| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.58 |
| Score | 0% | 52% |
| 0 lbs. | |
| 262.5 lbs. | |
| 131.25 lbs. | |
| 43.75 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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{75 lbs. \times 7 ft.}{4 ft.} \) = \( \frac{525 ft⋅lb}{4 ft.} \) = 131.25 lbs.
The measure of how much of the power put into a machine is turned into movement or force is called:
efficiency |
|
force multiplication |
|
power |
|
mechanical advantage |
The efficiency of a machine describes how much of the power put into the machine is turned into movement or force. A 100% efficient machine would turn all of the input power into output movement or force. However, no machine is 100% efficient due to friction, heat, wear and other imperfections that consume input power without delivering any output.
| 30 lbs. | |
| 60 lbs. | |
| 66 lbs. | |
| 61.5 lbs. |
This problem describes an inclined plane and, for an inclined plane, the effort force multiplied by the effort distance equals the resistance force multipied by the resistance distance:
Fede = Frdr
Plugging in the variables from this problem yields:
Fe x 5 ft. = 300 lbs. x 1 ft.
Fe = \( \frac{300 ft⋅lb}{5 ft.} \) = 60 lbs.
The principle of conservation of mechanical energy states that, as long as no other forces are applied, what will remain constant as an object falls?
total mechanical energy |
|
kinetic energy |
|
acceleration |
|
potential energy |
As an object falls, its potential energy is converted into kinetic energy. The principle of conservation of mechanical energy states that, as long as no other forces are applied, total mechanical energy (PE + KE) of the object will remain constant at all points in its descent.
| 20.63 lbs. | |
| 2 lbs. | |
| 225 lbs. | |
| 82.5 lbs. |
fAdA = fBdB + fCdC
For this problem, this equation becomes:
25 lbs. x 9 ft. = 30 lbs. x 2 ft. + fC x 8 ft.
225 ft. lbs. = 60 ft. lbs. + fC x 8 ft.
fC = \( \frac{225 ft. lbs. - 60 ft. lbs.}{8 ft.} \) = \( \frac{165 ft. lbs.}{8 ft.} \) = 20.63 lbs.