Answer to Question #180169 in Mechanics | Relativity for Menny

Question #180169

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.


1
Expert's answer
2021-04-13T06:31:58-0400

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

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

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

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

Fa=100;Fc=40;Fb=80;\vec F_a= -100;\vec F_c=-40;\vec F_b=-80;

let Vp and Vq support reactions\text{let } V_p \text{ and }V_q\text{ support reactions}

ΣMp=0;ΣMq=0;\Sigma M_p=0;\Sigma M_q=0;

ΣMp=FaAP+FcPC+FbPBVqPQ;\Sigma M_p = - F_a * AP + F_c * PC + F_b* PB - V_q * PQ ;

ΣMp=100+402+8054Vq\Sigma M_p= -100+40*2+80*5-4*V_q ;

Vq=95NV_q =95 N

ΣMq=FaAQFcPC+FBPB+VpPQ;\Sigma M_q=-F_a*AQ-F_c*PC+F_B*PB+V_p*PQ;

Vp=(1005402+801)4=125NV_p=\frac{ ( - 100 * 5 - 40 * 2 + 80 * 1)} { - 4} = 125N

Answer:Vp=125N;Vq=95NV_p=125 N;V_q = 95N


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