Question

Please help thank you

Answer 1

Answer:

$\theta \approx 59.036^{\circ}$, $T_{2} \approx 23.324\,N$

Explanation:

First we build the Free Body Diagram (please see first image for further details) associated with the mass, we notice that system consist of a three forces that form a right triangle (please see second image for further details): (i) The weight of the mass, (ii) two tensions.

The requested tension and angle can be found by the following trigonometrical and geometrical expressions:

$\theta = \tan^{-1} \frac{W}{T_{2}}$ (1)

$T_{1} = \sqrt{W^{2}+T_{2}^{2}}$ (2)

Where:

$W$ - Weight of the mass, measured in newtons.

$T_{1}$, $T_{2}$ - Tensions from the mass, measured in newtons.

If we know that $W = 20\,N$ and $T_{2} = 12\,N$, then the requested values are, respectively:

$\theta = \tan^{-1} \frac{20\,N}{12\,N}$

$\theta \approx 59.036^{\circ}$

$T_{2} = \sqrt{(20\,N)^{2}+(12\,N)^{2}}$

$T_{2} \approx 23.324\,N$