The horizontal force is necessary to hold the bag in the new position is 614 N
The given parameters are:
Start by calculating the angle between the rope, and the bag using:
[tex]sin(\theta) = \frac{d}{h}[/tex]
So, we have:
[tex]sin(\theta) = \frac{2.0m}{3.5m}[/tex]
[tex]sin(\theta) = 0.5714[/tex]
Take arc sin of both sides
[tex]\theta = sin^{-1}(0.5714)[/tex]
[tex]\theta = 34.8^o[/tex]
Start by calculating the tension on the rope using:
[tex]T = \frac{W}{cos(\theta)}[/tex]
Where W = mg.
So, we have:
[tex]T = \frac{mg}{cos(\theta)}[/tex]
This gives
[tex]T = \frac{90.0 \times 9.8}{cos(34.8)}[/tex]
[tex]T = \frac{882}{cos(34.8)}[/tex]
[tex]T = 1074.2[/tex]
The horizontal force is then calculated using:
[tex]F = Tsin(\theta)[/tex]
This gives
[tex]F = 1074.2 \times sin(34.8)[/tex]
This gives
[tex]F = 1074.2 \times \frac{2.0}{3.5}[/tex]
[tex]F = 613.8[/tex]
Approximate
[tex]F = 614[/tex]
Hence, the horizontal force is necessary to hold the bag in the new position is 614 N
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