# Equilibrium of forces ?

1. Jul 9, 2009

### harimakenji

1. The problem statement, all variables and given/known data
A uniform thin straight bar AE is at rest inside a hemisphere in the configuration, under the assumption the friction between the bar and the hemisphere is negligible. This configuration is possible as long as the length of the bar remains within a limited range. The center of the hemisphere is on the vertical plane containing the two points A and B. The upper plane BC of the hemisphere is kept horizontal. The direction AD and BD mean the direction of the force acting on the bar from the hemisphere at point A and that on the bar at point B respectively. DG is the direction of the force of gravity acting on the bar, where G is the center of gravity of the bar. The angle ABC = $$\theta$$ means the angle between the bar and the horizontal line, and angle ABD = $$\alpha$$ , angle BAD = $$\beta$$

a. find the value of $$\alpha$$
b. find the relation between $$\beta$$ and $$\theta$$
c. is the case $$\theta$$ = $$\pi$$/4 possible or impossible ?
d. is the case $$\theta$$ = 5$$\pi$$/24 posible or impossible ?
e. in case of $$\theta$$ = $$\pi$$/6, find the suitable ratio of the length of the bar to the diameter of the hemisphere

2. Relevant equations

3. The attempt at a solution
i'm not sure about my work...

a. because $$\alpha$$ is the angle of normal force, i think $$\alpha$$ = 90o

b. the normal reaction at A is perpendicular to the tangent at that point so OA is the radius of the sphere. OB is also the radius so $$\beta$$ = $$\theta$$

c. $$\theta$$ = $$\pi$$/4 is impossible because AOB must be right angle.

d. $$\theta$$ = 5$$\pi$$/24 is impossible because the total angle of ADB will not be 180o

e. don't know how to start....

thanks in advance

Last edited: Jul 9, 2009
2. Jul 10, 2009

### songoku

i think the answers for first four questions are right.

sorry can't help for the last one since i don't know either.

3. Jul 23, 2009

### harimakenji

thx for verifying my answer

anybody can give me a clue for the last one?

thanks in advance

4. Jul 23, 2009

### harimakenji

hello...

sorry to bump up but i really need help here

tq

5. Jul 25, 2009

### harimakenji

anyone can help with the last question?

6. Jul 26, 2009

### rl.bhat

Do you mean the ration of AE/BC ?

7. Jul 26, 2009

### harimakenji

Yes

Have any clues, rl.bhat?

thanks

EDIT : the answer is 2 / sqrt 3

8. Jul 26, 2009

### rl.bhat

DG is perpendicular to BC.
There fore angle DGA = 120 degrees.
Angle DOB = 60 degrees. Hence angle ODG = 30 degrees.
Since BC = AD. applying sine rule in the triangle ADG
AG/sin 30 = AD / sin 120
And AE = 2*AG.
Now find the ratio AE / BC

9. Jul 26, 2009

### harimakenji

I get it now

THANKS RL.BHAT !!!!

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