| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.26 |
| Score | 0% | 65% |
| 0.3 | |
| 0.9 | |
| 0.45 | |
| 3.3 |
Mechanical advantage (MA) is the ratio by which effort force relates to resistance force. If both forces are known, calculating MA is simply a matter of dividing resistance force by effort force:
MA = \( \frac{F_r}{F_e} \) = \( \frac{9 ft.}{30.0 ft.} \) = 0.3
In this case, the mechanical advantage is less than one meaning that each unit of effort force results in just 0.3 units of resistance force. However, a third class lever like this isn't designed to multiply force like a first class lever. A third class lever is designed to multiply distance and speed at the resistance by sacrificing force at the resistance. Different lever styles have different purposes and multiply forces in different ways.
When all forces acting on a system cancel each other out, this is called:
equilibrium |
|
potential energy |
|
stasis |
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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.
A box is resting on a smooth floor. Static friction is present:
at all times |
|
when an attempt is made to move the box |
|
only if normal force is present |
|
if the coefficient of friction is greater than one |
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).
| 7 | |
| 21 | |
| 10.5 | |
| 10 |
The mechanical advantage (MA) of an inclined plane is the effort distance divided by the resistance distance. In this case, the effort distance is the length of the ramp and the resistance distance is the height of the green box:
MA = \( \frac{d_e}{d_r} \) = \( \frac{14 ft.}{2 ft.} \) = 7
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 |
|
810 lbs |
|
45 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