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Goshia [24]
3 years ago
7

1. A first class lever in statistics equilibrium has a 75 Ib resistance force and a 20 Ib effort force. The lever's effort force

is located 5.2 ft from the fulcrum. What's the actual mechanical advantage of the system?
2. Using static equilibrium calculations, calculate the length from the fulcrum to the resistance force.
Engineering
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

1. 3.75

2. 1.39 ft

Explanation:

The mechanical advantage is the ratio of the resistance force to the effort force.

MA = 75 lb / 20 lb

MA = 3.75

The sum of the torques about the fulcrum equals zero.

∑τ = Iα

(20 lb) (5.2 ft) − (75 lb) x = 0

x = 1.39 ft

Round as needed.

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5 0
3 years ago
In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical
stealth61 [152]

Answer:

Explanation:

We know that Drag forceF_D

  F_D=\dfrac{1}{2}C_D\rho AV^2

Where

             C_D is the drag force constant.

                 A is the projected area.

                V is the velocity.

                ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

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6 0
4 years ago
A 400 kg machine is placed at the mid-span of a 3.2-m simply supported steel (E = 200 x 10^9 N/m^2) beam. The machine is observe
Alenkinab [10]

Answer:

moment of inertia = 4.662 * 10^6 mm^4

Explanation:

Given data :

Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 \pi f = 58.434 rad/secs

also Wm = \sqrt{\frac{g}{t} }  ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = \frac{wl^3}{48EI}

insert the value of t into equation 1

Wm^2 = \frac{g*48*E*I}{WL^3}   make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = \frac{WL^3* Wn^2}{48*g*E}

I = \frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}  = 4.662 * 10^6 mm^4

6 0
3 years ago
Is a water proof material used around tubs and
Leviafan [203]
I think it’s hard board
4 0
3 years ago
Read 2 more answers
A damped harmonic oscillator consists of a mass on a spring, with a damping force proportional to the speed of the block. If the
Murrr4er [49]

Answer:

TOTAL ENERGY = 0.74 j

Explanation:

Given data:

spring constant is 350 N/m

m = 0.24 kg

b = 0.41 kg/s

A = 0.075 M

\omega = \sqrt{\frac{k}{m}}

             = \sqrt{\frac{350}{0.24}}

y = e^{\frac{-b}{2m} t} A cos(\omega t)

  =e^{\frac{-0.41}{2*0.24} t} cos (\sqrt{\frac{350}{0.24}} t) *0.075

after one full cycle, mass will be at extreme point, hence K,E = 0

But total energy remain same

y after one full cycle is\

y =e^{\frac{-0.41}{2*0.24} 2\pi \sqrt{\frac{0.24}{350}}} cos (2\pi*0.075)

y = 0.06517 m

y = 6.517 cm

total energy = \frac{1}{2} k y^2

= \frac{1}{2}* 350* 0.06517^2 = 0.742 J

8 0
4 years ago
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