Answer to Question #180169 in Mechanics | Relativity for Menny

Question

A uniform beam 6.0m and weighing 40N at the center,rests on two supports, P and Q placed 1.0m from each end of the beam. Weights of 100N and 80N are hung from the ends of the beam near P and Q respectively. Calculate the reactions at the supports P and Q.

Answer 1

$\text{Let A and B these are the ends of the beam and }$

$\text{point C is the middle of the beam, then}$

$AP = 1;PC=2;PQ=4;PB = 5;$

$\text{By the condition of the problem:}$

$\vec F_a= -100;\vec F_c=-40;\vec F_b=-80;$

$\text{let } V_p \text{ and }V_q\text{ support reactions}$

$\Sigma M_p=0;\Sigma M_q=0;$

$\Sigma M_p = - F_a * AP + F_c * PC + F_b* PB - V_q * PQ ;$

$\Sigma M_p= -100+40*2+80*5-4*V_q$ ;

$V_q =95 N$

$\Sigma M_q=-F_a*AQ-F_c*PC+F_B*PB+V_p*PQ;$

$V_p=\frac{ ( - 100 * 5 - 40 * 2 + 80 * 1)} { - 4} = 125N$

Answer:$V_p=125 N;V_q = 95N$