Find Position Vector of Point for Half Torque Opposite Direction

  • Thread starter Thread starter bingkei
  • Start date Start date
  • Tags Tags
    Vector
Click For Summary
SUMMARY

The discussion focuses on determining the position vector of a point where the torque caused by a force of (7.00 + 6.00) N on a particle located at (9.00 + 10.00) m is half as large in magnitude and in the opposite direction. The torque is calculated to be -16, and the participant attempted to find a solution by dividing the torque by 2. However, the participant's calculation of 6.57 for the y-coordinate was incorrect, as it did not adhere to the requirement that the new position must lie on the y-axis.

PREREQUISITES
  • Understanding of torque and its calculation
  • Familiarity with vector positions in physics
  • Knowledge of dot products in vector mathematics
  • Basic principles of force and motion
NEXT STEPS
  • Review the principles of torque and its relationship with force and position vectors
  • Study the concept of vector projections and their applications in physics
  • Learn how to correctly apply the dot product in torque calculations
  • Explore examples of torque in different coordinate systems, particularly focusing on the y-axis
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone involved in solving vector-related problems in torque and force analysis.

bingkei
Messages
1
Reaction score
0
A particle is located at the vector position =(9.00 + 10.00) m and a force exerted on it is given by =(7.00 + 6.00) N. Consider another point about which the torque caused by this force on this particle will be in the opposite direction and half as large in magnitude.
Multiple such points can exist.
Only one such point can lie on the y-axis
Determine the position vector of such a point
_____ j m




2. Homework Equations :
dot products



3. The Attempt at a Solution :
I've taken the torque (-16) and divided by 2, kept F as (7,6) and kept x of the position vector...I got 6.57 for the y but it was wrong.
 
Physics news on Phys.org
Hi bingkei,

bingkei said:
A particle is located at the vector position =(9.00 + 10.00) m and a force exerted on it is given by =(7.00 + 6.00) N. Consider another point about which the torque caused by this force on this particle will be in the opposite direction and half as large in magnitude.
Multiple such points can exist.
Only one such point can lie on the y-axis
Determine the position vector of such a point
_____ j m




2. Homework Equations :
dot products



3. The Attempt at a Solution :
I've taken the torque (-16) and divided by 2, kept F as (7,6) and kept x of the position vector

If you keep the x coordinate from the original position vector, then your new position will not be on the y-axis.

...I got 6.57 for the y but it was wrong.
 

Similar threads

Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
1
Views
10K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
5
Views
10K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
7K
  • · Replies 5 ·
Replies
5
Views
6K