Find Position Vector of Point for Half Torque Opposite Direction

[bingkei]

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.

 

[alphysicist]

Hi bingkei,

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.

 

[kerimek]

I would approach this problem by first understanding the concept of torque and its relationship with force and position. Torque is the product of force and the perpendicular distance from the point of application of the force to the axis of rotation. In this case, the torque is given by the cross product of the force vector and the position vector of the particle.

Let's denote the position vector of the particle as r, and the force vector as F. We are given r = (9,10) m and F = (7,6) N. The torque caused by this force on the particle is given by T = r x F = (9,10) x (7,6) = -16 k Nm.

Now, we are asked to find the position vector of a point where the torque caused by this force will be in the opposite direction and half as large in magnitude. This means that the new torque will be T' = -8 k Nm. We can find the position vector of this point by using the dot product of the position vector and the force vector.

Let's denote the position vector of the new point as r'. We know that T' = r' x F = -8 k Nm. Taking the dot product of both sides with F, we get r' . F = -8 Nm. Substituting the values of r and F, we get (9,10) . (7,6) = -8. Solving for r', we get r' = (-8/13, -8/13) m.

This means that the position vector of the new point is (-8/13, -8/13) m. This point lies on the line joining the origin and the given position vector (9,10) m. It is also the point where the torque caused by the force (7,6) N will be in the opposite direction and half as large in magnitude.

As stated in the problem, there can be multiple such points, but only one can lie on the y-axis. This point can be found by setting the x-component of the position vector to 0, which gives us r' = (0, -8/13) m.

In summary, the position vector of the point where the torque caused by the force (7,6) N will be in the opposite direction and half as large in magnitude is (-8/