The discussion revolves around calculating the vector components of a force vector given its magnitude and direction, specifically focusing on the implications of measuring the angle from the positive y-axis instead of the x-axis.
The discussion is active, with participants providing guidance on the importance of sketching the vector and measuring angles correctly. There is an acknowledgment of the need to adjust calculations based on the angle's reference point, but no consensus has been reached regarding the final approach.
Participants note that the length of the vector is always nonnegative, but at least one component may be negative depending on the angle used. The original poster's confusion about the angle's measurement from the y-axis versus the x-axis is a central point of discussion.
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