Add together these vectors: V1= 60 Newtons at 45 degrees N of W; V2= 5Newtons at 30 degrees S of E
Is it [email protected] S of E??
Yes
Hannah and Tammy please show how you arrive at your answer.
To add two vectors, you need to first resolve them into their horizontal and vertical components, and then add the components individually.
Let's start with V1.
V1 has a magnitude of 60 Newtons and is at an angle of 45 degrees North of West.
To resolve it into horizontal and vertical components, we use trigonometry.
The horizontal component (V1x) can be found using cosine:
V1x = V1 * cos(angle)
= 60 * cos(45)
≈ 42.42 N
The vertical component (V1y) can be found using sine:
V1y = V1 * sin(angle)
= 60 * sin(45)
≈ 42.42 N
Now let's move on to V2.
V2 has a magnitude of 5 Newtons and is at an angle of 30 degrees South of East.
Again, we'll resolve it into horizontal and vertical components.
The horizontal component (V2x) can be found using cosine:
V2x = V2 * cos(angle)
= 5 * cos(30)
≈ 4.33 N
The vertical component (V2y) can be found using sine:
V2y = V2 * sin(angle)
= 5 * sin(30)
≈ 2.5 N
Now that we have the horizontal and vertical components for both vectors, we can add them individually to get the resultant vector.
The horizontal component of the resultant vector (Rx) is the sum of the horizontal components of V1 and V2:
Rx = V1x + V2x
= 42.42 + 4.33
= 46.75 N
The vertical component of the resultant vector (Ry) is the sum of the vertical components of V1 and V2:
Ry = V1y + V2y
= 42.42 + 2.5
= 44.92 N
To find the magnitude (R) and direction (angle) of the resultant vector, we can use the Pythagorean theorem and inverse trigonometric functions.
The magnitude of the resultant vector (R) is given by:
R = sqrt(Rx^2 + Ry^2)
= sqrt((46.75)^2 + (44.92)^2)
≈ 64.08 N
The direction (angle) of the resultant vector can be found using the inverse tangent function:
angle = atan(Ry/Rx)
= atan(44.92/46.75)
≈ 45.41 degrees
Therefore, the resultant vector is approximately 64.08 Newtons at 45.41 degrees North of West.
Based on the calculations, it seems like the answer you provided, 62.25 N at 40.50 degrees South of East, is not correct.