| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.01 |
| Score | 0% | 60% |
Two or more pulleys used together are called:
third-class lever |
|
block and tackle |
|
gears |
|
wheel and axle |
Two or more pulleys used together constitute a block and tackle which, unlike a fixed pulley, does impart mechanical advantage as a function of the number of pulleys that make up the arrangement. So, for example, a block and tackle with three pulleys would have a mechanical advantage of three.
| 5.83 lbs. | |
| 17.5 lbs. | |
| 2 lbs. | |
| 4.38 lbs. |
To balance this lever the torques on each side of the fulcrum must be equal. Torque is weight x distance from the fulcrum so the equation for equilibrium is:
Rada = Rbdb
where a represents the left side of the fulcrum and b the right, R is resistance (weight) and d is the distance from the fulcrum.Solving for Ra, our missing value, and plugging in our variables yields:
Ra = \( \frac{R_bd_b}{d_a} \) = \( \frac{20 lbs. \times 7 ft.}{8 ft.} \) = \( \frac{140 ft⋅lb}{8 ft.} \) = 17.5 lbs.
The principle of moments defines equilibrium in terms of:
energy |
|
torque |
|
power |
|
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.
A box is resting on a smooth floor. Static friction is present:
if the coefficient of friction is greater than one |
|
at all times |
|
when an attempt is made to move the box |
|
only if normal force is present |
For any given surface, the coefficient of static friction is higher than the coefficient of kinetic friction. More force is required to initally get an object moving than is required to keep it moving. Additionally, static friction only arises in response to an attempt to move an object (overcome the normal force between it and the surface).
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?
810 lbs |
|
0.83 lbs |
|
90 lbs |
|
45 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