| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.71 |
| Score | 0% | 54% |
| 0 | |
| None of these is correct | |
| 331.6 | |
| 200 |
Torque involves a perpendicular force applied to a lever arm that moves around a center of rotation. Increasing the length of the lever arm will do which of the following?
decrease applied force |
|
increase torque |
|
increase applied force |
|
decrease torque |
Torque measures force applied during rotation: τ = rF. Torque (τ, the Greek letter tau) = the radius of the lever arm (r) multiplied by the force (F) applied. Radius is measured from the center of rotation or fulcrum to the point at which the perpendicular force is being applied. The resulting unit for torque is newton-meter (N-m) or foot-pound (ft-lb).
Two gears are connected and the smaller gear drives the larger gear. The speed of rotation will __________ and the torque will __________.
decrease, increase |
|
increase, increase |
|
increase, decrease |
|
decrease, decrease |
Connected gears of different numbers of teeth are used together to change the rotational speed and torque of the input force. If the smaller gear drives the larger gear, the speed of rotation will be reduced and the torque will increase. If the larger gear drives the smaller gear, the speed of rotation will increase and the torque will be reduced.
A screw is most like which of the following other simple machines?
first-class lever |
|
block and tackle |
|
inclined plane |
|
wheel and axle |
A screw is an inclined plane wrapped in ridges (threads) around a cylinder. The distance between these ridges defines the pitch of the screw and this distance is how far the screw advances when it is turned once. The mechanical advantage of a screw is its circumference divided by the pitch.
| 56.9 lbs. | |
| 51.8 lbs. | |
| 54.8 lbs. | |
| 17.3 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 17 ft. = 440 lbs. x 2 ft.
Fe = \( \frac{880 ft⋅lb}{17 ft.} \) = 51.8 lbs.