Answer to Question #254376 in Mechanics | Relativity for Neil

Question

The turnbuckle is tightened until the tension in the cable AB equals 2.4 kN. Determine the vector expression for the tension T as a force acting on member AB. Also find the magnitude of the projection of T along the line AC.

Answer 1

To find the tension along any given axis, we need to find the unit vector along that given axis. It also helps to find the magnitude of the tension generated in the other given cable or wire attached to the system.

the position vector AB:

$\vec{r}_{AB}=2i+j-5k$ m

$r_{AB}=\sqrt{4+1+25}=\sqrt{30}$ m

unit vector of AB:

$\vec{u}=\frac{\vec{r}_{AB}}{r_{AB}}=\frac{2i+j-5k}{\sqrt{30}}$

 vector expression for the tension:

$\vec{T}=T\vec{u}=2.4\cdot \frac{2i+j-5k}{\sqrt{30}}=0.9i+0.4j-2.2k$ kN

the position vector AC:

$\vec{r}_{AC}=2i-2j-5k$ m

$r_{AC}=\sqrt{4+4+25}=\sqrt{33}$ m

unit vector of AC:

$\vec{u}=\frac{\vec{r}_{AC}}{r_{AC}}=\frac{2i-2j-5k}{\sqrt{33}}$

 the projection of tension along with the line AC:

$T_{AC}=\vec{T}\cdot\vec{u}_{AC}=(2.4\cdot \frac{2i-2j-5k}{\sqrt{33}})\cdot(0.9i+0.4j-2.2k)=2.1$ kN