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2 years ago

What is the IMA of this pulley belt system if the diameter of the input

Engineering

1 answer:

Stella [2.4K]2 years ago

8 0

Answer:

2.8

Explanation:

The ideal mechanical advantage of the pulley IMA = D'/D where D' = diameter of output pulley = 7 inches and D = diameter of input pulley = 2.5 inches

So, IMA = D'/D

= 7/2.5

= 2.8

So, the ideal mechanical advantage of the pulley IMA = 2.8

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A plate clutch is used to connect a motor shaft running at 1500rpm to shaft 1. The motor is rated at 4 hp. Using a service facto

vazorg [7]

Answer:

$(M_t)_{rated}=61.11lb-in$

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = $4 \times 0.7457$ =2.9828 KW

service factor(k)= 2.75

now,

$KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }$

$M_t= 18,989.09 \ N-mm= 168.06 lb-in$

torque rating

$(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in$

4 0

3 years ago

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is

romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

For cylinder AB:

Let Length of AB = 12 in

$c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\$

$\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}$

For cylinder BC:

Let Length of BC = 18 in

$c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\$

$\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}$

$2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}$

$T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in$

b) Maximum shear stress in BC

$\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi$

Maximum shear stress in AB

$\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi$

8 0

3 years ago

Select four items that an industrial engineer must obtain in order to practice in the field.

alex41 [277]

Answer:

Professional engineering license

Bachelor's degree

Computer science classes

job recommendations

3 0

2 years ago

Read 2 more answers

2. A F-22 Raptor has just climbed through an altitude of 9,874 m at 1,567 kph when a disk

BabaBlast [244]

The pressure difference across the sensor housing will be "95 kPa".

According to the question, the values are:

Altitude,

9874

Speed,

1567 kph

Pressure,

122 kPa

The temperature will be:

→ $T = 15.04-[0.00649(9874)]$

→ $= 15.04-64.082$

→ $= -49.042^{\circ} C$

now,

→ $P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}$

→ $= 27.074$

hence,

→ The pressure differential will be:

= $122-27$

= $95 \ kPa$

Thus the above solution is correct.

Learn more about pressure difference here:

brainly.com/question/15732832

3 0

3 years ago

Which of the following allows team members to visualize a design model from a variety of perspectives?

julsineya [31]

Answer: from what i know im pretty sure its isometrics or sketches im certain its sketches but not 100%

Explanation: A sketch is a rapidly executed freehand drawing that is not usually intended as a finished work. A sketch may serve a number of purposes: it might record something that the artist sees, it might record

8 0

3 years ago

Read 2 more answers