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maxonik [38]
3 years ago
5

The cable connecting the winch A to point B on the railroad car in figure is wound in at the constant rate of 2 m. When θ = 60◦

, determine (a) the velocity of B and ˙θ; and (b) the acceleration of B and ¨θ. Neglect the radius of the winch.​
Engineering
1 answer:
Crank3 years ago
7 0

Answer:

This question comprises two independent parts, (a) and (b) (a) In the diagram below, the cable connecting the winch A with point B on the railway carriage is being wound in at a constant rate of 2 m/s. Determine, for the instant in time when the angle θ= 60° (i) The velocity vector of the railway carriage (ii) The acceleration vector of the railway carriage The radius of the winch can be considered negligible. (2 marks) (3 marks) (b) The diagram below shows two rotating ‘driving links, OA and BC, connected by a telescoping link AB. Determine the angular velocity vector of the telescoping link AB when it is in the position shown if the driving links have the angular velocities indicated in the diagram. 150 2 rad/s 45 165 Dimensions in millimeters 60 0 2 rad/s

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*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t
riadik2000 [5.3K]

Rubber block is not shown. I have attached an image of it.

Answer:

A) ε_x = 0.0075

B) ε_y = 0.00375

C) γ_xy = 0.0122 rad

Explanation:

We are given;

δ = 0.03 in

L = 4 in

ν_r = 0.5

θ = 89.3° = 89.3π/180 rad

Let's calculate ε_x in the direction of axis x

Thus, ε_x = δ/L = 0.03/4 = 0.0075

Let's calculate ε_y in the direction of axis y;

ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375

Now, shear strain is angle between π/2 rad surfaces at that point.

Thus,

γ_xy =  π/2 - θ = π/2 - 89.3π/180

γ_xy = π(0.003889) = 0.0122 rad

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3 years ago
A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas
ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × 10^{5} Pa

P (gauge ) = 2 × 10^{5} Pa

so from idea gas equation

\frac{P1*V1}{T1} = \frac{P2*V2}{T2}   ................1

so {P2} = \frac{P1*T2}{T1}

{P2} = \frac{2*10^5*350}{300}

P2 = 2.33 × 10^{5} Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × 10^{5}  - 1.0 × 10^{5}

gauge pressure = 1.33 × 10^{5} Pa

so gauge pressure is 133 kPa

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Answer:

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andreev551 [17]

Answer:

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Explanation:

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Answer:

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