Weight Challenge

Good day,

Today in Physics, we were asked to solve a challenge for a mystery weight suspended by two cords which provided the tension.

The diagram was similar to the one below:

This resulted in the use of our formula SohCahToa.

Sin= opposite/hypotenuse

Cos=adjacent/hypotenuse

Tan=opposite/ adjacent

This resulted in a diagram resembling the one below:

The corresponding free body diagram resembles the one below:

The calculations to accompany this diagram are:

Ft1y + Ft2y= Fg  

Sin( beta)= Opp/Hypotenuse

Sin38=Ft1y/ 0.9

0.9n*sin(38)=Ft1y

Ft1y= 0.5540953277934n

Sin( beta)= Opp/Hypotenuse

sin (70)= Ft2y/2.2n

2.2n*sin(70)=Ft2y

Ft2y= 2.0673237657292n

Due to Ft1y + Ft2y= Fg 

Then, 0.5540953277934n+2.0673237657292n=Fg

Thus, Fg=2.6214190935226n