promotion image of download ymail app
Promoted

Force F = (-9.0 N)i + (5.0 N)j acts on a particle with position vector r = (-1.0 m)i + (2.0 m)j?

(a) What is the torque on the particle about the origin?

(b) What is the angle between the directions of r and F? (If there is no torque, enter 0.)

I got i and j for part A, but I got 13 as my k value, and that is not right. Please help

1 Answer

Relevance
  • Anonymous
    8 years ago
    Favorite Answer

    Torque is defined a r x F. Since both vectors are in the same plane (i-hat and j-hat), we know that the cross product will be in the k-hat direction.

    Drawing out the vectors and using the right hand rule tells us that the resulting vector will be positive and the definition of cross product says the magnitude will be |T| = |r||F|sin(theta) where theta is the angle formed between r and F.

    Weird that it asks the torque before the angle, but whatever.

    Angle can be found by making little triangles out of the magnitudes of the vectors and taking the inverse tangent.

    theta = angle formed by r minus angle formed by F

    theta = arctan(2/1) - arctan(5/9)

    theta = .6

    total torque is

    T = |r||F|sin(.6)

    |r| = sqrt( (-1)^2 + (2)^2 )

    |F| = sqrt( (-9)^2 + (5)^2 )

    T = 23.022 * .565

    (a) T = 12.999 k-hat

    (b) theta = .6

    I got the same (a) as you and I refuse to believe we're both wrong out of stubbornness and pride, so write an angry letter to your teacher and site internet stranger (me) as support for your argument.

    Source(s): in college for physics truuuuust me
    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.