Answer: 215.15 N
Explanation:
If we draw a free body diagram of the mass we will have the following:
[tex]\sum{F_{x}}=-Tcos\theta + F=0[/tex] (1)
[tex]\sum{F_{y}}=Tsin\theta - mg=0[/tex] (2)
Where [tex]T[/tex] is the tension force of the rope, [tex]m=28 kg[/tex] the mass, [tex]g=9.8 m/s^{2}[/tex] the acceleration due gravity and [tex]mg[/tex] is the weight.
On the other hand, we can calculate [tex]\theta[/tex] as follows:
[tex]cos\theta=\frac{s}{l}[/tex]
[tex]\theta=cos^{-1}(\frac{s}{l})[/tex]
Where [tex]s=11.1 m[/tex] and [tex]l=18 m[/tex]
[tex]\theta=cos^{-1}(\frac{11.1 m}{18 m})[/tex]
[tex]\theta=51.9\°[/tex] (3)
Now, we firstly need to find [tex]T[/tex] from (2):
[tex]T=\frac{mg}{sin\theta}[/tex] (4)
[tex]T=\frac{(28 kg)(9.8 m/s^{2})}{sin(51.9\°)}[/tex]
[tex]T=348.69 N[/tex] (5)
Substituting (5) in (1):
[tex]F=Tcos\theta[/tex] (6)
[tex]F=348.69 N cos(51.9\°)[/tex]
Finally:
[tex]F=215.15 N[/tex]
