Question

example3 a steel pipe weighting 10 kn is suspended from couple of steel - wire ropes (see figure). try to determine internal force of the steel - wire ropes.

Explanation:

Step1: Analyze horizontal - force equilibrium

In the horizontal direction, $\sum X = 0$. Since the two - rope system is symmetric, the horizontal components of the internal forces in the two ropes balance each other, so $N_1\sin30^{\circ}=N_2\sin30^{\circ}$, which gives $N_1 = N_2$.

Step2: Analyze vertical - force equilibrium

In the vertical direction, $\sum Y = 0$. The upward vertical components of the internal forces in the two ropes balance the weight of the steel pipe. The weight of the steel pipe is $G = 10kN$. The vertical component of the force in each rope is $N_i\cos30^{\circ}$ ($i = 1,2$). So $10-2N_1\cos30^{\circ}=0$.

Step3: Solve for $N_1$

We have the equation $10 - 2N_1\cos30^{\circ}=0$. Rearranging it to solve for $N_1$, we get $2N_1\cos30^{\circ}=10$, and $N_1=\frac{10}{2\cos30^{\circ}}=\frac{10}{2\times\frac{\sqrt{3}}{2}}=\frac{10}{\sqrt{3}}\approx5.78kN$. Since $N_1 = N_2$, $N_2 = 5.78kN$ as well.

Answer:

$N_1 = 5.78kN$, $N_2 = 5.78kN$