| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.69 |
| Score | 0% | 54% |
Force of friction due to kinetic friction is __________ the force of friction due to static friction.
lower than |
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higher than |
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opposite |
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the same as |
The formula for force of friction (Ff) is the same whether kinetic or static friction applies: Ff = μFN. To distinguish between kinetic and static friction, μk and μs are often used in place of μ.
A a seesaw / teeter-totter is an example of which of the following?
first-class lever |
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inclined plane |
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third-class lever |
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second-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.
| 8 | |
| 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 8 and the output radius (where the resistance is being applied) is 7 for a mechanical advantage of \( \frac{8}{7} \) = 1.14
| 245 ft. | |
| 18.38 ft. | |
| 6.13 ft. | |
| 5 ft. |
To balance this lever the torques at the green box and the blue arrow must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:
Rada = Rbdb
where a represents the green box and b the blue arrow, R is resistance (weight/force) and d is the distance from the fulcrum.Solving for db, our missing value, and plugging in our variables yields:
db = \( \frac{R_ad_a}{R_b} \) = \( \frac{35 lbs. \times 7 ft.}{40 lbs.} \) = \( \frac{245 ft⋅lb}{40 lbs.} \) = 6.13 ft.
| 0 ft⋅lb | |
| 9900 ft⋅lb | |
| 12 ft⋅lb | |
| 1 ft⋅lb |