A crawler crane has two tracks each sits on a patch of ground 15 feet long by 2 feet wide what is the area of ground that is in
1 answer:
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Answer:
the rate of heat loss is 2.037152 W
Explanation:
Given data
stainless steel K = 16 W
diameter (d1) = 10 cm
so radius (r1) = 10 /2 = 5 cm = 5 ×
radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 ×
temperature = 25°C
surface heat transfer coefficient = 6 6 W
outside air temperature = 15°C
To find out
the rate of heat loss
Solution
we know current is pass in series from temperature = 25°C to 15°C
first pass through through resistance R1 i.e.
R1 = ( r2 - r1 ) / 4 × r1 × r2 × K
R1 = ( 5.2 - 5 ) / 4
× 5 × 5.2 × 16 ×
R1 = 3.825 ×
same like we calculate for resistance R2 we know i.e.
R2 = 1 / ( h × area )
here area = 4 r2²
area = 4 (5.2 ×
)² = 0.033979
so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )
R2 = 4.90499
now we calculate the heat flex rate by the initial and final temp and R1 and R2
i.e.
heat loss = T1 -T2 / R1 + R2
heat loss = 25 -15 / 3.825 × + 4.90499
heat loss = 2.037152 W
Answer:
N_c = 3.03 N
F = 1.81 N
Explanation:
Given:
- The attachment missing from the question is given:
- The given expressions for the radial and θ direction of motion:
r = 0.5*θ
θ = 0.5*t^2 ...... (correction for the question)
- Mass of peg m = 0.5 kg
Find:
a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.
b) Determine the magnitude of the normal force of the slot on the peg.
Solution:
- Determine the expressions for radial kinematics:
dr/dt = 0.5*dθ/dt
d^2r/dt^2 = 0.5*d^2θ/dt^2
- Similarly the expressions for θ direction kinematics:
dθ/dt = t
d^2θ/dt^2 = 1
- Evaluate each at time t = 2 s.
θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°
r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2
- Evaluate the angle ψ between radial and horizontal direction:
tan Ψ = r / (dr/dθ) = 1 / 0.5
Ψ = 63.43°
- Develop a free body diagram (attached) and the compute the radial and θ acceleration:
a_r = d^2r / dt^2 - r * dθ/dt
a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2
a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)
a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2
- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:
Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r
θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ
Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:
N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5
N_c = 3.03 N
F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5
F = 1.81 N
