Question

QUESTION 5(two part question)Calculate FA and FB. right end, left end, for the beam shown in Fig. 9-56. The downward forces represent the weights of machineryon the beam. Assume the beam is uniform and has a mass of 280 kg. (a) FA. No units on answer.MQUESTION 6Calculate FA and FB, right end, left end, for the beam shown in Fig. 9-56. The downward forces represent the weights of machinery on the beam.Assume the beam is uniform and has a mass of 280 kg. (b) FB. No units on answer.

Answer 1

Take into account that the sum of the torques must be equal to zero:

$\Sigma\tau=0$

Based on the given image, and by applying the sum of torques around the point of force FA, you can obtain FB, as follow:

$\begin{gathered} \Sigma\tau_A=-4300N\cdot2.0m-3100N\cdot6.0m-2200N\cdot9.0m+F_B\cdot10.0m=0 \\ F_B\cdot10.0m=8600Nm+18600Nm+19800Nm \\ F_B\cdot10.0m=47000Nm \\ F_B=\frac{47000Nm}{10.0m} \\ F_B=4700N \end{gathered}$

where you have taken into account that clockwise torques are negative, and counterclockwise torques are positive.

Then, FB = 4700N

Next, by using the Newton's second law, consider that the sum of all forces must be equal to zero. In this way you obtain FA, as follow:

$\begin{gathered} \Sigma F=F_A-4300N-3100N-2200N+F_B-M\cdot g=0_{} \\ F_A-9600N+F_B-(280\operatorname{kg})(\frac{9.8m}{s^2})=0 \\ F_A=12344N-F_B \\ F_A=12344N-4700N \\ F_A=7644N \end{gathered}$

where you have used the weight of the beam by inlcuding the expression Mg, with M the mass of the beam and g the acceleration gravitational constant.

Then, FA = 7644N