| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.22 |
| Score | 0% | 64% |
The advantage of using a third-class lever is that it increases:
the distance traveled by the load |
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the force applied to the load |
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the speed of the load |
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the mechanical advantage of the lever |
A third-class lever is used to increase distance traveled by an object in the same direction as the force applied. The fulcrum is at one end of the lever, the object at the other, and the force is applied between them. This lever does not impart a mechanical advantage as the effort force must be greater than the load but does impart extra speed to the load. Examples of third-class levers are shovels and tweezers.
Friction between two or more solid objects that are not moving relative to each other is called:
dynamic friction |
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static friction |
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kinetic friction |
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gravitational friction |
Static friction is friction between two or more solid objects that are not moving relative to each other. An example is the friction that prevents a box on a sloped surface from sliding farther down the surface.
When all forces acting on a system cancel each other out, this is called:
stasis |
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potential energy |
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rest |
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equilibrium |
When a system is stable or balanced (equilibrium) all forces acting on the system cancel each other out. In the case of torque, equilibrium means that the sum of the anticlockwise moments about a center of rotation equal the sum of the clockwise moments.
Collinear forces:
pass through a common point |
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act along the same line of action |
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are unrelated to each other |
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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.
| 0.05 ft. | |
| 0.2 ft. | |
| 0.07 ft. | |
| 0.4 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 da, our missing value, and plugging in our variables yields:
da = \( \frac{R_bd_b}{R_a} \) = \( \frac{15 lbs. \times 1 ft.}{75 lbs.} \) = \( \frac{15 ft⋅lb}{75 lbs.} \) = 0.2 ft.