| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.92 |
| Score | 0% | 58% |
A wedge converts force applied to its blunt end into force __________ its inclined surface.
perpendicular to |
|
along |
|
opposite to |
|
parallel to |
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 advantage of using a third-class lever is that it increases:
the force applied to the load |
|
the distance traveled by the load |
|
the speed of the load |
|
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.
| 30 lbs. | |
| 15 lbs. | |
| 0 lbs. | |
| 7.5 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 Ra, our missing value, and plugging in our variables yields:
Ra = \( \frac{R_bd_b}{d_a} \) = \( \frac{60 lbs. \times 3 ft.}{6 ft.} \) = \( \frac{180 ft⋅lb}{6 ft.} \) = 30 lbs.
| 17 lbs. | |
| 8.75 lbs. | |
| 17.5 lbs. | |
| 5.83 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 = \( \frac{R_ad_a}{d_b} \) = \( \frac{35 lbs. \times 1 ft.}{2 ft.} \) = \( \frac{35 ft⋅lb}{2 ft.} \) = 17.5 lbs.
The force exerted on an object due to gravity is called:
mass |
|
weight |
|
density |
|
potential energy |
Mass is an intrinsic property of matter and does not vary. Weight is the force exerted on the mass of an object due to gravity and a specific case of Newton's Second Law of Motion. Replace force with weight and acceleration with acceleration due to gravity on Earth (g) and the result is the formula for weight: W = mg or, substituting for g, weight equals mass multiplied by 9.8 m/s2.