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How to Calculate Beta Angle for Pulley B in a Frictional Belt and Pulley System?
AI Thread Summary
To calculate the beta angle for pulley B in a frictional belt and pulley system, it is derived from the contact angle of the belt with the pulleys. The calculation involves subtracting the given angle of 60 degrees from 180 degrees, resulting in beta being 120 degrees. This approach assumes that the arcs from both pulleys' circles, represented as blue and yellow arcs, will sum to 360 degrees. The reference point for the angle is crucial, as it shifts from the top of the first pulley to the bottom of the second pulley, maintaining parallelism between the pulleys. Understanding these geometric relationships is essential for accurate calculations in pulley systems.
- #2
masterflex
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I was thinking about it and drawing circles and lines. Can you tell me if this is a general rule (please look at attachment where I draw circles): when lines are drawn (simulating a rubber band around 2 pulleys), adding blue arcs from both circles will always equal 360 degrees? The blue arc from one circle, and the yellow arc from the other circle are equal in terms of the contact angle, yes?
Attachments
- #3
Double A
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They get beta = 120 from the given angle of 60 degrees.
180 - 60 = 120 degrees
The 60 degrees is referancing from the top of the pulley. To change the referance point to the bottom of the second pulley they performed the above calculation.
Also, since their referance point is perpendicular to the belt on both pullies they are parallel to each other so the angle referances can then be translated from the larger pulley to the smaller pulley.
180 - 60 = 120 degrees
The 60 degrees is referancing from the top of the pulley. To change the referance point to the bottom of the second pulley they performed the above calculation.
Also, since their referance point is perpendicular to the belt on both pullies they are parallel to each other so the angle referances can then be translated from the larger pulley to the smaller pulley.
Last edited:
- #4
masterflex
- 16
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cool, thx.
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are;
Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction.
Must be LESS than 1 kg.
NO parachutes.
NO glue or Tape can touch the egg.
MUST be able to take egg out in less than 1 minute.
Grade A large eggs will be used.
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