At a construction site, there are constant arguments and conflicts amongst workers of different contractors and sub-contractors.
1 answer:
You might be interested in
To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.
By definition the entropy change would be defined as
Using the Boyle equation we have
Where,
= Specific heat at constant pressure
= Initial temperature of gas
= Final temprature of gas
R = Universal gas constant
= Initial specific Volume of gas
= Final specific volume of gas
According to the statement, it is an isothermal process and the tank is therefore rigid
The equation would turn out as
<em>Therefore the entropy change of the ideal gas is 0</em>
Into the surroundings we have that
Where,
Q = Heat Exchange
T = Temperature in the surrounding
Replacing with our values we have that
<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>
The complete question is;
The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.
The image of this system is attached.
Answer:
Velocity = 11.8 ft/s
Explanation:
Since the wheel at A rotates about a fixed axis, then;
v_c = ω•r_c
r_c is 4.5 in. Let's convert it to ft.
So, r_c = 4.5/12 ft = 0.375 ft
Thus;
v_c = 0.375ω
Now the mass moment of inertia about of wheel A about it's mass centre is given as;
I_a = m•(k_a)²
The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug
Also, let's convert ka from inches to ft.
So, ka = 6/12 = 0.5
So,I_a = 1.5528 × 0.5²
I_a = 0.388 slug.ft²
The kinetic energy of the system would be;
T = Ta + Tc
Where; Ta = ½•I_a•ω²
And Tc = ½•m_c•(v_c)²
So, T = ½•I_a•ω² + ½•m_c•(v_c)²
Now, m_c is given as 200 lb.
Converting to slug, we have;
m_c = (200/32.2) slugs
Plugging in the relevant values, we have;
T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)
This now gives;
T = 0.6307 ω²
The system is initially at rest at T1 = 0.
Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.
Whereas at B, M does positive work and at C, W_c does negative work.
When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π
While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b
Where, rb = 3 in = 3/12 ft = 0.25 ft
ra = 7.5 in = 7.5/12 ft = 0.625 ft
So, θ_a = (0.25/0.625) × 10π
θ_a = 4π
Thus, we can say that the crate will have am upward displacement through a distance;
s_c = r_c × θ_a = 0.375 × 4π
s_c = 1.5π ft
So, the work done by M is;
U_m = M × θ_b
U_m = 50lb × 10π
U_m = 500π
Also,the work done by W_c is;
U_Wc = -W_c × s_c = -200lb × 1.5π
U_Wc = -300π
From principle of work and energy;
T1 + (U_m + U_Wc) = T
Since T1 is zero as stated earlier,
Thus ;
0 + 500π - 300π = 0.6307 ω²
0.6307ω² = 200π
ω² = 200π/0.6307
ω² = 996.224
ω = √996.224
ω = 31.56 rad/s
We earlier derived that;v_c = 0.375ω
Thus; v_c = 0.375 × 31.56
v_c = 11.8 ft/s
