The figure shows a structure subject to a force of 60 kN and a torque of 30 kN.m. Determine the horizontal (left-to-right orient
ed) and vertical (bottom-up oriented) force modulus acting on point A and also the tension developed on the BC rope. The blue arrow indicates the "direction of rotation" caused in the structure by the action of the torque in relation to point A.
1 answer:
Answer:
Fₓ = 21.9 kN
Fᵧ = 84.3 kN
T = 32.7 kN
Explanation:
Draw a free body diagram (assuming the weight of the structure is included in the 60 kN force).
There are vertical and horizontal reaction forces at A (Fᵧ and Fₓ), and a tension force T at B pulling down along the rope.
The length of BC is √(2.7² + 3²) = √16.29. Using similar triangles, the vertical and horizontal components of the tension force are:
Tᵧ = 3 T / √16.29 ≈ 0.743 T
Tₓ = 2.7 T / √16.29 ≈ 0.669 T
Sum of moments about A in the counterclockwise direction:
∑τ = Iα
Tᵧ (1 m) + Tₓ (3 m) − 60 kN (1 m) − 30 kNm = 0
Tᵧ + 3 Tₓ = 90 kN
0.743 T + 3 (0.669 T) = 90 kN
2.750 T = 90 kN
T = 32.7 kN
Sum of forces in the +x direction:
∑F = ma
Fₓ − Tₓ = 0
Fₓ = Tₓ
Fₓ = 0.669 T
Fₓ = 21.9 kN
Sum of forces in the +y direction:
∑F = ma
Fᵧ − Tᵧ − 60 kN= 0
Fᵧ = Tᵧ + 60 kN
Fᵧ = 0.743 T + 60 kN
Fᵧ = 84.3 kN
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Answer:
a) T_1 = 938 K , T_2 = 931 K
b) To prevent softening of the materials, which would occur below their melting points, the reactor should not be operated much above:
q = 3*10^8 W/m^3
Explanation:
Given:
- See the attachment for the figure for this question.
- Melting point of Thorium T_th = 2000 K
- Melting point of Thorium T_g = 2300 K
Find:
a) If the thermal energy is uniformly generated in the fuel element at a rate q = 10^8 W/m^3 then what are the temperatures T_1 and T_2 at the inner and outer surfaces, respectively, of the fuel element?
b) Compute and plot the temperature distribution in the composite wall for selected values of q. What is the maximum allowable value of q.
Solution:
part a)
- The outer surface temperature of the fuel, T_2, may be determined from the rate equation:
q*A_th = T_2 - T_inf / R'_total
Where,
A_th: Area of the thorium section
T_inf: The temperature of coolant = 600 K
R'_total: The resistance per unit length.
- Calculate the resistance per unit length R' from thorium surface to coolant:
R'_total = Ln(r_3/r_2) / 2*pi*k_g + 1 / 2*pi*r_3*h
Plug in values:
R'_total = Ln(14/11) / 2*pi*3 + 1 / 2*pi*0.014*2000
R'_total = 0.0185 mK / W
- And the heat rate per unit length may be determined by applying an energy balance to a control surface about the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that:
q' = q*A_th = q*pi*(r_2^2 - r_1^2)
q' = 10^8*pi*(0.011^2 - 0.008^2) = 17,907 W / m
Hence,
T_2 = q' * R'_total + T_inf
T_2 = 17,907*0.0185 + 600
T_2 = 931 K
- With zero heat flux at the inner surface of the fuel element, We will apply the derived results for boundary conditions as follows:
T_1 = T_2 + (q*r_2^2/4*k_th)*( 1 - (r_1/r_2)^2) - (q*r_1^2/2*k_th)*Ln(r_2/r_1)
Plug values in:
T_1 = 931+(10^8*0.011^2/4*57)*( 1 - (.8/1.1)^2) - (10^8*0.008^2/2*57)*Ln(1.1/.8)
T_1 = 931 + 25 - 18 = 938 K
part b)
The temperature distributions may be obtained by using the IHT model for one-dimensional, steady state conduction in a hollow tube. For the fuel element (q > 0), an adiabatic surface condition is prescribed at r_1 while heat transfer from the outer surface at r_2 to the coolant is governed by the thermal resistance:
R"_total = 2*pi*r_2*R'_total
R"_total = 2*pi*0.011*0.0185 = 0.00128 m^2K/W
- For the graphite ( q = 0 ), the value of T_2 obtained from the foregoing solution is prescribed as an inner boundary condition at r_2, while a convection condition is prescribed at the outer surface (r_3).
- For 5*10^8 < q and q > 5*10^8, the distributions are given in attachment.
The graphs obtained:
- The comparatively large value of k_t yields small temperature variations across the fuel element, while the small value of k_g results in large temperature variations across the graphite.
Operation at q = 5*10^8 W/^3 is clearly unacceptable, since the melting points of thorium and graphite are exceeded and approached, respectively. To prevent softening of the materials, which would occur below their melting points, the reactor should not be operated much above:
q = 3*10^8 W/m^3
Answer: D.) 39,200 J
Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height. In this case m = 400 kg, g = 9.8, h = 10 m thus:
P.E.= 39,200 Joules
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When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.
The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.
Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.
If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,
Therefore,
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