| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.34 |
| Score | 0% | 67% |
What type of load acts on a relatively small area of a structure?
non-uniformly distributed load |
|
impact load |
|
concentrated load |
|
dynamic load |
A concentrated load acts on a relatively small area of a structure, a static uniformly distributed load doesn't create specific stress points or vary with time, a dynamic load varies with time or affects a structure that experiences a high degree of movement, an impact load is sudden and for a relatively short duration and a non-uniformly distributed load creates different stresses at different locations on a structure.
| 80% | |
| 20% | |
| 160% | |
| 0% |
| 320 lbs. | |
| 323 lbs. | |
| 106 lbs. | |
| 352 lbs. |
The mechanical advantage (MA) of a block and tackle pulley is equal to the number of times the effort force changes direction. An easy way to count how many times the effort force changes direction is to count the number of ropes that support the resistance which, in this problem, is 8. With a MA of 8, a 40 lbs. effort force could lift 40 lbs. x 8 = 320 lbs. resistance.
| 6.2 lbs. | |
| 7.2 lbs. | |
| 5 lbs. | |
| 79.17 lbs. |
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 6 and the output radius (where the resistance is being applied) is 5 for a mechanical advantage of \( \frac{6}{5} \) = 1.2
MA = \( \frac{load}{effort} \) so effort = \( \frac{load}{MA} \) = \( \frac{95 lbs.}{1.2} \) = 79.17 lbs.
| 0.97 ft. | |
| 175 ft. | |
| 3.89 ft. | |
| 0 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 db, our missing value, and plugging in our variables yields:
db = \( \frac{R_ad_a}{R_b} \) = \( \frac{35 lbs. \times 5 ft.}{45 lbs.} \) = \( \frac{175 ft⋅lb}{45 lbs.} \) = 3.89 ft.