Problem

The arched pipe has a mass of 80 kg and rests on the surface of the platform for which the...

The arched pipe has a mass of 80 kg and rests on the surface of the platform for which the coefficient of static friction is μs = 0.3. Determine the greatest angular acceleration α of the platform, starting from rest when θ = 45°, without causing the pipe to slip on the platform.

Step-by-Step Solution

Solution 1

Schematic diagram of the pipe is shown,

Picture 1

Free body diagram of the pipe is as shown,

Picture 4

Consider the free body diagram of the arched pipe, write the equation of equilibrium in horizontal direction:

\(\mu_{s} N_{A}+\mu_{s} N_{B}=m R \alpha \sin \theta\)

Here, \(N_{A}\) and \(N_{B}\) are the normal reactions at \(A\) and \(B\) respectively, \(\mu_{s}\) is the coefficient of static friction, \(m\) mass of the arched beam, a is angular acceleration and \(\theta\) is angle made by the arm with horizontal axis.

Substitute, \(0.3\) for \(\mu_{s}, 80 \mathrm{~kg}\) for \(m, 1\) for \(R\) and \(45^{\circ}\) for \(\theta\).

\(0.3 N_{A}+0.3 N_{B}=80 \times 1(\alpha) \sin 45^{\circ}\)

\(0.3 N_{A}+0.3 N_{B}-56.56(\alpha)=0\)

Divide both sides with \(0.3 .\)

\(N_{A}+N_{B}-188.54(\alpha)=0 \ldots \ldots(1)\)

Write the equation of equilibrium in vertical-direction

\(N_{A}+N_{B}-m g=m R \alpha \cos \theta\)

Substitute, \(80 \mathrm{~kg}\) for \(m, 9.81 \mathrm{~m} / \mathrm{s}^{2}\) for \(g, 1\) for \(\mathrm{R}\) and \(45^{\circ}\) for \(\theta\) in equation (2)

\(N_{A}+N_{B}-80 \times 1 \times 9.81=80 \times 1(\alpha) \cos 45^{\circ}\)

\(N_{A}+N_{B}-56.56(\alpha)=784.8 \ldots \ldots .(2)\)

Subtract equation (2) from equation (1).

\(\begin{aligned}-188.54(\alpha)-(-56.56(\alpha)) &=-784.8 \\-188.54(\alpha)+56.56(\alpha) &=-784.8 \\ \alpha &=5.946 \mathrm{~m} / \mathrm{s}^{2} \end{aligned}\)

Therefore, the angular acceleration is, \(5.946 \frac{\mathrm{rad}}{\mathrm{s}^{2}}\)

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