Solving vectors using component method

AI Thread Summary
The discussion focuses on solving vector addition using the component method. Participants calculate the x and y components for three vectors: A (175 km at 30 degrees north of east), B (153 km at 20 degrees west of north), and C (195 km west). Clarifications are made regarding the angles for each vector, emphasizing the importance of consistent angle measurement from the positive x-axis. Ultimately, the correct approach to determining the resultant vector's magnitude and direction is confirmed, resolving initial confusion about vector C's contribution. Understanding vector direction and component calculation is essential for accurate results in vector addition.
yety124
Messages
2
Reaction score
0

Homework Statement


Add the following vectors using component method:
Vector A=175km; 30 degrees north of east
Vector B=153km; 20 degrees west of north
Vector C=195km; West
a)Find x and y component of each vector
b)Solve for the magnitude of resultant vector
c) Solve for direction(angle)

2. The attempt at a solution
So i these are what i got
Vector A: x=175cos 30...x=151.56 y=175sin 30...y=87.5
Vector B: x=153cos 20...x=-143.77 y=153sin 20...y=52.33
(are these correct?)
so after getting this i added all the x's and y's but then i realized that the question had said that vector C goes 195km west, so i now got confused what to do next.

please help thank you for your time.
 
Physics news on Phys.org
There is nothing special about "in a westerly direction". A westerly vector contributes a component D.cos 180o in the x-direction, and D.sin 0o in the y-direction.
 
well that was stupid of me, thanks for the reply finally solved it :D
 
If \theta is measured counterclockwise from the positive x-axis then a vector of length r and angle \theta has components rcos(\theta) and r sin(\theta). Strictly speaking, you are free to choose the "positive x-axis" any way you want as long as you are consistent but the usual convention is that the positive x-axis points East.

For the first vector you are given that r= 175km and the directon is 30 degrees north of east. "north of east" is counterclockwise from east so the angle is \theta= 30 degrees.

For the second vector you are given that r= 153km and the direction is 20 degrees west of north. West is clockwise of north but north itself is 90 degrees clockwise of east. The angle is \theta= 90+20= 110 degrees.

For the third vector you are given that r= 195km and the direction is west. West is exactly opposite east so the angle is \theta= 180 degrees.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top