axis on the tip of the plank(the bottom tip) *clockwise*

* a = half the lenght of the plank (assuming the plank is uniform)
*(a/2)(sin[theta/2]) = draw a triangle from the center of the ball to the tip of the plank, we then obtain sin(theta/2) = x/(a/2)
*sin[theta] come from the resultant of N(plank)y

= W(cos[theta + beta])(a) - N(plank)(a/2)(sin[theta])(sin[theta/2]) = 0

subtitude N(plank) and switch the n(plank) thingy to the other side

W(cos[theta + beta])(a) = 2w(sin[beta]) / (sin[theta]) * (a/2)(sin[theta/2])

simplify

cos(theta + beta) = sin(beta)*sin(theta/2)

...*poof! a miracle happen*

2sin^2(theta/2)*cos{theta + beta) = sin(beta)

T.T somebody help me...i think i mess up on the angle D:
 
  • #3
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:X no one helping me >.<

i really having a hard time figuring out the angle. T.T
 

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