Finding the Position Vector of a Point to Achieve Half Torque of a Particle

[Th3Proj3ct]

Homework Statement

A particle is located at the vector position =(5.00i + 7.00j) m and a force exerted on it is given by =(3.00i + 2.00j) N.
(a) What is the torque acting on the particle about the origin?

(b) 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. Select the following conditions that are true.
No such point can exist.
Only one such point can exist.
Multiple such points can exist.
No such a point can lie on the y-axis.
Only one such point can lie on the y-axis.
Multiple such points can lie on the y-axis.

Determine the position vector of such a point.
[only the J position]

Homework Equations

dot, product, Torque = F*r

The Attempt at a Solution

Part A i got, by the cross product: 10k-21k=-11k
Part B i got 3 and 5, multiply points can exist, but only 1 on the Y-axis.
Yet for Part C I'm lost, I assume the i position is the same (3) so the i would have to be a value to make the total torque 5.5 (half of 11). When i put in 1.357for j; dot product would give: 15k-9.5k=5.5k, which is half the magnitude, and opposite direction. But apparently I am doing something wrong. Any help please?

 

[anathema]

Hello, thank you for your post. It seems like you have a solid understanding of the concepts involved in this problem. Let's go through your solution for part C and see where you may have gone wrong.

You correctly identified that the torque acting on the particle is -11k. This means that the torque must be balanced by an equal and opposite torque in order for the particle to remain in equilibrium. This can only occur if there is a point where the force and the position vectors are perpendicular to each other.

To find this point, we can use the dot product of the force and position vectors. We know that the dot product of two perpendicular vectors is equal to zero, so we can set up the following equation:

(3i + 2j) * (xi + yj) = 0

Expanding this out, we get:

3x + 2y = 0

This means that the point (x,y) must satisfy this equation. We can also use the fact that the magnitude of the torque at this point is half of the original torque, so we can set up another equation:

(3i + 2j) * (xi + yj) = -5.5k

Expanding this out, we get:

3x + 2y = -5.5

We now have a system of two equations with two unknowns. Solving for x and y, we get:

x = -1.1
y = 1.65

This means that the point (-1.1, 1.65) satisfies both equations and would be the position vector of the point where the torque is in the opposite direction and half the magnitude. Note that this point does not lie on the y-axis, as it has a non-zero x value. I hope this helps clarify any confusion you had. Good luck with your studies!

 

[ogb p]

Your approach for Part A and Part B seems correct. For Part C, you are on the right track but your calculation for the position vector of the point is not quite accurate. To find the position vector, you need to divide the torque by the force, not the other way around. So the correct calculation would be (5.5k)/(3i+2j) = 1.833i + 1.222j. This position vector would give a torque of 5.5k, which is half the magnitude of the original torque and in the opposite direction. This position vector lies on the line passing through the origin and the original position vector of the particle, and it is the only point on that line that satisfies the given conditions.