| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.28 |
| Score | 0% | 66% |
The force required to initally get an object moving is __________ the force required to keep it moving.
lower than |
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opposite |
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higher than |
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the same as |
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).
A screw is most like which of the following other simple machines?
wheel and axle |
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first-class lever |
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inclined plane |
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block and tackle |
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.
Which of these is the formula for force?
F = a/m |
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F = ma |
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F = m/a |
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F = am2 |
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.
| 9.64 ft. | |
| 2.41 ft. | |
| 0 ft. | |
| 38.57 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 = \( \frac{R_bd_b}{R_a} \) = \( \frac{75 lbs. \times 9 ft.}{70 lbs.} \) = \( \frac{675 ft⋅lb}{70 lbs.} \) = 9.64 ft.
| 1.5 ft. | |
| 7 ft. | |
| 6 ft. | |
| 4.5 ft. |
fAdA = fBdB
For this problem, the equation becomes:
30 lbs. x 4 ft. = 80 lbs. x dB
dB = \( \frac{30 \times 4 ft⋅lb}{80 lbs.} \) = \( \frac{120 ft⋅lb}{80 lbs.} \) = 1.5 ft.