| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.85 |
| Score | 0% | 57% |
For any given surface, the coefficient of static friction is ___________ the coefficient of kinetic friction.
lower than |
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equal to |
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higher than |
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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).
| -1 | |
| 1 | |
| 0.88 | |
| 1.14 |
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 7 and the output radius (where the resistance is being applied) is 8 for a mechanical advantage of \( \frac{7}{8} \) = 0.88
Friction resists movement in a direction __________ to the movement.
normal |
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perpendicular |
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parallel |
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opposite |
Friction resists movement. Kinetic (also called sliding or dynamic) friction resists movement in a direction opposite to the movement. Because it opposes movement, kinetic friction will eventually bring an object to a stop. An example is a rock that's sliding across ice.
A fixed pulley is useful for which of the following?
multiplying the input distance |
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multiplying the input force |
|
changing the direction of the input force |
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changing the direction of the output force |
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.
| 70 lbs. | |
| 52.45 lbs. | |
| 10 lbs. | |
| 10.01 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 7 for a mechanical advantage of \( \frac{10}{7} \) = 1.43
MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{75 lbs.}{1.43} \) = 52.45 lbs.