If a force ? f = 5.0 y n acts on the particle at this instant, what is the torque about the origin?

Answer:

The torque about the origin is 5.0 N.

Explanation:

To calculate the torque (τ) about the origin, you need to know both the force (F) and the position vector (r) from the origin to the point where the force is applied. The formula for torque is:

τ = r × F

Where:

- τ is the torque vector.

- r is the position vector.

- F is the force vector.

In this case, you're given the force vector F = 5.0 y N, where "y" represents the unit vector in the y-direction.

Assuming that the particle is located at a position where the position vector r is given as (x, y), you need to express the position vector in terms of unit vectors. Since the force is applied along the y-axis, you can write the position vector as:

r = x i + y j

Now, you can calculate the torque:

τ = r × F

τ = (x i + y j) × (5.0 y N)

When you take the cross product, the component of the position vector along the direction of the force will contribute to the torque. In this case, only the y-component of the position vector (y j) is along the direction of the force (5.0 y N).

So, the torque about the origin is:

τ = y (5.0 y N)

Now, you can calculate the magnitude of the torque:

|τ| = |y (5.0 y N)| = 5.0 |y|^2 N

The magnitude of the unit vector "y" is 1, so:

|τ| = 5.0 N

The torque about the origin is 5.0 N.