Chapter 7, Problem 7.154RP

Knowing that the turnbuckle has been tightened until the tension in wire AD is 850 N, determine the internal forces at the point indicated: 7.154 Point J.

To determine

The internal forces.

Answer to Problem 7.154RP

The internal forces are $V=400\text{N}$, $F=750\text{N}$ and the moment of force is $130\text{N}\cdot \text{m}$.

Explanation of Solution

The free body diagram is given in figure 1.

From the diagram, net force along horizontal and vertical directions is zero. Write the equilibrium condition for the forces.

$V=\left(\frac{CD-AB}{\sqrt{{\left(CD-AB\right)}^2+B{C}^2}}\right)T_{AD}$

$F=\left(\frac{BC}{\sqrt{{\left(CD-AB\right)}^2+B{C}^2}}\right)T_{AD}$

Here, $T_{AD}$ is the tension in wire AD, V is the internal force along horizontal direction and F is the force along vertical direction.

The net moment at point J is zero. Therefore, the equilibrium condition is,

$M=V\left(BJ\right)+F\left(AB\right)$

Here, M is the moment of internal force.

Conclusion:

Substitute 280 mm for CD, 120 mm for AB, 300 mm for BC and 850 N for $T_{AD}$ in the expression for V.

$\begin{array}{c}V=\left(\frac{280\text{mm}-120\text{mm}}{\sqrt{{\left(280\text{mm}-120\text{mm}\right)}^2+{\left(300\text{mm}\right)}^2}}\right)\left(850\text{N}\right)\\ =400\text{N}\end{array}$

Substitute 280 mm for CD, 120 mm for AB, 300 mm for BC and 850 N for $T_{AD}$ in the expression for F.

$\begin{array}{c}F=\left(\frac{300\text{mm}}{\sqrt{{\left(280\text{mm}-120\text{mm}\right)}^2+{\left(300\text{mm}\right)}^2}}\right)\left(850\text{N}\right)\\ =750\text{N}\end{array}$

Substitute 400 N for V, 750 N for F, 100 mm for BJ and 120 mm for AB in the expression for M.

$\begin{array}{c}M=\left(\text{400}\text{N}\right)\left(100\text{mm}\right)+\left(750\text{N}\right)\left(120\text{mm}\right)\\ =130000\text{N}\cdot \text{mm}\\ =130\text{N}\cdot \text{m}\end{array}$

Therefore, the internal forces are $V=400\text{N}$, $F=750\text{N}$ and the moment of force is $130\text{N}\cdot \text{m}$.