The horizontal and vertical components of the tension at the given point in the swing are 281.6 N and 351 N respectively.
Given data:
The length of chain is, L = 2.5 m.
The magnitude of tension on each chain is, T = 450 N.
Distance above the lowest point is, d = 55 cm = 0.55 m.
In problem, first we need to obtain the angle of inclination made by string horizontally.
So, the angle inclined by the string with horizontal is given as,
[tex]cos \theta =\dfrac{L-d}{L}\\\\cos \theta =\dfrac{2.5-0.55}{2.5}\\\\\theta = cos^{-1}(\dfrac{1.95}{2.5})\\\\\theta=38.74^{\circ}[/tex]
Now, the horizontal component of tension force acting on the string is,
[tex]T_{H}=T \times cos \theta\\T_{H}=450 \times cos 38.74\\T_{H}=281.6 \;\rm N[/tex]
And, the vertical component of tension force acting on the string is,
[tex]T_{V}=T \times sin \theta\\T_{V}=450 \times sin 38.74\\T_{V}=351 \;\rm N[/tex]
Thus, the horizontal and vertical components of the tension at this point in the swing are 281.6 N and 351 N respectively.
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