The length of a vector is always nonnegative, but in this case, at least one of the components will be negative. As already suggested, draw a sketch of the vector.rashad764 said:Does the length of become F negative as well as the angle?
Don't you take the sin and cos of the angle, 70=x/cos57.1 then multiply 70(cos 57.1)fresh_42 said:Have you drawn an image? Or more straight forward: how big is the angle measured form the positive x-axis as it is usually done? Do you know the formula for the components, given the angle and the length?
Yes, but your angle is wrong. The angle for these formulas is measured from the x-axis, but you have a number measured from the y-axis.rashad764 said:Don't you take the sin and cos of the angle, 70=x/cos57.1 then multiply 70(cos 57.1)
How would I measure it from the x axisfresh_42 said:Yes, but your angle is wrong. The angle for these formulas is measured from the x-axis, but you have a number measured from the y-axis.
Do I draw the vector 57.1 degrees from x-axis then measure the differencefresh_42 said:Sketch it, then you will see. How big is the difference between the two measurements? Of course you could as well calculate with the given angle, but then you will have to adjust the formulas to the new situation. This leaves us with the question: How is ##x= 70N \cdot \cos(57.1°)## found?
If it were from the x-axis, then the angle would be 57.1°rashad764 said:57.1∘counterclockwise from positive y−axis
the angle from the x-axis is 147.1fresh_42 said:You draw the vector at 57.1 degrees from the y-axis (counterclockwise, as given) and next measure the angle from the x-axis and put this new angle into the formulas. Or you draw a triangle with the given data and calculate the side lengths of this triangle plus adjust the signs according to the drawing.
Right. And this is the angle your formulas are made for. Otherwise you would have had to use the triangle in the second quadrant (with different formulas) and use a positive y value and a negative x value.rashad764 said:the angle from the x-axis is 147.1
To calculate the magnitude of a vector, you can use the Pythagorean theorem, which states that the magnitude (or length) of a vector is equal to the square root of the sum of the squares of its components. In other words, if a vector has components x and y, its magnitude is equal to sqrt(x^2 + y^2).
The magnitude of a vector refers to its length or size, while the direction of a vector refers to the angle that the vector makes with a reference axis. Both magnitude and direction are needed to fully describe a vector in two-dimensional space.
To find the components of a vector, you can use trigonometric functions. If the magnitude of the vector is given as M and the direction as θ, the x-component can be calculated as M*cos(θ) and the y-component as M*sin(θ).
No, a vector cannot have a negative magnitude. Magnitude is always a positive value that represents the length or size of the vector.
A vector can be represented using its components in the form of an ordered pair (x, y), where x is the x-component and y is the y-component. This ordered pair is often referred to as the vector's position vector.