| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.64 |
| Score | 0% | 53% |
The principle of moments defines equilibrium in terms of:
torque |
|
power |
|
energy |
|
speed |
According to the principle of moments, you can maintain equilibrium if the moments (forces) tending to clockwise rotation are equal to the moments tending to counterclockwise rotation. Another name for these moments of force is torque.
Collinear forces:
pass through a common point |
|
act in a common plane |
|
are unrelated to each other |
|
act along the same line of action |
Collinear forces act along the same line of action, concurrent forces pass through a common point and coplanar forces act in a common plane.
| 3740 \( \frac{ft⋅lb}{s} \) | |
| 11220 \( \frac{ft⋅lb}{s} \) | |
| 10 \( \frac{ft⋅lb}{s} \) | |
| 14960 \( \frac{ft⋅lb}{s} \) |
Assuming force applied remains constant, which of the following will result in more work being done?
moving the object with more speed |
|
moving the object with more acceleration |
|
increasing the coefficient of friction |
|
moving the object farther |
Work is accomplished when force is applied to an object: W = Fd where F is force in newtons (N) and d is distance in meters (m). Thus, the more force that must be applied to move an object, the more work is done and the farther an object is moved by exerting force, the more work is done.
| 192.5 lbs. | |
| 0 lbs. | |
| 16.04 lbs. | |
| 64.17 lbs. |
fAdA = fBdB + fCdC
For this problem, this equation becomes:
45 lbs. x 11 ft. = 55 lbs. x 2 ft. + fC x 6 ft.
495 ft. lbs. = 110 ft. lbs. + fC x 6 ft.
fC = \( \frac{495 ft. lbs. - 110 ft. lbs.}{6 ft.} \) = \( \frac{385 ft. lbs.}{6 ft.} \) = 64.17 lbs.