Question

A string runs up and to the left in thex⁢yplane, making an angle of 39 degrees to the vertical. Determine the unit vector that points along the string.

Answer 1

To find the unit vector for the string making a 39-degree angle with the vertical, it involves using trigonometric functions to find the x and y components, resulting in the vector v = -cos(51°)i + sin(51°)j.

Step-by-step explanation:

To determine the unit vector that points along the string, which makes an angle of 39 degrees to the vertical in the xy-plane, we need to find its components along the x- and y-axes. The string makes an angle δ = 39 degrees with the vertical, therefore with the horizontal (x-axis) it makes an angle of 51 degrees (90 degrees - 39 degrees) because 90 degrees is the total angle between vertical and horizontal lines.

Given the angle with the horizontal, we find the components of the unit vector using trigonometric functions. The x-component (horizontal) will be cos(51°) and the y-component (vertical) will be sin(51°). Since the string points up and to the left, the x-component will be negative (-cos(51°)), and the y-component will be positive (sin(51°)).

The unit vector along the string is therefore v = -cos(51°)i + sin(51°)j, where i and j are the unit vectors in the direction of the x-axis and y-axis, respectively.

Answer 2

`\vec v=-0.63\hat i+0.78\hat j`

Step-by-step explanation:

Vectors

It's an element that has both magnitude and direction. It can be geometrically represented as a directed line segment whose length is the magnitude and the arrow points in its direction. There are several ways to mathematically represent a vector. The most commonly used are in the polar and rectangular coordinates. In the polar form, a vector has a magnitude and an angle measured respect to the horizontal right direction. It's written as . The rectangular representation has the two coordinates (x,y) measured in the known plane xy. The conversion between both systems is

`x=rcos\theta`

`y=rsin\theta`

A representation of the vector is shown in the figure below. The angle of
`39^o` is measured respect to the vertical, we must add
`90^o` to make it compliant with the standard expressions of vectors. We have r=1,
`\theta=39^o+90^o=129^o`

`x=(1)cos129^o=-0.63`

`y=(1)sin129^o=0.78`

The unit vector is

`\vec v=<-0.63,0.78>`

Or equivalently

`\boxed {\vec v=-0.63\hat i+0.78\hat j}`