Problem 002-mm | Method of Members | Engineering Mechanics Review at MATHalino

Problem 002-mm
Members AB and BC shown in Fig. F-002(MM) are pinned together at point B, and are pinned to the floor at points A and C. The structure supports a pulley at point B with which, a person is hoisting a 2.0 kN load. Member BC has a weight of 1.6 kN, which may be considered to act at its center, while AB is made of strong-light material and has negligible weight. Determine the value of the external support reactions at A and C.

Solution 002-mm

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AB is an axial member while BC is a three-force member. Thus,

$A_x = R_A \cos 37^\circ$

$A_y = R_A \sin 37^\circ$

$\Sigma M_C = 0$

$A_x(1) + A_y(8) = 4.0(4) + 1.6(2)$

$R_A \cos 37^\circ + (R_A \sin 37^\circ)(8) = 19.2$

$5.6132R_A = 19.2$

$R_A = 3.42 \, \text{ kN}$ answer

$\Sigma F_H = 0$

$C_x = A_x$

$C_x = R_A \cos 37^\circ$

$C_x = 3.42 \cos 37^\circ$

$C_x = 2.73 \, \text{ kN}$ answer

$\Sigma F_V = 0$

$C_y + A_y = 4.0 + 1.6$

$C_y + R_A \sin 37^\circ = 5.6$

$C_y + 3.42 \sin 37^\circ = 5.6$

$C_y = 3.54 \, \text{ kN}$ answer

Problem 002-mm | Method of Members

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