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 130P from Chapter 8 from Hibbeler's Engineering Mechanics.

Problem 130P

Chapter:
Problem:
The handcart has wheels with a diameter of 6 in. If a crate...

Step-by-Step Solution

Step 1

We are given the diameter of the handcart wheel is $d = 6\;{\rm{in}}$, the weight of the crate is $W = 1500\;{\rm{lb}}$, and the coefficient of the rolling resistance is $a = 0.04\;{\rm{in}}$.


We are asked the force P that must be applied to the handle to overcome the rolling resistance.


 
Step 2

The free body diagram of the system is:

Images


We have the negligible weight of the cart, therefore the weight of the overall system is $W = 1500\;{\rm{lb}}$.

We have the normal reaction acting over the system is N.

We have the angle of force applied to the handle from horizontal is $\theta $.


 
Step 3

The forces acting over the system in vertical direction by equilibrium condition is,

\begin{array}{c} \Sigma {F_y} = 0\\ W + P\sin \theta - N = 0\\ N = W + P\sin \theta \end{array}
 
Step 4

The force P that must be applied to the handle to overcome the rolling resistance is,

\begin{array}{l} P\cos \theta = \frac{{Na}}{R}\\ P\cos \theta = \frac{{\left( {W + P\sin \theta } \right)a}}{{\left( {\frac{D}{2}} \right)}} \end{array}

Here, R is the radius of the handcart wheel.


 
Step 5

Substitute the values in the above expression.

\begin{array}{c} P\left( {\frac{4}{5}} \right) = \frac{{\left[ {1500\;{\rm{lb}} + P\left( {\frac{3}{5}} \right)} \right]\left( {0.04\;{\rm{in}}} \right)}}{{\left( {\frac{{6\;{\rm{in}}}}{2}} \right)}}\\ 2.4P = 60\;{\rm{lb}} + {\rm{0}}{\rm{.024}}P\\ P = 25.3\;{\rm{lb}} \end{array}