Question #250338

The force system shown consist of a Couple C and the 4 Forces. The resultant of this system is 500kNm counter-clockwise couple. a. What is the value of P? b. What is the value of Q? c. What is the value of C?


1
Expert's answer
2021-10-14T18:13:57-0400




Rx=iFxi=0=>1213Q+45P+80kN=0R_x=\sum_iF_{xi}=0=>-\dfrac{12}{13}Q+\dfrac{4}{5}P+80kN=0

Ry=iFyi=0=>513Q+35P20kN=0R_y=\sum_iF_{yi}=0=>-\dfrac{5}{13}Q+\dfrac{3}{5}P-20kN=0

Solving these equations simultaneously gives


1213Q+45P+80kN=0-\dfrac{12}{13}Q+\dfrac{4}{5}P+80kN=0

313Q+15P=513Q35P-\dfrac{3}{13}Q+\dfrac{1}{5}P=\dfrac{5}{13}Q-\dfrac{3}{5}P

Q=1310PQ=\dfrac{13}{10}P

65P+45P+80kN=0-\dfrac{6}{5}P+\dfrac{4}{5}P+80kN=0

P=200kN,Q=260kNP=200kN, Q=260kN

Given that CR=500kNmC^{R}=500kN\cdot m counterclockwise, and choosing point AA as the moment center, we have

CR=iMAiC^{R}=\sum_iM_{Ai}

=>500kNm=20kN(3m)C+80kN(4m)=>500kN\cdot m=-20kN(3m)-C+80kN(4m)

+35P(6m)+45P(6m)+\dfrac{3}{5}P(6m)+\dfrac{4}{5}P(6m)

Substitute P=200kNP=200kN


C=500kNm60kNm+320kNmC=-500kN\cdot m-60kN\cdot m+320kN\cdot m

+1920kNm+1920kN\cdot m

C=1680kNmC=1680kN\cdot m

Because the values for P,Q,P, Q, and CC are positive, each force acts in the direction shown in the figure.


P=200kN,Q=260kNP=200kN, Q=260kN

C=1680kNmC=1680kN\cdot m

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