| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.23 |
| Score | 0% | 65% |
A fixed pulley has a mechanical advantage of:
2 |
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-1 |
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1 |
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0 |
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 steering wheel of a car is an example of which type of simple machine?
fixed pulley |
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wheel and axle |
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block and tackle |
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first-class lever |
A wheel and axle uses two different diameter wheels mounted to a connecting axle. Force is applied to the larger wheel and large movements of this wheel result in small movements in the smaller wheel. Because a larger movement distance is being translated to a smaller distance, force is increased with a mechanical advantage equal to the ratio of the diameters of the wheels. An example of a wheel and axle is the steering wheel of a car.
| 8 | |
| 1.3 | |
| 4 | |
| 12 |
The mechanical advantage of a gear train is its gear ratio. The gear ratio (Vr) 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 three gears so the equation becomes:
Vr = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) = \( \frac{24}{16} \) \( \frac{16}{6} \) = \( \frac{24}{6} \) = 4
| 1.33 | |
| 3 | |
| 0.75 | |
| 1 |
The mechanical advantage of a wheel and axle is the input radius divided by the output radius:
MA = \( \frac{r_i}{r_o} \)
In this case, the input radius (where the effort force is being applied) is 4 and the output radius (where the resistance is being applied) is 3 for a mechanical advantage of \( \frac{4}{3} \) = 1.33
| 52.5 lbs. | |
| 105 lbs. | |
| 315 lbs. | |
| 26.25 lbs. |
To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:
Rada = Rbdb
where a represents the left side of the fulcrum and b the right, R is resistance (weight) and d is the distance from the fulcrum.Solving for Rb, our missing value, and plugging in our variables yields:
Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{70 lbs. \times 9 ft.}{6 ft.} \) = \( \frac{630 ft⋅lb}{6 ft.} \) = 105 lbs.