Question

Dee is on a swing in the playground. The chains are 2.5 m long, and the tension in each chain is 450 N when Dee is 55 cm above the lowest point of her swing. Tension is a vector directed along the chain, measured in newtons, abbreviated N. What are the horizontal and vertical components of the tension at this point in the swing?

Short answer

The vertical component of the tension is 351 N, and the horizontal component of the tension is 4.5 N.

Step-by-step solution

Write the given information

Length of the chain, L = 2.5 m
The tension in each chain, T=450 N
The height of swing from lowest point, d = 55cm = 0.55 m

To determine the horizontal and vertical components of the tension in the chain 

Firstly, determine the angle between the vertical and the chain.
From the figure, the angle θ is expressed as

$$\begin{gathered} \cos \theta =\left(\frac{L-d}{L}\right) \\ \theta ={\cos }^{-1}\left(\frac{L-d}{L}\right) \\ \theta ={\cos }^{-1}\left(\frac{2.5-0.55}{2.5}\right) \\ \theta ={\cos }^{-1}\left(\frac{2.5-0.55}{2.5}\right) \\ \theta =38.7° \end{gathered}$$

Now, the vertical component of the tension is given by
T_y= T cos θ
= 450 (0.78)
= 351N
Thus, the vertical component of the tension is 351 N

Similarly, the horizontal component of the tension in the chain is given by
T_x= T sin θ
= 450 (0.01)
= 4.5 N
Thus, the horizontal component of the tension is 4.5 N