| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.30 |
| Score | 0% | 66% |
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| 2 |
The mechanical advantage (MA) of an inclined plane is the effort distance divided by the resistance distance. In this case, the effort distance is the length of the ramp and the resistance distance is the height of the green box:
MA = \( \frac{d_e}{d_r} \) = \( \frac{14 ft.}{7 ft.} \) = 2
Which of the following represents how much two materials resist sliding across each other?
normal friction |
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coefficient of friction |
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static friction |
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kinetic friction |
Coefficient of friction (μ) represents how much two materials resist sliding across each other. Smooth surfaces like ice have low coefficients of friction while rough surfaces like concrete have high μ.
Collinear forces:
pass through a common point |
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act in a common plane |
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are unrelated to each other |
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act along the same line of action |
Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.
A a seesaw / teeter-totter is an example of which of the following?
second-class lever |
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inclined plane |
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first-class lever |
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third-class lever |
A first-class lever is used to increase force or distance while changing the direction of the force. The lever pivots on a fulcrum and, when a force is applied to the lever at one side of the fulcrum, the other end moves in the opposite direction. The position of the fulcrum also defines the mechanical advantage of the lever. If the fulcrum is closer to the force being applied, the load can be moved a greater distance at the expense of requiring a greater input force. If the fulcrum is closer to the load, less force is required but the force must be applied over a longer distance. An example of a first-class lever is a seesaw / teeter-totter.
| 15 lbs. | |
| 20 lbs. | |
| 47.5 lbs. | |
| 12 lbs. |
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 10 and the output radius (where the resistance is being applied) is 5 for a mechanical advantage of \( \frac{10}{5} \) = 2.0
MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{95 lbs.}{2.0} \) = 47.5 lbs.