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 6FP from Chapter 5 from Hibbeler's Engineering Mechanics.

Problem 6FP

Chapter:
Problem:
Determine the reactions at the smooth contact points A, B and C...

Step-by-Step Solution

Step 1

We are given that the force applied by the person is $F = 250\;{\rm{N}}$.


We are asked to calculate the reaction force on points A, B and C.

Images


We have the distance between points A and B is ${d_1} = 0.15\;{\rm{m}}$.


We have the distance between point B and C is ${d_2} = 0.2\;{\rm{m}}$.


We have the distance between point C and arm is ${d_3} = 0.4\;{\rm{m}}$.


 
Step 2

The free body diagram of the system is shown as,

Images


 
Step 3

Applying the force equilibrium equation along x-axis:

\[\begin{array}{c} \sum {F_x} = 0\\ {F_C}\cos 60^\circ - F\cos 30^\circ = 0 \end{array}\]
 
Step 4

Substitute the known values in the equation:

\[\begin{array}{c} {F_C}\left( {0.5} \right) - \left( {250\;{\rm{N}}} \right)\left( {0.866} \right) = 0\\ {F_C} = \frac{{216.5\;{\rm{N}}}}{{0.5}}\\ = 433\;{\rm{N}} \end{array}\]
 
Step 5

Applying the moment of forces at point B:

\[ - {F_A}\sin 30^\circ {d_1} - {F_C}{d_2} + F\cos 30^\circ \left( {{d_2} + {d_3}} \right) = 0\]
 
Step 6

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} - {F_A}\left( {0.5} \right)\left( {0.15\;{\rm{m}}} \right) - \\ \left( {433\;{\rm{N}}} \right)\left( {0.2\;{\rm{m}}} \right) + \\ \left( {250\;{\rm{N}}} \right)\left( {0.866} \right)\left( {0.2\;{\rm{m}} + 0.4\;{\rm{m}}} \right) \end{array} \right\} = 0\\ {F_A}\left( {0.075\;{\rm{m}}} \right) = 129.9\;{\rm{N}} \cdot {\rm{m}} - 86.6\;{\rm{N}} \cdot {\rm{m}}\\ {F_A} = \frac{{\left( {43.3\;{\rm{N}} \cdot {\rm{m}}} \right)}}{{\left( {0.075\;{\rm{m}}} \right)}}\\ = 577.33\;{\rm{N}} \end{array}\]
 
Step 7

Applying the force equilibrium equation along y-axis:

\[ - F\sin 30^\circ + {F_C}\sin 60^\circ + {F_B} - {F_A} = 0\]
 
Step 8

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} - \left( {250\;{\rm{N}}} \right)\left( {0.5} \right) + \\ \left( {433\;{\rm{N}}} \right)\left( {0.866} \right) + \\ {F_B} - 577.33\;{\rm{N}} \end{array} \right\} = 0\\ {F_B} = 125\;{\rm{N}} - 374.98\;{\rm{N}} + 577.33\;{\rm{N}}\\ = 327.35\;{\rm{N}} \end{array}\]