| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.40 |
| Score | 0% | 68% |
| 7.5 | |
| 1.7 | |
| 4.5 | |
| 1.5 |
The gear ratio (Vr) of a gear train is the product of the gear ratios between the pairs of meshed gears. Let N represent the number of teeth for each gear:
Vr = \( \frac{N_1}{N_2} \) \( \frac{N_2}{N_3} \) \( \frac{N_3}{N_4} \) ... \( \frac{N_n}{N_{n+1}} \)
In this problem, we have only two gears so the equation becomes:Vr = \( \frac{N_1}{N_2} \) = \( \frac{24}{16} \) = 1.5
Drag is a type of:
work |
|
kinetic energy |
|
friction |
|
potential energy |
Drag is friction that opposes movement through a fluid like liquid or air. The amount of drag depends on the shape and speed of the object with slower objects experiencing less drag than faster objects and more aerodynamic objects experiencing less drag than those with a large leading surface area.
If the handles of a wheelbarrow are 3 ft. from the wheel axle, what force must you exert to lift the handles if it's carrying a 270 lb. load concentrated at a point 0.5 ft. from the axle?
45 lbs |
|
0.83 lbs |
|
810 lbs |
|
90 lbs |
This problem describes a second-class lever and, for a second class lever, 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 3 ft. = 270 lbs x 0.5 ft
Fe = 135 ft-lb. / 3 ft
Fe = 45 lbs
| 0ft⋅lb | |
| 800ft⋅lb | |
| 6400ft⋅lb | |
| 1600 ft⋅lb |
Which of the following represents the force a surface exerts when an object presses against it?
normal force |
|
mass |
|
friction |
|
counter force |
Normal force (FN) represents the force a surface exerts when an object presses against it.