| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.45 |
| Score | 0% | 69% |
What type of load is sudden and for a relatively short duration?
impact load |
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dynamic load |
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non-uniformly distributed load |
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concentrated load |
A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.
A wedge is most similar to what other type of simple machine?
inclined plane |
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first-class lever |
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third-class lever |
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second-class lever |
The wedge is a moving inclined plane that is used to lift, hold, or break apart an object. A wedge converts force applied to its blunt end into force perpendicular to its inclined surface. In contrast to a stationary plane where force is applied to the object being moved, with a wedge the object is stationary and the force is being applied to the plane. Examples of a wedge include knives and chisels.
The force required to initally get an object moving is __________ the force required to keep it moving.
opposite |
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lower than |
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higher than |
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the same as |
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).
| 7.7 | |
| 6 | |
| 7 | |
| 8.5 |
The mechanical advantage (MA) of a wedge is its length divided by its thickness:
MA = \( \frac{l}{t} \) = \( \frac{35 in.}{5 in.} \) = 7
| 78 lbs. | |
| 300 lbs. | |
| 19.5 lbs. | |
| 6.5 lbs. |
fAdA = fBdB + fCdC
For this problem, this equation becomes:
25 lbs. x 12 ft. = 35 lbs. x 3 ft. + fC x 10 ft.
300 ft. lbs. = 105 ft. lbs. + fC x 10 ft.
fC = \( \frac{300 ft. lbs. - 105 ft. lbs.}{10 ft.} \) = \( \frac{195 ft. lbs.}{10 ft.} \) = 19.5 lbs.