| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.14 |
| Score | 0% | 63% |
When all forces acting on a system cancel each other out, this is called:
stasis |
|
potential energy |
|
equilibrium |
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rest |
When a system is stable or balanced (equilibrium) all forces acting on the system cancel each other out. In the case of torque, equilibrium means that the sum of the anticlockwise moments about a center of rotation equal the sum of the clockwise moments.
| 2.25 ft. | |
| 6.75 ft. | |
| 20.25 ft. | |
| 13.5 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 = \( \frac{R_bd_b}{R_a} \) = \( \frac{45 lbs. \times 9 ft.}{60 lbs.} \) = \( \frac{405 ft⋅lb}{60 lbs.} \) = 6.75 ft.
| 11.25 lbs. | |
| 8.44 lbs. | |
| 16.88 lbs. | |
| 33.75 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 = \( \frac{R_bd_b}{d_a} \) = \( \frac{45 lbs. \times 3 ft.}{4 ft.} \) = \( \frac{135 ft⋅lb}{4 ft.} \) = 33.75 lbs.
| 16.88 lbs. | |
| 22.5 lbs. | |
| 2.81 lbs. | |
| 5.63 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 Rb, our missing value, and plugging in our variables yields:
Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{15 lbs. \times 3 ft.}{8 ft.} \) = \( \frac{45 ft⋅lb}{8 ft.} \) = 5.63 lbs.
Assuming force applied remains constant, which of the following will result in more work being done?
moving the object with more acceleration |
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moving the object with more speed |
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increasing the coefficient of friction |
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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.