Question

A diagonal bar CD was welded to the AB beam supported at its ends. Two perpendicular inertias, each with the value P, are applied to the bar ends, providing a pair of forces. Moreover, a vertical force Q acts on the beam. Determine the reaction of supports A and B. Do not let go of the dead weight of the beam and the bar.

Given: P = 200N, Q = 100N, l = 2m, r = 1m, a = 0.5m

Correct answer: RBY = -75N, RA = 175N

a. RBY = -75N, RA = 175N
b. RBY = 75N, RA = -175N
c. RBY = -125N, RA = 125N
d. RBY = 125N, RA = -125N

Answer 1

The question involves calculating the reaction forces at the supports A and B of a beam in static equilibrium under the action of two perpendicular forces P and a vertical force Q, by considering the sum of forces and torques to be zero.

Step-by-step explanation:

The problem in question deals with static equilibrium where forces and torques are balanced. The pair of forces applied to the diagonal bar CD must create an equal and opposite reaction at the supports A and B to satisfy equilibrium conditions. Each inertia with value P applies a force, and there is an additional vertical force Q. To determine the reaction at the supports, one must consider the rotational equilibrium about one of the supports which involves calculating torques. Given the problem states the correct answer as RBY = -75N, RA = 175N, we can infer that the total downward force equals the total upward force, and the sum of the moments around the pivot also equals zero. The negative value of RBY suggests that it is a reaction in the opposite direction of assumed positive direction.