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puteri [66]
2 years ago
14

The assembly consists of two blocks A and B, which have a mass of 20 kg and 30 kg, respectively. Determine the distance B must d

escend in order for A to achieve a speed of 3 m>s starting from rest.

Engineering
2 answers:
lozanna [386]2 years ago
7 0

Answer:

ΔδB = 5.7 m

Explanation:

Given

mA = 20 Kg

mB = 30 Kg

vA = 3 m/s

vA0 = vB0 = 0 m/s (the blocks released from rest)

Since the cable length (L) is constant

3*δA + δB = L

3*ΔδA = - ΔδB

If we apply

d(3*δA + δB)/dt = dL/dt

3*vA + vB = 0   ⇒    3*vA = - vB

vB = -3*(3 m/s)    ⇒    vB = - 9 m/s

We apply Conservation of Energy

TA0 + TB0 + vA0 + vB0 = TA + TB + vA + vB

0 + 0 + 0 + 0 = 0.5*mA*vA² + 0.5*mB*vB² + WA*ΔδA + WB*ΔδB

0 = 0.5*mA*vA² + 0.5*mB*(-3*vA)² + WA*(ΔδB/3) - WB*ΔδB

0 = 0.5*20 Kg*(3 m/s)² + 0.5*30 Kg*(-9 m/s)² + (20 Kg*9.81 m/s²)*(ΔδB/3) - (30 Kg*9.81 m/s²)*ΔδB

⇒   ΔδB = 5.7 m

The pic shown can help us to understand the question.

Sedaia [141]2 years ago
6 0

Answer:

5.70 m

Explanation:

3sₐ + sB = L

3∆sₐ = - ∆sB

3vₐ = -vB

3*3 = -vB

vB = -9 ms⁻²

T₁ + V₁ = T₂ + V₂

(0 + 0) + (0 + 0) =  1/2(20)(3)² + 1/2(30)(-9)² + 20*9.81*(sB/3) + 30*9.81* sB

0 =  90  +  1215  - 228.9sB

-1305 = - 228.9sB

sB = 1305/228.9 = 5.70 m

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Saturated steam coming off the turbine of a steam power plant at 40°C condenses on the outside of a 3-cm-outer-diameter, 35-m-lo
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Answer:

93.57 KJ/s

Explanation:

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3 years ago
The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu
marishachu [46]

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

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3 years ago
Consider a plane composite wall that is composed of two materials of thermal conductivities kA = 0.1 W/m*K and kB = 0.04 W/m*K a
nadya68 [22]

Answer:

q=39.15 W/m²

Explanation:

We know that

Thermal resistance due to conductivity given as

R=L/KA

Thermal resistance due to heat transfer coefficient given as

R=1/hA

Total thermal resistance

R_{th}=\dfrac{L_A}{AK_A}+\dfrac{L_B}{AK_B}+\dfrac{1}{Ah_1}+\dfrac{1}{Ah_2}+\dfrac{1}{Ah_3}

Now by putting the values

R_{th}=\dfrac{0.01}{0.1A}+\dfrac{0.02}{0.04A}+\dfrac{1}{10A}+\dfrac{1}{20A}+\dfrac{1}{0.3A}

R_{th}=4.083/A\ K/W

We know that

Q=ΔT/R

Q=\dfrac{\Delta T}{R_{th}}

Q=A\times \dfrac{200-40}{4.086}

So heat transfer per unit volume is 39.15 W/m²

q=39.15 W/m²

4 0
3 years ago
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