Physics: Vector Help - Magnitude & Angle Calculations

  • Thread starter Thread starter buffgilville
  • Start date Start date
  • Tags Tags
    Physics Vector
AI Thread Summary
To find the equilibrant vector of a resultant vector measuring 5 units at an angle of 23 degrees, it is necessary to determine that the equilibrant has the same magnitude but an angle of 203 degrees, as it is opposite in direction. For the vector with a magnitude of 8.73 units at an angle of 155 degrees, the x-component is calculated using the formula F*cos(x) and the y-component using F*sin(x). The confusion surrounding counter-clockwise angles is clarified: counter-clockwise angles are considered positive while clockwise angles are negative. Understanding these conventions is crucial for accurately calculating vector components and magnitudes. Proper application of these principles will lead to correct vector analysis.
buffgilville
Messages
91
Reaction score
0
Can someone please help me with this?
1) A resultant vector is 5 units long and makes an angle of 23 degrees measured counter-clockwise with respect to the positive x-axis. What are the magnitude and angle (measured counter-clockwise with respect to the positive x-axis) of the equilibrant vector?

2) Find the x- and y-components of the vector whose magnitude is 8.73 units making an angle of 155 degrees measured counter-clockwise with respect to the positive x-axis.
 
Physics news on Phys.org
These kinds of questions have already been answered, look at some other posts.

Think of the fact that the x-component of a vector of magnitude F and angle x with the x-axis is
equal to F*cos (x). The y-component is F*sin(x)

Once the components are given you can work the other way around and determin the magnitude of the vector as F = sqrt(x² + y²)

regards
marlon
 
But what does the counter-clockwise mean? Does it mean that the angle is negative? I'm confused.
 
Last edited:
Clockwise means a negative angle, and counter-clockwise means a positive angle.
 
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 .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top