# Resultant Vector using Components

• kukumaluboy
In summary, a man pushing a toy car caused it to undergo 2 displacements, with the first having a magnitude of 150 cm and making an angle of 120 degrees with the positive x axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35 degrees at the positive x axis. The second displacement was found to have a magnitude of 196 cm and a direction of -14.65 degrees.
kukumaluboy

## Homework Statement

A man pushing a toy car causes it to undergo 2 displacements. The first has a magnitude of 150 cm and makes an angle of 120degrees with the positive x axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35 degrees at the positive x axis. Find the magnitude and direction of the second displacement.

## The Attempt at a Solution

Displacement A + Displacement B = Displacement R
B = R - A

R = Rx + Ry
R = 140cos35i + 140sin35j

180° - 120° = 60
A = Ay + Ax
A = 150sin60j - 150cos60i (2nd Quad?? Not too sure if need to put negative here)

Hence:
B = Rx + Ry - (Ay + Ax)
B = 140cos35i + 140sin35j - (150sin60j - 150cos60i)
B = (140cos35 +150cos60)i + (140sin35 - 150sin60)j

B = sqrt( Bi^2 + Bj^2 )
B = 196

To find Direction
tetha = taninverse( (140sin35 - 150sin60)/ (140cos35 +150cos60) )
tetha = -14.65 degrees ?? weird??

kukumaluboy said:

## Homework Statement

A man pushing a toy car causes it to undergo 2 displacements. The first has a magnitude of 150 cm and makes an angle of 120degrees with the positive x axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35 degrees at the positive x axis. Find the magnitude and direction of the second displacement.

## The Attempt at a Solution

Displacement A + Displacement B = Displacement R
B = R - A

R = Rx + Ry
R = 140cos35i + 140sin35j

180° - 120° = 60
A = Ay + Ax
A = 150sin60j - 150cos60i (2nd Quad?? Not too sure if need to put negative here)
Yes. cos(120)= -cos(60). If you are not certain of the sign leave it as cos(120) and sin(120).

Hence:
B = Rx + Ry - (Ay + Ax)
B = 140cos35i + 140sin35j - (150sin60j - 150cos60i)
B = (140cos35 +150cos60)i + (140sin35 - 150sin60)j

B = sqrt( Bi^2 + Bj^2 )
B = 196
Yes, that is correct.

To find Direction
tetha = taninverse( (140sin35 - 150sin60)/ (140cos35 +150cos60) )
tetha = -14.65 degrees ?? weird??
What is weird about it? Did you draw a picture?

1 person
Yes i just drew it. So -14.65 means theta is 360 - 14.65? I suck at math

## 1. What is a resultant vector using components?

A resultant vector using components is a way to represent the total or net effect of multiple individual vectors. It involves breaking down each vector into its horizontal and vertical components and then adding or subtracting those components to calculate the resultant vector.

## 2. How do you find the magnitude of a resultant vector using components?

The magnitude of a resultant vector using components can be found using the Pythagorean theorem. You square the horizontal and vertical components of the individual vectors, add them together, and then take the square root of the sum to find the magnitude of the resultant vector.

## 3. Can the direction of a resultant vector using components be negative?

Yes, the direction of a resultant vector using components can be negative. This indicates that the resultant vector is in the opposite direction of the positive direction of the axis used to represent it.

## 4. How is a resultant vector using components different from a resultant vector using the polygon method?

A resultant vector using components is calculated by breaking down vectors into their components and using mathematical operations to determine the resultant vector. On the other hand, a resultant vector using the polygon method involves drawing vectors to scale and using geometric methods to determine the resultant vector.

## 5. Can a resultant vector using components be used in three-dimensional space?

Yes, a resultant vector using components can be used in three-dimensional space. In this case, each vector would have three components (horizontal, vertical, and depth), and the resultant vector would have three components as well. The magnitude and direction of the resultant vector would be calculated using the same methods as in two-dimensional space.

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