Please help ASAP and show all work. Thanks

1. An industrial water shutoff valve is designed to operate with 10 lb of effort force. The valve will encounter 100 lb of resistance force applied to a 2 in. diameter axle.
--a. What is the required actual mechanical advantage of the system?
--b. What is the required wheel diameter to overcome the resistance force?

2. A worker on a zip line crew lifts participants weighing approximately 200 lb several times a day from the ground 20 feet below. A block and tackle system is designed with 50 lb of effort force designed to lift the materials.
--a. What is the required actual mechanical advantage?
--b. How many supporting strands will be needed in the pulley system?

3. A simple gear train is composed of three gears. Gear A is the driver and has 10 teeth, gear B has 8 teeth, and gear C has 20 teeth.
--a. If the output is at C, what is the gear ratio?
--b. If gear A rotates at 60 rpm, how fast is gear C rotating?
--c. If the output of torque at gear C is 150 ftlb, what is the input torque at gear A?

4. Look at the picture provided.
--a. What class of lever is shown in the figure? (How do you know?)
--b. How much effort force is needed to balance the 100 lb load?

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;

[tex]AMA = \dfrac{F_R}{F_E}[/tex]

Where;

[tex]F_R[/tex] = The resistance force magnitude

[tex]F_E[/tex] = The effort force magnitude

1. a. We have;

[tex]F_R[/tex] = 10 lb.

[tex]F_E[/tex] = 100 lb.

[tex]AMA = \dfrac{100 \ lb}{10 \ lb} = 10[/tex]

b. [tex]Mechanical \ advantage, \ M.A. = \dfrac{Distance \ moved \ by \ load}{Distance \ moved \ by \ effort}[/tex]

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

[tex]M.A. = 10 = \dfrac{\pi \cdot D}{\pi \times 2}[/tex]

∴ 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

[tex]The \ gear \ ratio = \dfrac{The \ number \ of \ teeth \ on \ the \ driven \ gear}{The \ number \ of \ teeth \ on \ the \ driver \ gear}[/tex]

The driver gear = The input gear

Therefore, we have;

[tex]The \ gear \ ratio = \dfrac{20 \ teeth}{10 \ teeth} = 2[/tex]

The gear ratio = 2

-b [tex]The \ gear \ ratio = \dfrac{The \ driver \ gear\ speed}{The \ driven \ gear\ speed}[/tex]

Therefore, we have;

[tex]The \ gear \ ratio = 2 = \dfrac{60 \ rpm}{Gear\ C \, speed}[/tex]

[tex]Gear\ C \, speed = \dfrac{60 \ rpm}{2} = 30 \ rpm[/tex]

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

[tex]The \ gear \ ratio = \dfrac{Driven \ gear \ torque}{Driver \ gear \ torque}[/tex]

Therefore;

[tex]2 = \dfrac{150 \ ft \cdot lb}{Driver \ gear \ torque}[/tex]

[tex]Driver \ gear \ torque = \dfrac{150 \ ft \cdot lb}{2} = 75 \ ft \cdot lb[/tex]

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.

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Could anyone answer this, please? It's about solid mechanics. I will give you 100 points!!! It's due at midnight.