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 1RP from Chapter 17 from Hibbeler's Engineering Mechanics.

Problem 1RP

Chapter:
Problem:
The handcart has a mass of 200 kg and center of mass at G...

Step-by-Step Solution

Step 1

We are given the mass of handcart $m = 200\;{\rm{kg}}$, force $P = 50\;{\rm{N}}$ and the height at which force exerts is $h = 0.5\,{\rm{m}}$.


We are asked to determine the normal reactions at each of the wheels at A and B.


 
Step 2

The free body diagram of the picture is as follows:

Images


Here, $W$ is the weight of the cart, ${N_A}$ and ${N_B}$ are the normal force andg is the gravitational acceleration.


 
Step 3

To find the force in y-direction, we will use the relation,

\[\begin{array}{c} \Sigma {F_y} = 0\\ - W + 2\left( {{N_A} + {N_B}} \right) - P\sin 60^\circ = 0 \end{array}\]

On plugging the values in the above relation, we get,

\[\begin{array}{c} - \left( {200\;{\rm{kg}} \times 9.81\,{\rm{m/}}{{\rm{s}}^2}} \right) + 2\left( {{N_A} + {N_B}} \right) - \left( {50\;{\rm{N}}} \right)\sin 60^\circ = 0\\ 2\left( {{N_A} + {N_B}} \right) = \left( {1962 + 43.3} \right)\;{\rm{N}}\\ {N_A} = 1002.651 - {N_B} \end{array}\] … (1)
 
Step 4

To find the moment about G, we will use the relation,

\[\begin{array}{c} \Sigma {M_G} = 0\\ - 2{N_A}{d_1} + 2{N_B}{d_2} + P\cos 60^\circ {d_3} - P\sin 60^\circ {d_4} = 0 \end{array}\]

Here, ${d_1}$ is the distance from point A to G whose value is ${d_1} = 0.3\,{\rm{m}}$, ${d_2}$ is the distance from point B to G whose value is ${d_2} = 0.2\,{\rm{m}}$, ${d_3}$ is the distance from point at which vertical force is acting to G whose value is ${d_3} = \left( {0.5\;{\rm{m}} - 0.2\;{\rm{m}}} \right) = 0.3\;{\rm{m}}$ and ${d_4}$ is the distance from point at which horizontal force is acting to G whose value is ${d_4} = \left( {0.4\;{\rm{m}} + 0.2\;{\rm{m}}} \right) = 0.6\;{\rm{m}}$.


On plugging the values in the above relation, we get,

\[\begin{array}{c} - 2{N_A}\left( {0.3\;{\rm{m}}} \right) + 2{N_B}\left( {0.2\;{\rm{m}}} \right) + \left( {50\;{\rm{N}}} \right)\cos 60^\circ \left( {0.3\;{\rm{m}}} \right) - \left( {50\;{\rm{N}}} \right)\sin 60^\circ \left( {0.6\;{\rm{m}}} \right) = 0\\ - 0.6{N_A} + 0.4{N_B} + 7.5 - 25.98 = 0\\ - 0.6\left( {1002.651 - {N_B}} \right) + 0.4{N_B} = 18.48\\ {N_B} = 620.07\;{\rm{N}} \end{array}\]

On plugging the values in equation (1), we get:

\[\begin{array}{c} {N_A} = 1002.651 - \left( {620.07\;{\rm{N}}} \right)\\ {N_A} = 382.58\;{\rm{N}} \end{array}\]