| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.91 |
| Score | 0% | 58% |
| 0 ft⋅lb | |
| 12 ft⋅lb | |
| 108 ft⋅lb | |
| -33 ft⋅lb |
| 4.91 ft. | |
| 1.64 ft. | |
| 19.64 ft. | |
| 270 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 db, our missing value, and plugging in our variables yields:
db = \( \frac{R_ad_a}{R_b} \) = \( \frac{30 lbs. \times 9 ft.}{55 lbs.} \) = \( \frac{270 ft⋅lb}{55 lbs.} \) = 4.91 ft.
Sam can do 50 ft. lb. of work in 2 minutes and 5 seconds. What would Sam have to do to increase his power output?
do the work in 2 minutes |
|
do 25 ft. lb. of work in 2 minutes 5 seconds |
|
do the work in 3 minutes |
|
do 100 ft. lb. of work in 4 minutes 12 seconds |
Power is the rate of doing work or \(\frac{W}{t}\). To increase power, increase the work being done in the same amount of time or do the same amount of work in less time.
The mechanical advantage of a third class lever is always:
less than one |
|
greater than one |
|
equal to one |
|
not equal to one |
A third class lever is designed to multiply distance and speed at the expense of effort force. Because the effort force is greater than the resistance, the mechanical advantage of a third class lever is always less than one.
An example of a third class lever is a broom. The fulcrum is at your hand on the end of the broom, the effort force is your other hand in the middle, and the resistance is at the bottom bristles. The effort force of your hand in the middle multiplies the distance and speed of the bristles at the bottom but at the expense of producing a brushing force that's less than the force you're applying with your hand.
| 21.88 lbs. | |
| 10 lbs. | |
| 262.5 lbs. | |
| 87.5 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 Ra, our missing value, and plugging in our variables yields:
Ra = \( \frac{R_bd_b}{d_a} \) = \( \frac{75 lbs. \times 7 ft.}{6 ft.} \) = \( \frac{525 ft⋅lb}{6 ft.} \) = 87.5 lbs.