document.write( "Question 1182845: A uniform beam 3m long weighs 100N. Loads of 50 N and 150 N are placed on the
\n" ); document.write( "beam at points which are ½ m and 2½ m, respectively, from its left end. If the beam is kept in a horizontal position by supports at its two ends, find these reactions. \n" ); document.write( "

Algebra.Com's Answer #812972 by ikleyn(52781) $\"\"$ $\"About$

You can put this solution on YOUR website!
.
\n" ); document.write( "A uniform beam 3m long weights 100 N. Loads of 50 N and 150 N are placed on the
\n" ); document.write( "beam at points which are ½ m and 2½ m, respectively, from its left end.
\n" ); document.write( "If the beam is kept in a horizontal position by supports at its two ends, find these reactions.
\n" ); document.write( "~~~~~~~~~~~~~~\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "

\r\n" );
document.write( "Let x be the reaction force at the left end support, and\r\n" );
document.write( "\r\n" );
document.write( "Let y be the reaction force at the right end support.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Then we have two equations.\r\n" );
document.write( "\r\n" );
document.write( "One equation says that the total weight is at equilibrium\r\n" );
document.write( "\r\n" );
document.write( "    x + y = 100 + 50 + 150  newtons,  \r\n" );
document.write( "\r\n" );
document.write( "or\r\n" );
document.write( "\r\n" );
document.write( "    x + y = 300.            (1)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The second equation says that the total rotational moments about the central point of the beam is zero \r\n" );
document.write( "(no rotation).  The distances from the central points to the loads BOTH are 1.00 m.\r\n" );
document.write( "The weight of the beam is distributed uniformly and therefore creates ZERO rotational moment, so we can forget about it. \r\n" );
document.write( "THEREFORE, the equation of rotational moment takes the form\r\n" );
document.write( "\r\n" );
document.write( "     -x*1.5 + 50*1 + y*1.5 - 150*1 = 0,     ( ! take into account the signs of the moments ! )\r\n" );
document.write( "\r\n" );
document.write( "or\r\n" );
document.write( "\r\n" );
document.write( "     -1.5x + 1.5y = 150*1 - 50*1 \r\n" );
document.write( "\r\n" );
document.write( "     -1.5x + 1.5y = 100\r\n" );
document.write( "\r\n" );
document.write( "        -x + y    =  66 .     (2)\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Equations (1) and (2) are easy to solve.  They imply  y = 183.33 N,  x = 116.67 N,  which is your answer.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "ANSWER.  The reaction forces are  116.67 N (left end at x= 0)  and  183.33 N  (right end at x= 3 m).\r\n" );
document.write( "

\r
\n" ); document.write( "\n" ); document.write( "Solved.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );