The flatbed truck carries a large section of circular pipe secured only by the two fixed blocks A and B of height h. The truck i
2 answers:
Answer:
The maximum speed for which the pipe will be restrained is 18.05 m/s.
Explanation:
We note that the
Securing blocks have height, h = 0.1 m
The radius of travel, rho = 60 m
The radius of the circular pipe, R = 0.8 m
We note that the weight of the pipe is acting at the centroid of the pipe
Therefore, for the pipe to slip, it has to climb the wedge h
Taking moments about h, we have
mg × R sin α = ma × R cos α
a/g = Tan α
But a =
Therefore = g tanα
Since height of the block, h = 0.1 m therefore,
R cos α = R - h
That is 0.8 cos α = 0.8 - 0.1 = 0.7
Therefore α = cos⁻¹ (0.7/0.8) = 28.96 °
From which V² = rho × g× tanα = 60 × 9.81 × tan 28.96
= 325.66 m²/s²
∴ V = √(325.66 m²/s²) = 18.05 m/s
Maximum speed = 18.05 m/s.
Answer:
The maximum speed is 18.1 m/s
Explanation:
the angle made by the line of action of reaction forces and normal reaction with the center is equal to:
If R=0.8 m and h=0.1 m, we have:
the equilibrium of forces acting in y-direction is zero and we have the following:
Clearing RB:
where RB is reaction force on the ring and m is the mass of circular ring
the equilibrium of forces acting in n-direction is equal to:
where p is the radius and v is the speed. if RB=mg/cosα
Replacing values:
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