| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.04 |
| Score | 0% | 61% |
Specific gravity is a comparison of the density of an object with the density of:
air |
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oil |
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water |
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carbon |
Specific gravity is the ratio of the density of equal volumes of a substance and water and is measured by a hyrdometer.
| 0 ft. | |
| 1.5 ft. | |
| 4.5 ft. | |
| 13.5 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 = \( \frac{R_ad_a}{R_b} \) = \( \frac{45 lbs. \times 1 ft.}{10 lbs.} \) = \( \frac{45 ft⋅lb}{10 lbs.} \) = 4.5 ft.
A a seesaw / teeter-totter is an example of which of the following?
second-class lever |
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third-class lever |
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first-class lever |
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inclined plane |
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.
| 140 lbs. | |
| 1260 lbs. | |
| 315 lbs. | |
| 35 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 = \( \frac{R_ad_a}{d_b} \) = \( \frac{70 lbs. \times 9 ft.}{2 ft.} \) = \( \frac{630 ft⋅lb}{2 ft.} \) = 315 lbs.
Which of the following will increase the mechanical advantage of this inclined plane?
lower the force acting at the blue arrow |
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shorten the ramp |
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lengthen the ramp |
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increase the force acting at the blue arrow |
The mechanical advantage (MA) of an inclined plane is the effort distance divided by the resistance distance. In order to increase mechanical advantage, this ratio must increase which means making the effort distance longer and this can be accomplished by lengthening the length of the ramp.