| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.66 |
| Score | 0% | 73% |
| 45 ft. | |
| 360 ft. | |
| 2 ft. | |
| 1 ft. |
Win = Wout
Feffort x deffort = Fresistance x dresistance
In this problem, the effort work is 180 ft⋅lb and the resistance force is 90 lbs. and we need to calculate the resistance distance:
Win = Fresistance x dresistance
180 ft⋅lb = 90 lbs. x dresistance
dresistance = \( \frac{180ft⋅lb}{90 lbs.} \) = 2 ft.
The mechanical advantage of connected gears is proportional to which characteristic of the gears?
speed |
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circumference |
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number of teeth |
|
diameter |
The mechanical advantage (amount of change in speed or torque) of connected gears is proportional to the number of teeth each gear has. Called gear ratio, it's the ratio of the number of teeth on the larger gear to the number of teeth on the smaller gear. For example, a gear with 12 teeth connected to a gear with 9 teeth would have a gear ratio of 4:3.
A fixed pulley has a mechanical advantage of:
-1 |
|
2 |
|
0 |
|
1 |
A fixed pulley is used to change the direction of a force and does not multiply the force applied. As such, it has a mechanical advantage of one. The benefit of a fixed pulley is that it can allow the force to be applied at a more convenient angle, for example, pulling downward or horizontally to lift an object instead of upward.
The force required to initally get an object moving is __________ the force required to keep it moving.
higher than |
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lower than |
|
the same as |
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opposite |
For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface).
| 5 | |
| 10 | |
| 11 | |
| 4.5 |
The gear ratio (Vr) of a gear train is the product of the gear ratios between the pairs of meshed gears. Let N represent the number of teeth for each gear:
Vr = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) \( \frac{N_3}{N_4} \) ... \( \frac{N_n}{N_{n+1}} \)
In this problem, we have only two gears so the equation becomes:Vr = \( \frac{N_1}{N_2} \) = \( \frac{20}{4} \) = 5