| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.02 |
| Score | 0% | 60% |
| 2 lbs. | |
| 6 lbs. | |
| 50 lbs. | |
| 3 lbs. |
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 Rb, our missing value, and plugging in our variables yields:
Rb = \( \frac{R_ad_a}{d_b} \) = \( \frac{10 lbs. \times 3 ft.}{5 ft.} \) = \( \frac{30 ft⋅lb}{5 ft.} \) = 6 lbs.
When all forces acting on a system cancel each other out, this is called:
stasis |
|
potential energy |
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rest |
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equilibrium |
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.
Which of the following represents how much two materials resist sliding across each other?
kinetic friction |
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coefficient of friction |
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static friction |
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normal friction |
Coefficient of friction (μ) represents how much two materials resist sliding across each other. Smooth surfaces like ice have low coefficients of friction while rough surfaces like concrete have high μ.
A truck is using a rope to pull a car. Tension in the rope is greatest in which of the following places?
near the truck |
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near the car |
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tension is equal in all parts of the rope |
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in the middle |
Tension is a force that stretches or elongates something. When a cable or rope is used to pull an object, for example, it stretches internally as it accepts the weight that it's moving. Although tension is often treated as applying equally to all parts of a material, it's greater at the places where the material is under the most stress.
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?
90 lbs |
|
0.83 lbs |
|
45 lbs |
|
810 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