| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.99 |
| Score | 0% | 60% |
Concurrent forces:
act along the same line of action |
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act in a common dimension |
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pass through a common point |
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act in a common plane |
Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.
Connected gears of different numbers of teeth are used together to change which of the following charasteristics of the input force?
rotational direction |
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force |
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energy |
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torque |
Connected gears of different numbers of teeth are used together to change the rotational speed and torque of the input force. If the smaller gear drives the larger gear, the speed of rotation will be reduced and the torque will increase. If the larger gear drives the smaller gear, the speed of rotation will increase and the torque will be reduced.
Which of the following is the formula for gravitational potential energy?
\(PE = { 1 \over 2} mg^2\) |
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\(PE = mg^2h\) |
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\(PE = { 1 \over 2} mv^2\) |
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\(PE = mgh\) |
Gravitational potential energy is energy by virtue of gravity. The higher an object is raised above a surface the greater the distance it must fall to reach that surface and the more velocity it will build as it falls. For gravitational potential energy, PE = mgh where m is mass (kilograms), h is height (meters), and g is acceleration due to gravity which is a constant (9.8 m/s2).
Which class of lever is used to increase force on an object in the same direction as the force is applied?
first |
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third |
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all of these |
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second |
A second-class lever is used to increase force on an object in the same direction as the force is applied. This lever requires a smaller force to lift a larger load but the force must be applied over a greater distance. The fulcrum is placed at one end of the lever and mechanical advantage increases as the object being lifted is moved closer to the fulcrum or the length of the lever is increased. An example of a second-class lever is a wheelbarrow.
| 4.8 ft. | |
| 1 ft. | |
| 1.2 ft. | |
| 1.6 ft. |
fAdA = fBdB
For this problem, the equation becomes:
15 lbs. x 8 ft. = 25 lbs. x dB
dB = \( \frac{15 \times 8 ft⋅lb}{25 lbs.} \) = \( \frac{120 ft⋅lb}{25 lbs.} \) = 4.8 ft.