- #1

- 19

- 0

## 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 apparantly im doing something wrong. Any help please?