Question

Please help ASAP and show all work. Thanks

Answer 1

Answer:

1. --a 10

--b 100 in.

2. --a 4

--b 4

3. --a 2

--b 30 rpm

--c 75 ft.·lb.

4. --a Second class lever

--b 50 lbs.

Explanation:

The Actual Mechanical Advantage, AMA, is given as follows;

$AMA = \dfrac{F_R}{F_E}$

Where;

$F_R$ = The resistance force magnitude

$F_E$ = The effort force magnitude

1. a. We have;

$F_R$ = 10 lb.

$F_E$ = 100 lb.

$AMA = \dfrac{100 \ lb}{10 \ lb} = 10$

b. $Mechanical \ advantage, \ M.A. = \dfrac{Distance \ moved \ by \ load}{Distance \ moved \ by \ effort}$

The diameter of the axle, d = 2 in.

Let 'D' represent the diameter of the wheel, we have;

The distance moved by the axle, c = π·d

The distance moved by the load, C = π·D

$M.A. = 10 = \dfrac{\pi \cdot D}{\pi \times 2}$

∴ 2 × 10 = D

D = 20 in.

The required wheel diameter to overcome the resistance force, D = 100 in.

2. --a The mass of the participants, m = 200 lb.

The depth of the ground of the participants = 20 feet

The effort force = 50 lb

Actual Mechanical Advantage, AMA = 200 lb./(50 lb.) = 4

--b. The number of strands of pulley needed ≈ The mechanical advantage = 4

3. The number of gears on Gear A = 10 teeth

The number of gears on Gear B = 8 teeth

The number of gears on Gear C = 20 teeth

-a. Given that the driver gear = Gear A

The output gear = Gear C

$The \ gear \ ratio = \dfrac{The \ number \ of \ teeth \ on \ the \ driven \ gear}{The \ number \ of \ teeth \ on \ the \ driver \ gear}$

The driver gear = The input gear

Therefore, we have;

$The \ gear \ ratio = \dfrac{20 \ teeth}{10 \ teeth} = 2$

The gear ratio = 2

-b $The \ gear \ ratio = \dfrac{The \ driver \ gear\ speed}{The \ driven \ gear\ speed}$

Therefore, we have;

$The \ gear \ ratio = 2 = \dfrac{60 \ rpm}{Gear\ C \, speed}$

$Gear\ C \, speed = \dfrac{60 \ rpm}{2} = 30 \ rpm$

-c The output (driven) gear torque at Gear C = 150 ft.·lb.

$The \ gear \ ratio = \dfrac{Driven \ gear \ torque}{Driver \ gear \ torque}$

Therefore;

$2 = \dfrac{150 \ ft \cdot lb}{Driver \ gear \ torque}$

$Driver \ gear \ torque = \dfrac{150 \ ft \cdot lb}{2} = 75 \ ft \cdot lb$

The input (driver) torque at Gear A = 75 ft·lb

4. -a Given that the load is between the effort and the fulcrum, we have;

The type of lever is a second class lever

-b The distance between the load and the fulcrum = 4 feet

The distance between the effort and the fulcrum = 8 feet

We have;

100 lbs × 4 ft. = Effort × 8 ft.

∴ Effort = 100 lbs × 4 ft./(8 ft.) = 50 lbs.

The effort = 50 lbs.