for the beam and loading shown, determine (a) the reaction at a, (b) the tension in cable bc. answer (a) 245…

Explanation:

Step1: Take moments about point A

Let the tension in cable BC be $T$. The sum of moments about point A is zero. The clock - wise moments are due to the applied loads and the anti - clockwise moment is due to the tension in the cable. The vertical distance from A to the line of action of the tension force is the horizontal distance from A to B. $\sum M_A=0$. The applied loads and their distances from A are:

• $15$ lb at a distance of $(6 + 8+8 + 6)=28$ in.
• $20$ lb at a distance of $(8 + 8+6)=22$ in.
• $35$ lb at a distance of $(8 + 6)=14$ in.
• $20$ lb at a distance of $6$ in.
• The tension $T$ in cable BC acts at a horizontal distance of $28$ in from A. $15\times28+20\times22 + 35\times14+20\times6=T\times28$

Step2: Calculate the left - hand side of the moment equation

$15\times28 = 420$, $20\times22=440$, $35\times14 = 490$, $20\times6=120$. The sum of the left - hand side is $420+440 + 490+120=1470$. So, $1470=T\times28$.

Step3: Solve for the tension T

$T=\frac{1470}{28}=140$ lb.

Step4: Calculate the vertical reaction at A

For vertical equilibrium, $\sum F_y = 0$. Let the vertical reaction at A be $R_A$. The vertical forces acting on the beam are the applied vertical loads and the vertical component of the tension in the cable (in this case, since the cable is horizontal, the vertical component of the tension is 0). $R_A=15 + 20+35 + 20+15$ $R_A = 245$ lb.

Answer:

(a) $245$ lb (b) $140$ lb