Question

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 is in a left turn of radius rho. Determine the maximum speed for which the pipe will be restrained. Use the values rho = 60 m, h = 0.1 m, and R = 0.8 m.

Answer 1

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 = $\frac{V^{2} }{rho}$

Therefore  $\frac{V^{2} }{rho}$ = 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 2

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:

$\alpha =cos^{-1}(\frac{R-h}{R})$

If R=0.8 m and h=0.1 m, we have:

$\alpha =cos^{-1}(\frac{0.8-0.1}{0.8})=29$

the equilibrium of forces acting in y-direction is zero and we have the following:

$Fy=0\\R_{B}^{2}cos\alpha -mg=0$

Clearing RB:

$R_{B}=\frac{mg}{cos\alpha }$

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:

$Fn=0\\R_{B}sin\alpha =m\frac{v^{2} }{p}$

where p is the radius and v is the speed. if RB=mg/cosα

$v^{2}=pgtan\alpha$

Replacing values:

$v^{2}=60*9.8*tan29\\v=18.1 m/s$