A force F =(8.700 N)i +(−9.000 N)j F→=(8.700 N)i +(−9.000 N)j is exerted on a particle located at r=(−8.400 m)i +(7.700 m)j r→=(−8.400 m)i +(7.700 m)j. What is the torque on the particle about the origin?

To find the torque on the particle about the origin, we need to calculate the cross product of the force vector and the position vector of the particle.

The cross product of two vectors, A and B, is given by the formula:

A x B = |A| |B| sin(θ) n

where |A| and |B| are the magnitudes of vectors A and B, θ is the angle between them, and n is the unit vector perpendicular to the plane formed by A and B in accordance with the right-hand rule.

In this case, the force vector F = (8.700 N)i + (-9.000 N)j and the position vector r = (-8.400 m)i + (7.700 m)j.

First, calculate the magnitude of vectors F and r:

|F| = √(8.700 N)^2 + (-9.000 N)^2 = √(75.69 + 81) = √156.69 ≈ 12.51 N

|r| = √((-8.400 m)^2 + (7.700 m)^2) = √(70.56 + 59.29) = √129.85 ≈ 11.39 m

Next, calculate the angle θ between the force vector F and the position vector r using the dot product:

F · r = |F| |r| cos(θ)

(8.700 N)(-8.400 m) + (-9.000 N)(7.700 m) = (12.51 N)(11.39 m) cos(θ)

Solving for cos(θ):

-73.08 - 69.30 = 142.3869 cos(θ)

-142.38 = 142.3869 cos(θ)

cos(θ) ≈ -1

Since cos(θ) ≈ -1, we know that the angle θ is 180 degrees or π radians.

Now, calculate the cross product:

F x r = |F| |r| sin(θ) n

F x r = (12.51 N)(11.39 m) sin(π) n

sin(π) = 0

Therefore, the torque on the particle about the origin is 0.