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 90P from Chapter 8 from Hibbeler's Engineering Mechanics.

Problem 90P

Chapter:
Problem:
The smooth beam is being hoisted using a rope that is wrapped...

Step-by-Step Solution

Step 1

We are given the coefficient of static friction between the rope and the ring as ${\mu _s} = 0.3$.


We are asked to determine the smallest angle $\theta $ for the equilibrium.


 
Step 2

We will draw a free body diagram of the joint $A$.

Images


Here, ${T_1}$ is the tension in the rope.


 
Step 3

We will resolve the forces in vertical direction.

\[\begin{array}{c} \sum {{F_y}} = 0\\ {\bf{T}} - {T_1}\cos \left( {\frac{\theta }{2}} \right) - {T_1}\cos \left( {\frac{\theta }{2}} \right) = 0\\ {\bf{T}} = 2{T_1}\cos \left( {\frac{\theta }{2}} \right)\;\;\;...\left( 1 \right) \end{array}\]
 
Step 4

The formula of the tensions in the ropes is given below.

\[\begin{array}{l} {T_2} = {T_1}{e^{{\mu _s}\beta }}\\ {\bf{T}} = {T_1}{e^{{\mu _s}\beta }}\;\;\;\;\;\;...\left( 2 \right) \end{array}\]

Here, $\beta = \frac{\theta }{2}$ is the angle made by the rope.


 
Step 5

We will convert the angle from degree to radian.

\[\begin{array}{l} \beta = \frac{\theta }{2} \times \left( {\frac{\pi }{{180^\circ }}} \right)\\ \beta = \left( {\frac{{\pi \theta }}{{360^\circ }}} \right) \end{array}\]
 
Step 6

Substitute the given value in the equation (2) to find the smallest angle.

\[\begin{array}{c} 2{T_1}\cos \left( {\frac{\theta }{2}} \right) = {T_1}{e^{0.3\left( {\frac{{\pi \theta }}{{360^\circ }}} \right)}}\\ 2\cos \left( {\frac{\theta }{2}} \right) = {e^{0.3\left( {\frac{{\pi \theta }}{{360^\circ }}} \right)}}\\ 2\cos \left( {\frac{\theta }{2}} \right) = {e^{\left( {2.6 \times {{10}^{ - 3}}} \right)\theta }}\\ \theta \approx 99.2^\circ \end{array}\]

So, the smallest angle is $99.2^\circ $.