| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.30 |
| Score | 0% | 66% |
| 0.29 ft. | |
| 0.04 ft. | |
| 0.07 ft. | |
| 0.21 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 da, our missing value, and plugging in our variables yields:
da = \( \frac{R_bd_b}{R_a} \) = \( \frac{5 lbs. \times 1 ft.}{70 lbs.} \) = \( \frac{5 ft⋅lb}{70 lbs.} \) = 0.07 ft.
Friction resists movement in a direction __________ to the movement.
parallel |
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perpendicular |
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normal |
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opposite |
Friction resists movement. Kinetic (also called sliding or dynamic) friction resists movement in a direction opposite to the movement. Because it opposes movement, kinetic friction will eventually bring an object to a stop. An example is a rock that's sliding across ice.
When it comes to force, mass and acceleration have what kind of relationship?
logarithmic |
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inverse |
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linear |
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exponential |
Newton's Second Law of Motion states that "The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object." This Law describes the linear relationship between mass and acceleration when it comes to force and leads to the formula F = ma or force equals mass multiplied by rate of acceleration.
A fixed pulley has a mechanical advantage of:
-1 |
|
0 |
|
1 |
|
2 |
A fixed pulley is used to change the direction of a force and does not multiply the force applied. As such, it has a mechanical advantage of one. The benefit of a fixed pulley is that it can allow the force to be applied at a more convenient angle, for example, pulling downward or horizontally to lift an object instead of upward.
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 |
|
810 lbs |
|
90 lbs |
|
0.83 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