Engineering Mechanics: Statics and Dynamics, 14th Edition

Engineering Mechanics: Statics and Dynamics, 14th Edition

Authors: Russell C. Hibbeler

ISBN-13: 978-0133915426

See our solution for Question 79P from Chapter 13 from Hibbeler's Engineering Mechanics.

Problem 79P

Chapter:
Problem:
The airplane, traveling at a constant speed of 50 m/s, is executing a...

Step-by-Step Solution

Step 1

We are given the constant speed of airplane as $v = 50\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$, the banking angle as $\theta = 15^\circ $ and the mass of a pilot as $m = 70\;{\rm{kg}}$.


We are asked to determine the radius of curvature $\rho $ of the turn and the normal force of the seat on the pilot.


 
Step 2

The free-body diagram of the airplane pilot can be drawn as:

Images


Here, N is the normal force.


 
Step 3

On balancing the forces in y-direction, we have:

\[\begin{array}{c} \sum {F_y} = m{a_y}\\ N\sin \theta - mg = m{a_y} \end{array}\]

As there is no acceleration in vertical direction, the value of ${a_y}$ will be zero. Hence, on substituting the values in the above expression, we get:

\[\begin{array}{c} N\sin 15^\circ - \left( {70\;{\rm{kg}}} \right) \times \left( {9.81\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}} \right) = m \times 0\\ N\sin 15^\circ = 686.7\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}\\ N = 2653.20\;{\rm{N}} \times \left( {\frac{{{{10}^{ - 3}}\;{\rm{kN}}}}{{1\;{\rm{N}}}}} \right)\\ N = 2.65\;{\rm{kN}} \end{array}\]
 
Step 4

On balancing the horizontal forces, we get:

\[N\cos \theta = \frac{{m{v^2}}}{\rho }\]

Substitute the values in the above expression, we get:

\[\begin{array}{c} \left( {2653.20\;{\rm{N}}} \right)\cos 15^\circ = \frac{{\left( {70\;{\rm{kg}}} \right) \times {{\left( {50\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}} \right)}^2}}}{\rho }\\ \left( {2562.8\;{\rm{N}}} \right) = \frac{{\left( {175000\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}} \right)}}{\rho }\\ \rho = 68.3\;{\rm{m}} \end{array}\]