Answer to Question #250338 in Other for asasasa

Question

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?

Answer 1

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

$$R_y=\sum_iF_{yi}=0=>-\dfrac{5}{13}Q+\dfrac{3}{5}P-20kN=0$$

Solving these equations simultaneously gives

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

$$-\dfrac{3}{13}Q+\dfrac{1}{5}P=\dfrac{5}{13}Q-\dfrac{3}{5}P$$

$$Q=\dfrac{13}{10}P$$

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

$$P=200kN, Q=260kN$$

Given that $C^{R}=500kN\cdot m$ counterclockwise, and choosing point $A$ as the moment center, we have

$$C^{R}=\sum_iM_{Ai}$$

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

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

Substitute $P=200kN$

$$C=-500kN\cdot m-60kN\cdot m+320kN\cdot m$$

$$+1920kN\cdot m$$

$$C=1680kN\cdot m$$

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

$$P=200kN, Q=260kN$$

$$C=1680kN\cdot m$$