| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.95 |
| Score | 0% | 59% |
| 1.31 ft. | |
| 2.63 ft. | |
| 10.5 ft. | |
| 5.25 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 = \( \frac{R_bd_b}{R_a} \) = \( \frac{15 lbs. \times 7 ft.}{40 lbs.} \) = \( \frac{105 ft⋅lb}{40 lbs.} \) = 2.63 ft.
| 345 | |
| 206.5 | |
| None of these is correct | |
| -89 |
| 3 | |
| 10 | |
| 7 | |
| 9 |
Mechanical advantage is resistance force divided by effort force:
MA = \( \frac{F_r}{F_e} \) = \( \frac{560 lbs.}{80 lbs.} \) = 7
Which class of lever is used to increase force on an object in the same direction as the force is applied?
all of these |
|
third |
|
first |
|
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.
| -3 | |
| 1.6 | |
| 5 | |
| 0.63 |
The mechanical advantage of a wheel and axle is the input radius divided by the output radius:
MA = \( \frac{r_i}{r_o} \)
In this case, the input radius (where the effort force is being applied) is 8 and the output radius (where the resistance is being applied) is 5 for a mechanical advantage of \( \frac{8}{5} \) = 1.6