Describe a gear train that would transform a counterclockwise input rotation to a counterclockwise output rotation where the dri
1 answer:
Answer:
For a gear train that would train that transform a counterclockwise input into a counterclockwise output such that the gear that is driven rotates three times when the driver rotates once, we have;
1) The number of gears in the gear train = 3 gears with an arrangement such that there is a gear in between the input and the output gear that rotates clockwise for the output gear to rotate counter clockwise
2) The speed ratio of the driven gear to the driver gear = 3
Therefore, we have;
Therefore, for a speed ratio of 3, the number of teeth of the driver gear, driving the output gear, must be 3 times, the number of teeth of the driven gear
Explanation:
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Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
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